Find minimum number of merge operations to make an array palindrome

Here given code implementation process.

/*
    C program for
    Find minimum number of merge operations to make an array palindrome
*/
#include <stdio.h>
// Display array elements
void printArray(int arr[], int n)
{
    for (int i = 0; i < n; ++i)
    {
        printf(" %d",arr[i]);
    }
}

void minMergePalindrome(int arr[], int n)
{   
    if(n<=0)
    {
        return;
    }

    // Auxiliary variables
    int operation = 0;

    int start = 0;

    int last = n-1;

  
    int temp[n];

    // Copy array element
    for (int i = 0; i < n; ++i)
    {
        temp[i] = arr[i];
    }

    while(start < last)
    {

        if(temp[start] == temp[last])
        {
            // When boundary element is similar
            // Change boundary position
            start ++;
            last --;
        }
        else if(temp[start] > temp[last])
        {
            // When left side boundary element is 
            // greater than right side boundary element.
            // Change last boundary -1 element 
            // by sum of last two boundary element.
            temp[last-1] = temp[last-1] + temp[last]; 
            
            // Reduce last boundary position
            last --;

            // increase operation 
            operation++; 
        }
        else
        {
            // Change left side next element is 
            // Sum of two starting boundary element.
            temp[start+1] = temp[start+1] + temp[start]; 

            // Increase start boundary position
            start ++;
            // Increase operation
            operation++;
        }
    }
   
    
    printf("\n Array  : ");

    printArray(arr, n);

    printf("\n Result : %d",operation);
}
int main(int argc, char const *argv[])
{
    int arr1[]= { 2, 1, 4, 3 , 5,  2, 1, 2};
    int arr2[]= {4, 3, 2, 3, 4, 7};
    int arr3[]= {1, 2, 1};
    int arr4[]= {1, 4, 1 , 5, 1};
    int arr5[]= {1, 2, 3, 1};
    // Test A
    int n = sizeof(arr1)/sizeof(arr1[0]);
    /*
        
        [2, 1 , 4 , 3 , 5 , 2, 1, 2]
         -  -                  -  -
        [Note boundary element 2, 1 is palindrome]
        
        2, 1 ,4 , 3 , [5 + 2] 1, 2
        -  -             ➀    -  -

        2, 1 [4 + 3] ,  7,   1, 2
                ➁

        2, 1,    7   , 7     1, 2  
                 -     - 
        [ 2  1 7  7 1 2 ] is palindrome   
        --------------------------------    
        Result : 2 operation

    */
    minMergePalindrome(arr1,n);


    // Test B
    n = sizeof(arr2)/sizeof(arr2[0]);
    /*
        
        [4   3   2   3   4   7]
        ---------------------
        [4 + 3   2   3   4   7]
           ➀ //operation

           7     2 + 3   4   7
                 -   -
                   ➁  //operation    
           7       5  +  4    7
                   -     -
                      ➂ //operation
           7          9       7
        
        [7 9 7] is palindrome
       -------------------------   
        Result = 3 operation
    */
    minMergePalindrome(arr2,n);


    // Test C
    n = sizeof(arr3)/sizeof(arr3[0]);
    /*
        
        [1, 2, 1]
         -  -  - 
        1 2 1  is palindrome
        Result = 0
    */
    minMergePalindrome(arr3,n);


    // Test D
    n = sizeof(arr4)/sizeof(arr4[0]);
    /*
        
        [1, 4, 1 , 5, 1]
         -  -  - 
        
        1, [4 + 1] , 5   1
              ➀  
        1,  5 , 5   1
           
        Result = 1
    */
    minMergePalindrome(arr4,n);


    // Test E
    n = sizeof(arr5)/sizeof(arr5[0]);
    /*
        
        [1 , [2 + 3], 1]
                ➀ 

        Result = 1
    */
    minMergePalindrome(arr5,n);
    return 0;
}

Output

 Array  :  2 1 4 3 5 2 1 2
 Result : 2
 Array  :  4 3 2 3 4 7
 Result : 3
 Array  :  1 2 1
 Result : 0
 Array  :  1 4 1 5 1
 Result : 1
 Array  :  1 2 3 1
 Result : 1
// Java program for
// Find minimum number of merge operations to make an array palindrome
public class PalindromeOperation
{
	// Display array elements
	public void printArray(int[] arr, int n)
	{
		for (int i = 0; i < n; ++i)
		{
			System.out.print(" " + arr[i]);
		}
	}
	public void minMergePalindrome(int[] arr, int n)
	{
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		int operation = 0;
		int start = 0;
		int last = n - 1;
		int[] temp = new int[n];
		// Copy array element
		for (int i = 0; i < n; ++i)
		{
			temp[i] = arr[i];
		}
		while (start < last)
		{
			if (temp[start] == temp[last])
			{
				// When boundary element is similar
				// Change boundary position
				start++;
				last--;
			}
			else if (temp[start] > temp[last])
			{
				// When left side boundary element is 
				// greater than right side boundary element.
				// Change last boundary -1 element 
				// by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last];
				// Reduce last boundary position
				last--;
				// increase operation 
				operation++;
			}
			else
			{
				// Change left side next element is 
				// Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start];
				// Increase start boundary position
				start++;
				// Increase operation
				operation++;
			}
		}
		System.out.print("\n Array : ");
		printArray(arr, n);
		System.out.print("\n Result : " + operation);
	}
	public static void main(String[] args)
	{
		PalindromeOperation task = new PalindromeOperation();
		int[] arr1 = {
			2 , 1 , 4 , 3 , 5 , 2 , 1 , 2
		};
		int[] arr2 = {
			4 , 3 , 2 , 3 , 4 , 7
		};
		int[] arr3 = {
			1 , 2 , 1
		};
		int[] arr4 = {
			1 , 4 , 1 , 5 , 1
		};
		int[] arr5 = {
			1 , 2 , 3 , 1
		};
		// Test A
		int n = arr1.length;
		/*
		        
		    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
		     -  -                  -  -
		    [Note boundary element 2, 1 is palindrome]
		    
		    2, 1 ,4 , 3 , [5 + 2] 1, 2
		    -  -             ➀    -  -

		    2, 1 [4 + 3] ,  7,   1, 2
		            ➁

		    2, 1,    7   , 7     1, 2  
		             -     - 
		    [ 2  1 7  7 1 2 ] is palindrome   
		    --------------------------------    
		    Result : 2 operation

		*/
		task.minMergePalindrome(arr1, n);
		// Test B
		n = arr2.length;
		/*
		    
		    [4   3   2   3   4   7]
		    ---------------------
		    [4 + 3   2   3   4   7]
		       ➀ //operation

		       7     2 + 3   4   7
		             -   -
		               ➁  //operation    
		       7       5  +  4    7
		               -     -
		                  ➂ //operation
		       7          9       7
		    
		    [7 9 7] is palindrome
		   -------------------------   
		    Result = 3 operation
		*/
		task.minMergePalindrome(arr2, n);
		// Test C
		n = arr3.length;
		/*
		        
		    [1, 2, 1]
		     -  -  - 
		    1 2 1  is palindrome
		    Result = 0
		*/
		task.minMergePalindrome(arr3, n);
		// Test D
		n = arr4.length;
		/*
		    
		    [1, 4, 1 , 5, 1]
		     -  -  - 
		    
		    1, [4 + 1] , 5   1
		          ➀  
		    1,  5 , 5   1
		       
		    Result = 1
		*/
		task.minMergePalindrome(arr4, n);
		// Test E
		n = arr5.length;
		/*
		    
		    [1 , [2 + 3], 1]
		            ➀ 

		    Result = 1
		*/
		task.minMergePalindrome(arr5, n);
	}
}

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
// Include header file
#include <iostream>

using namespace std;
// C++ program for
// Find minimum number of merge operations to make an array palindrome
class PalindromeOperation
{
	public:
		// Display array elements
		void printArray(int arr[], int n)
		{
			for (int i = 0; i < n; ++i)
			{
				cout << " " << arr[i];
			}
		}
	void minMergePalindrome(int arr[], int n)
	{
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		int operation = 0;
		int start = 0;
		int last = n - 1;
		int temp[n];
		// Copy array element
		for (int i = 0; i < n; ++i)
		{
			temp[i] = arr[i];
		}
		while (start < last)
		{
			if (temp[start] == temp[last])
			{
				// When boundary element is similar
				// Change boundary position
				start++;
				last--;
			}
			else if (temp[start] > temp[last])
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last];
				// Reduce last boundary position
				last--;
				// increase operation
				operation++;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start];
				// Increase start boundary position
				start++;
				// Increase operation
				operation++;
			}
		}
		cout << "\n Array : ";
		this->printArray(arr, n);
		cout << "\n Result : " << operation;
	}
};
int main()
{
	PalindromeOperation *task = new PalindromeOperation();
	int arr1[] = {
		2 , 1 , 4 , 3 , 5 , 2 , 1 , 2
	};
	int arr2[] = {
		4 , 3 , 2 , 3 , 4 , 7
	};
	int arr3[] = {
		1 , 2 , 1
	};
	int arr4[] = {
		1 , 4 , 1 , 5 , 1
	};
	int arr5[] = {
		1 , 2 , 3 , 1
	};
	// Test A
	int n = sizeof(arr1) / sizeof(arr1[0]);
	/*
	    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	     -  -                  -  -
	    [Note boundary element 2, 1 is palindrome]
	    
	    2, 1 ,4 , 3 , [5 + 2] 1, 2
	    -  -             ➀    -  -
	    2, 1 [4 + 3] ,  7,   1, 2
	            ➁
	    2, 1,    7   , 7     1, 2  
	             -     - 
	    [ 2  1 7  7 1 2 ] is palindrome   
	    --------------------------------    
	    Result : 2 operation
	*/
	task->minMergePalindrome(arr1, n);
	// Test B
	n = sizeof(arr2) / sizeof(arr2[0]);
	/*
	    [4   3   2   3   4   7]
	    ---------------------
	    [4 + 3   2   3   4   7]
	       ➀ //operation
	       7     2 + 3   4   7
	             -   -
	               ➁  //operation    
	       7       5  +  4    7
	               -     -
	                  ➂ //operation
	       7          9       7
	    
	    [7 9 7] is palindrome
	   -------------------------   
	    Result = 3 operation
	*/
	task->minMergePalindrome(arr2, n);
	// Test C
	n = sizeof(arr3) / sizeof(arr3[0]);
	/*
	    [1, 2, 1]
	     -  -  - 
	    1 2 1  is palindrome
	    Result = 0
	*/
	task->minMergePalindrome(arr3, n);
	// Test D
	n = sizeof(arr4) / sizeof(arr4[0]);
	/*
	    [1, 4, 1 , 5, 1]
	     -  -  - 
	    
	    1, [4 + 1] , 5   1
	          ➀  
	    1,  5 , 5   1
	       
	    Result = 1
	*/
	task->minMergePalindrome(arr4, n);
	// Test E
	n = sizeof(arr5) / sizeof(arr5[0]);
	/*
	    [1 , [2 + 3], 1]
	            ➀ 
	    Result = 1
	*/
	task->minMergePalindrome(arr5, n);
	return 0;
}

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
// Include namespace system
using System;
// Csharp program for
// Find minimum number of merge operations to make an array palindrome
public class PalindromeOperation
{
	// Display array elements
	public void printArray(int[] arr, int n)
	{
		for (int i = 0; i < n; ++i)
		{
			Console.Write(" " + arr[i]);
		}
	}
	public void minMergePalindrome(int[] arr, int n)
	{
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		int operation = 0;
		int start = 0;
		int last = n - 1;
		int[] temp = new int[n];
		// Copy array element
		for (int i = 0; i < n; ++i)
		{
			temp[i] = arr[i];
		}
		while (start < last)
		{
			if (temp[start] == temp[last])
			{
				// When boundary element is similar
				// Change boundary position
				start++;
				last--;
			}
			else if (temp[start] > temp[last])
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last];
				// Reduce last boundary position
				last--;
				// increase operation
				operation++;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start];
				// Increase start boundary position
				start++;
				// Increase operation
				operation++;
			}
		}
		Console.Write("\n Array : ");
		this.printArray(arr, n);
		Console.Write("\n Result : " + operation);
	}
	public static void Main(String[] args)
	{
		PalindromeOperation task = new PalindromeOperation();
		int[] arr1 = {
			2 , 1 , 4 , 3 , 5 , 2 , 1 , 2
		};
		int[] arr2 = {
			4 , 3 , 2 , 3 , 4 , 7
		};
		int[] arr3 = {
			1 , 2 , 1
		};
		int[] arr4 = {
			1 , 4 , 1 , 5 , 1
		};
		int[] arr5 = {
			1 , 2 , 3 , 1
		};
		// Test A
		int n = arr1.Length;
		/*
		    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
		     -  -                  -  -
		    [Note boundary element 2, 1 is palindrome]
		    
		    2, 1 ,4 , 3 , [5 + 2] 1, 2
		    -  -             ➀    -  -
		    2, 1 [4 + 3] ,  7,   1, 2
		            ➁
		    2, 1,    7   , 7     1, 2  
		             -     - 
		    [ 2  1 7  7 1 2 ] is palindrome   
		    --------------------------------    
		    Result : 2 operation
		*/
		task.minMergePalindrome(arr1, n);
		// Test B
		n = arr2.Length;
		/*
		    [4   3   2   3   4   7]
		    ---------------------
		    [4 + 3   2   3   4   7]
		       ➀ //operation
		       7     2 + 3   4   7
		             -   -
		               ➁  //operation    
		       7       5  +  4    7
		               -     -
		                  ➂ //operation
		       7          9       7
		    
		    [7 9 7] is palindrome
		   -------------------------   
		    Result = 3 operation
		*/
		task.minMergePalindrome(arr2, n);
		// Test C
		n = arr3.Length;
		/*
		    [1, 2, 1]
		     -  -  - 
		    1 2 1  is palindrome
		    Result = 0
		*/
		task.minMergePalindrome(arr3, n);
		// Test D
		n = arr4.Length;
		/*
		    [1, 4, 1 , 5, 1]
		     -  -  - 
		    
		    1, [4 + 1] , 5   1
		          ➀  
		    1,  5 , 5   1
		       
		    Result = 1
		*/
		task.minMergePalindrome(arr4, n);
		// Test E
		n = arr5.Length;
		/*
		    [1 , [2 + 3], 1]
		            ➀ 
		    Result = 1
		*/
		task.minMergePalindrome(arr5, n);
	}
}

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
package main
import "fmt"
// Go program for
// Find minimum number of merge operations to make an array palindrome

// Display array elements
func printArray(arr[] int, n int) {
	for i := 0 ; i < n ; i++ {
		fmt.Print(" ", arr[i])
	}
}
func minMergePalindrome(arr[] int, n int) {
	if n <= 0 {
		return
	}
	// Auxiliary variables
	var operation int = 0
	var start int = 0
	var last int = n - 1
	var temp = make([] int, n)
	// Copy array element
	for i := 0 ; i < n ; i++ {
		temp[i] = arr[i]
	}
	for (start < last) {
		if temp[start] == temp[last] {
			// When boundary element is similar
			// Change boundary position
			start++
			last--
		} else if temp[start] > temp[last] {
			// When left side boundary element is
			// greater than right side boundary element.
			// Change last boundary -1 element
			// by sum of last two boundary element.
			temp[last - 1] = temp[last - 1] + temp[last]
			// Reduce last boundary position
			last--
			// increase operation
			operation++
		} else {
			// Change left side next element is
			// Sum of two starting boundary element.
			temp[start + 1] = temp[start + 1] + temp[start]
			// Increase start boundary position
			start++
			// Increase operation
			operation++
		}
	}
	fmt.Print("\n Array : ")
	printArray(arr, n)
	fmt.Print("\n Result : ", operation)
}
func main() {

	var arr1 = [] int { 2, 1, 4, 3 , 5,  2, 1, 2}
	var arr2 = [] int { 4, 3, 2, 3, 4, 7 }
	var arr3 = [] int { 1, 2, 1 }
	var arr4 = [] int { 1, 4, 1 , 5, 1 }
	var arr5 = [] int { 1, 2, 3, 1 }
	// Test A
	var n int = len(arr1)
	/*
	    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	     -  -                  -  -
	    [Note boundary element 2, 1 is palindrome]
	    
	    2, 1 ,4 , 3 , [5 + 2] 1, 2
	    -  -             ➀    -  -
	    2, 1 [4 + 3] ,  7,   1, 2
	            ➁
	    2, 1,    7   , 7     1, 2  
	             -     - 
	    [ 2  1 7  7 1 2 ] is palindrome   
	    --------------------------------    
	    Result : 2 operation
	*/
	minMergePalindrome(arr1, n)
	// Test B
	n = len(arr2)
	/*
	    [4   3   2   3   4   7]
	    ---------------------
	    [4 + 3   2   3   4   7]
	       ➀ //operation
	       7     2 + 3   4   7
	             -   -
	               ➁  //operation    
	       7       5  +  4    7
	               -     -
	                  ➂ //operation
	       7          9       7
	    
	    [7 9 7] is palindrome
	   -------------------------   
	    Result = 3 operation
	*/
	minMergePalindrome(arr2, n)
	// Test C
	n = len(arr3)
	/*
	    [1, 2, 1]
	     -  -  - 
	    1 2 1  is palindrome
	    Result = 0
	*/
	minMergePalindrome(arr3, n)
	// Test D
	n = len(arr4)
	/*
	    [1, 4, 1 , 5, 1]
	     -  -  - 
	    
	    1, [4 + 1] , 5   1
	          ➀  
	    1,  5 , 5   1
	       
	    Result = 1
	*/
	minMergePalindrome(arr4, n)
	// Test E
	n = len(arr5)
	/*
	    [1 , [2 + 3], 1]
	            ➀ 
	    Result = 1
	*/
	minMergePalindrome(arr5, n)
}

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
<?php
// Php program for
// Find minimum number of merge operations to make an array palindrome
class PalindromeOperation
{
	// Display array elements
	public	function printArray($arr, $n)
	{
		for ($i = 0; $i < $n; ++$i)
		{
			echo(" ".$arr[$i]);
		}
	}
	public	function minMergePalindrome($arr, $n)
	{
		if ($n <= 0)
		{
			return;
		}
		// Auxiliary variables
		$operation = 0;
		$start = 0;
		$last = $n - 1;
		$temp = array_fill(0, $n, 0);
		// Copy array element
		for ($i = 0; $i < $n; ++$i)
		{
			$temp[$i] = $arr[$i];
		}
		while ($start < $last)
		{
			if ($temp[$start] == $temp[$last])
			{
				// When boundary element is similar
				// Change boundary position
				$start++;
				$last--;
			}
			else if ($temp[$start] > $temp[$last])
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				$temp[$last - 1] = $temp[$last - 1] + $temp[$last];
				// Reduce last boundary position
				$last--;
				// increase operation
				$operation++;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				$temp[$start + 1] = $temp[$start + 1] + $temp[$start];
				// Increase start boundary position
				$start++;
				// Increase operation
				$operation++;
			}
		}
		echo("\n Array : ");
		$this->printArray($arr, $n);
		echo("\n Result : ".$operation);
	}
}

function main()
{
	$task = new PalindromeOperation();
	$arr1 = array(2, 1, 4, 3, 5, 2, 1, 2);
	$arr2 = array(4, 3, 2, 3, 4, 7);
	$arr3 = array(1, 2, 1);
	$arr4 = array(1, 4, 1, 5, 1);
	$arr5 = array(1, 2, 3, 1);
	// Test A
	$n = count($arr1);
	/*
	    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	     -  -                  -  -
	    [Note boundary element 2, 1 is palindrome]
	    
	    2, 1 ,4 , 3 , [5 + 2] 1, 2
	    -  -             ➀    -  -
	    2, 1 [4 + 3] ,  7,   1, 2
	            ➁
	    2, 1,    7   , 7     1, 2  
	             -     - 
	    [ 2  1 7  7 1 2 ] is palindrome   
	    --------------------------------    
	    Result : 2 operation
	*/
	$task->minMergePalindrome($arr1, $n);
	// Test B
	$n = count($arr2);
	/*
	    [4   3   2   3   4   7]
	    ---------------------
	    [4 + 3   2   3   4   7]
	       ➀ //operation
	       7     2 + 3   4   7
	             -   -
	               ➁  //operation    
	       7       5  +  4    7
	               -     -
	                  ➂ //operation
	       7          9       7
	    
	    [7 9 7] is palindrome
	   -------------------------   
	    Result = 3 operation
	*/
	$task->minMergePalindrome($arr2, $n);
	// Test C
	$n = count($arr3);
	/*
	    [1, 2, 1]
	     -  -  - 
	    1 2 1  is palindrome
	    Result = 0
	*/
	$task->minMergePalindrome($arr3, $n);
	// Test D
	$n = count($arr4);
	/*
	    [1, 4, 1 , 5, 1]
	     -  -  - 
	    
	    1, [4 + 1] , 5   1
	          ➀  
	    1,  5 , 5   1
	       
	    Result = 1
	*/
	$task->minMergePalindrome($arr4, $n);
	// Test E
	$n = count($arr5);
	/*
	    [1 , [2 + 3], 1]
	            ➀ 
	    Result = 1
	*/
	$task->minMergePalindrome($arr5, $n);
}
main();

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
// Node JS program for
// Find minimum number of merge operations to make an array palindrome
class PalindromeOperation
{
	// Display array elements
	printArray(arr, n)
	{
		for (var i = 0; i < n; ++i)
		{
			process.stdout.write(" " + arr[i]);
		}
	}
	minMergePalindrome(arr, n)
	{
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		var operation = 0;
		var start = 0;
		var last = n - 1;
		var temp = Array(n).fill(0);
		// Copy array element
		for (var i = 0; i < n; ++i)
		{
			temp[i] = arr[i];
		}
		while (start < last)
		{
			if (temp[start] == temp[last])
			{
				// When boundary element is similar
				// Change boundary position
				start++;
				last--;
			}
			else if (temp[start] > temp[last])
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last];
				// Reduce last boundary position
				last--;
				// increase operation
				operation++;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start];
				// Increase start boundary position
				start++;
				// Increase operation
				operation++;
			}
		}
		process.stdout.write("\n Array : ");
		this.printArray(arr, n);
		process.stdout.write("\n Result : " + operation);
	}
}

function main()
{
	var task = new PalindromeOperation();
	var arr1 = [2, 1, 4, 3, 5, 2, 1, 2];
	var arr2 = [4, 3, 2, 3, 4, 7];
	var arr3 = [1, 2, 1];
	var arr4 = [1, 4, 1, 5, 1];
	var arr5 = [1, 2, 3, 1];
	// Test A
	var n = arr1.length;
	/*
	    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	     -  -                  -  -
	    [Note boundary element 2, 1 is palindrome]
	    
	    2, 1 ,4 , 3 , [5 + 2] 1, 2
	    -  -             ➀    -  -
	    2, 1 [4 + 3] ,  7,   1, 2
	            ➁
	    2, 1,    7   , 7     1, 2  
	             -     - 
	    [ 2  1 7  7 1 2 ] is palindrome   
	    --------------------------------    
	    Result : 2 operation
	*/
	task.minMergePalindrome(arr1, n);
	// Test B
	n = arr2.length;
	/*
	    [4   3   2   3   4   7]
	    ---------------------
	    [4 + 3   2   3   4   7]
	       ➀ //operation
	       7     2 + 3   4   7
	             -   -
	               ➁  //operation    
	       7       5  +  4    7
	               -     -
	                  ➂ //operation
	       7          9       7
	    
	    [7 9 7] is palindrome
	   -------------------------   
	    Result = 3 operation
	*/
	task.minMergePalindrome(arr2, n);
	// Test C
	n = arr3.length;
	/*
	    [1, 2, 1]
	     -  -  - 
	    1 2 1  is palindrome
	    Result = 0
	*/
	task.minMergePalindrome(arr3, n);
	// Test D
	n = arr4.length;
	/*
	    [1, 4, 1 , 5, 1]
	     -  -  - 
	    
	    1, [4 + 1] , 5   1
	          ➀  
	    1,  5 , 5   1
	       
	    Result = 1
	*/
	task.minMergePalindrome(arr4, n);
	// Test E
	n = arr5.length;
	/*
	    [1 , [2 + 3], 1]
	            ➀ 
	    Result = 1
	*/
	task.minMergePalindrome(arr5, n);
}
main();

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
#  Python 3 program for
#  Find minimum number of merge operations to make an array palindrome
class PalindromeOperation :
	#  Display list elements
	def printArray(self, arr, n) :
		i = 0
		while (i < n) :
			print(" ", arr[i], end = "")
			i += 1
		
	
	def minMergePalindrome(self, arr, n) :
		if (n <= 0) :
			return
		
		#  Auxiliary variables
		operation = 0
		start = 0
		last = n - 1
		temp = [0] * (n)
		i = 0
		#  Copy list element
		while (i < n) :
			temp[i] = arr[i]
			i += 1
		
		while (start < last) :
			if (temp[start] == temp[last]) :
				#  When boundary element is similar
				#  Change boundary position
				start += 1
				last -= 1
			elif (temp[start] > temp[last]) :
				#  When left side boundary element is
				#  greater than right side boundary element.
				#  Change last boundary -1 element
				#  by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last]
				#  Reduce last boundary position
				last -= 1
				#  increase operation
				operation += 1
			else :
				#  Change left side next element is
				#  Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start]
				#  Increase start boundary position
				start += 1
				#  Increase operation
				operation += 1
			
		
		print("\n Array : ", end = "")
		self.printArray(arr, n)
		print("\n Result : ", operation, end = "")
	

def main() :
	task = PalindromeOperation()
	arr1 = [2, 1, 4, 3, 5, 2, 1, 2]
	arr2 = [4, 3, 2, 3, 4, 7]
	arr3 = [1, 2, 1]
	arr4 = [1, 4, 1, 5, 1]
	arr5 = [1, 2, 3, 1]
	#  Test A
	n = len(arr1)
	#    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	#     -  -                  -  -
	#    [Note boundary element 2, 1 is palindrome]
	#    2, 1 ,4 , 3 , [5 + 2] 1, 2
	#    -  -             ➀    -  -
	#    2, 1 [4 + 3] ,  7,   1, 2
	#            ➁
	#    2, 1,    7   , 7     1, 2  
	#             -     - 
	#    [ 2  1 7  7 1 2 ] is palindrome   
	#    --------------------------------    
	#    Result : 2 operation
	task.minMergePalindrome(arr1, n)
	#  Test B
	n = len(arr2)
	#    [4   3   2   3   4   7]
	#    ---------------------
	#    [4 + 3   2   3   4   7]
	#       ➀ //operation
	#       7     2 + 3   4   7
	#             -   -
	#               ➁  //operation    
	#       7       5  +  4    7
	#               -     -
	#                  ➂ //operation
	#       7          9       7
	#    [7 9 7] is palindrome
	#   -------------------------   
	#    Result = 3 operation
	task.minMergePalindrome(arr2, n)
	#  Test C
	n = len(arr3)
	#    [1, 2, 1]
	#     -  -  - 
	#    1 2 1  is palindrome
	#    Result = 0
	task.minMergePalindrome(arr3, n)
	#  Test D
	n = len(arr4)
	#    [1, 4, 1 , 5, 1]
	#     -  -  - 
	#    1, [4 + 1] , 5   1
	#          ➀  
	#    1,  5 , 5   1
	#    Result = 1
	task.minMergePalindrome(arr4, n)
	#  Test E
	n = len(arr5)
	#    [1 , [2 + 3], 1]
	#            ➀ 
	#    Result = 1
	task.minMergePalindrome(arr5, n)

if __name__ == "__main__": main()

Output

 Array :   2  1  4  3  5  2  1  2
 Result :  2
 Array :   4  3  2  3  4  7
 Result :  3
 Array :   1  2  1
 Result :  0
 Array :   1  4  1  5  1
 Result :  1
 Array :   1  2  3  1
 Result :  1
#  Ruby program for
#  Find minimum number of merge operations to make an array palindrome
class PalindromeOperation 
	#  Display array elements
	def printArray(arr, n) 
		i = 0
		while (i < n) 
			print(" ", arr[i])
			i += 1
		end

	end

	def minMergePalindrome(arr, n) 
		if (n <= 0) 
			return
		end

		#  Auxiliary variables
		operation = 0
		start = 0
		last = n - 1
		temp = Array.new(n) {0}
		i = 0
		#  Copy array element
		while (i < n) 
			temp[i] = arr[i]
			i += 1
		end

		while (start < last) 
			if (temp[start] == temp[last]) 
				#  When boundary element is similar
				#  Change boundary position
				start += 1
				last -= 1
			elsif (temp[start] > temp[last]) 
				#  When left side boundary element is
				#  greater than right side boundary element.
				#  Change last boundary -1 element
				#  by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last]
				#  Reduce last boundary position
				last -= 1
				#  increase operation
				operation += 1
			else
 
				#  Change left side next element is
				#  Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start]
				#  Increase start boundary position
				start += 1
				#  Increase operation
				operation += 1
			end

		end

		print("\n Array : ")
		self.printArray(arr, n)
		print("\n Result : ", operation)
	end

end

def main() 
	task = PalindromeOperation.new()
	arr1 = [2, 1, 4, 3, 5, 2, 1, 2]
	arr2 = [4, 3, 2, 3, 4, 7]
	arr3 = [1, 2, 1]
	arr4 = [1, 4, 1, 5, 1]
	arr5 = [1, 2, 3, 1]
	#  Test A
	n = arr1.length
	#    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	#     -  -                  -  -
	#    [Note boundary element 2, 1 is palindrome]
	#    2, 1 ,4 , 3 , [5 + 2] 1, 2
	#    -  -             ➀    -  -
	#    2, 1 [4 + 3] ,  7,   1, 2
	#            ➁
	#    2, 1,    7   , 7     1, 2  
	#             -     - 
	#    [ 2  1 7  7 1 2 ] is palindrome   
	#    --------------------------------    
	#    Result : 2 operation
	task.minMergePalindrome(arr1, n)
	#  Test B
	n = arr2.length
	#    [4   3   2   3   4   7]
	#    ---------------------
	#    [4 + 3   2   3   4   7]
	#       ➀ //operation
	#       7     2 + 3   4   7
	#             -   -
	#               ➁  //operation    
	#       7       5  +  4    7
	#               -     -
	#                  ➂ //operation
	#       7          9       7
	#    [7 9 7] is palindrome
	#   -------------------------   
	#    Result = 3 operation
	task.minMergePalindrome(arr2, n)
	#  Test C
	n = arr3.length
	#    [1, 2, 1]
	#     -  -  - 
	#    1 2 1  is palindrome
	#    Result = 0
	task.minMergePalindrome(arr3, n)
	#  Test D
	n = arr4.length
	#    [1, 4, 1 , 5, 1]
	#     -  -  - 
	#    1, [4 + 1] , 5   1
	#          ➀  
	#    1,  5 , 5   1
	#    Result = 1
	task.minMergePalindrome(arr4, n)
	#  Test E
	n = arr5.length
	#    [1 , [2 + 3], 1]
	#            ➀ 
	#    Result = 1
	task.minMergePalindrome(arr5, n)
end

main()

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
// Scala program for
// Find minimum number of merge operations to make an array palindrome
class PalindromeOperation()
{
	// Display array elements
	def printArray(arr: Array[Int], n: Int): Unit = {
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr(i));
			i += 1;
		}
	}
	def minMergePalindrome(arr: Array[Int], n: Int): Unit = {
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		var operation: Int = 0;
		var start: Int = 0;
		var last: Int = n - 1;
		var temp: Array[Int] = Array.fill[Int](n)(0);
		var i: Int = 0;
		// Copy array element
		while (i < n)
		{
			temp(i) = arr(i);
			i += 1;
		}
		while (start < last)
		{
			if (temp(start) == temp(last))
			{
				// When boundary element is similar
				// Change boundary position
				start += 1;
				last -= 1;
			}
			else if (temp(start) > temp(last))
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				temp(last - 1) = temp(last - 1) + temp(last);
				// Reduce last boundary position
				last -= 1;
				// increase operation
				operation += 1;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				temp(start + 1) = temp(start + 1) + temp(start);
				// Increase start boundary position
				start += 1;
				// Increase operation
				operation += 1;
			}
		}
		print("\n Array : ");
		printArray(arr, n);
		print("\n Result : " + operation);
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: PalindromeOperation = new PalindromeOperation();
		var arr1: Array[Int] = Array(2, 1, 4, 3, 5, 2, 1, 2);
		var arr2: Array[Int] = Array(4, 3, 2, 3, 4, 7);
		var arr3: Array[Int] = Array(1, 2, 1);
		var arr4: Array[Int] = Array(1, 4, 1, 5, 1);
		var arr5: Array[Int] = Array(1, 2, 3, 1);
		// Test A
		var n: Int = arr1.length;
		/*
		    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
		     -  -                  -  -
		    [Note boundary element 2, 1 is palindrome]
		    
		    2, 1 ,4 , 3 , [5 + 2] 1, 2
		    -  -             ➀    -  -
		    2, 1 [4 + 3] ,  7,   1, 2
		            ➁
		    2, 1,    7   , 7     1, 2  
		             -     - 
		    [ 2  1 7  7 1 2 ] is palindrome   
		    --------------------------------    
		    Result : 2 operation
		*/
		task.minMergePalindrome(arr1, n);
		// Test B
		n = arr2.length;
		/*
		    [4   3   2   3   4   7]
		    ---------------------
		    [4 + 3   2   3   4   7]
		       ➀ //operation
		       7     2 + 3   4   7
		             -   -
		               ➁  //operation    
		       7       5  +  4    7
		               -     -
		                  ➂ //operation
		       7          9       7
		    
		    [7 9 7] is palindrome
		   -------------------------   
		    Result = 3 operation
		*/
		task.minMergePalindrome(arr2, n);
		// Test C
		n = arr3.length;
		/*
		    [1, 2, 1]
		     -  -  - 
		    1 2 1  is palindrome
		    Result = 0
		*/
		task.minMergePalindrome(arr3, n);
		// Test D
		n = arr4.length;
		/*
		    [1, 4, 1 , 5, 1]
		     -  -  - 
		    
		    1, [4 + 1] , 5   1
		          ➀  
		    1,  5 , 5   1
		       
		    Result = 1
		*/
		task.minMergePalindrome(arr4, n);
		// Test E
		n = arr5.length;
		/*
		    [1 , [2 + 3], 1]
		            ➀ 
		    Result = 1
		*/
		task.minMergePalindrome(arr5, n);
	}
}

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1
import Foundation;
// Swift 4 program for
// Find minimum number of merge operations to make an array palindrome
class PalindromeOperation
{
	// Display array elements
	func printArray(_ arr: [Int], _ n: Int)
	{
		var i: Int = 0;
		while (i < n)
		{
			print(" ", arr[i], terminator: "");
			i += 1;
		}
	}
	func minMergePalindrome(_ arr: [Int], _ n: Int)
	{
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		var operation: Int = 0;
		var start: Int = 0;
		var last: Int = n - 1;
		var temp: [Int] = Array(repeating: 0, count: n);
		var i: Int = 0;
		// Copy array element
		while (i < n)
		{
			temp[i] = arr[i];
			i += 1;
		}
		while (start < last)
		{
			if (temp[start] == temp[last])
			{
				// When boundary element is similar
				// Change boundary position
				start += 1;
				last -= 1;
			}
			else if (temp[start] > temp[last])
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last];
				// Reduce last boundary position
				last -= 1;
				// increase operation
				operation += 1;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start];
				// Increase start boundary position
				start += 1;
				// Increase operation
				operation += 1;
			}
		}
		print("\n Array : ", terminator: "");
		self.printArray(arr, n);
		print("\n Result : ", operation, terminator: "");
	}
}
func main()
{
	let task: PalindromeOperation = PalindromeOperation();
	let arr1: [Int] = [2, 1, 4, 3, 5, 2, 1, 2];
	let arr2: [Int] = [4, 3, 2, 3, 4, 7];
	let arr3: [Int] = [1, 2, 1];
	let arr4: [Int] = [1, 4, 1, 5, 1];
	let arr5: [Int] = [1, 2, 3, 1];
	// Test A
	var n: Int = arr1.count;
	/*
	    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	     -  -                  -  -
	    [Note boundary element 2, 1 is palindrome]
	    
	    2, 1 ,4 , 3 , [5 + 2] 1, 2
	    -  -             ➀    -  -
	    2, 1 [4 + 3] ,  7,   1, 2
	            ➁
	    2, 1,    7   , 7     1, 2  
	             -     - 
	    [ 2  1 7  7 1 2 ] is palindrome   
	    --------------------------------    
	    Result : 2 operation
	*/
	task.minMergePalindrome(arr1, n);
	// Test B
	n = arr2.count;
	/*
	    [4   3   2   3   4   7]
	    ---------------------
	    [4 + 3   2   3   4   7]
	       ➀ //operation
	       7     2 + 3   4   7
	             -   -
	               ➁  //operation    
	       7       5  +  4    7
	               -     -
	                  ➂ //operation
	       7          9       7
	    
	    [7 9 7] is palindrome
	   -------------------------   
	    Result = 3 operation
	*/
	task.minMergePalindrome(arr2, n);
	// Test C
	n = arr3.count;
	/*
	    [1, 2, 1]
	     -  -  - 
	    1 2 1  is palindrome
	    Result = 0
	*/
	task.minMergePalindrome(arr3, n);
	// Test D
	n = arr4.count;
	/*
	    [1, 4, 1 , 5, 1]
	     -  -  - 
	    
	    1, [4 + 1] , 5   1
	          ➀  
	    1,  5 , 5   1
	       
	    Result = 1
	*/
	task.minMergePalindrome(arr4, n);
	// Test E
	n = arr5.count;
	/*
	    [1 , [2 + 3], 1]
	            ➀ 
	    Result = 1
	*/
	task.minMergePalindrome(arr5, n);
}
main();

Output

 Array :   2  1  4  3  5  2  1  2
 Result :  2
 Array :   4  3  2  3  4  7
 Result :  3
 Array :   1  2  1
 Result :  0
 Array :   1  4  1  5  1
 Result :  1
 Array :   1  2  3  1
 Result :  1
// Kotlin program for
// Find minimum number of merge operations to make an array palindrome
class PalindromeOperation
{
	// Display array elements
	fun printArray(arr: Array < Int > , n: Int): Unit
	{
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr[i]);
			i += 1;
		}
	}
	fun minMergePalindrome(arr: Array < Int > , n: Int): Unit
	{
		if (n <= 0)
		{
			return;
		}
		// Auxiliary variables
		var operation: Int = 0;
		var start: Int = 0;
		var last: Int = n - 1;
		val temp: Array < Int > = Array(n)
		{
			0
		};
		var i: Int = 0;
		// Copy array element
		while (i < n)
		{
			temp[i] = arr[i];
			i += 1;
		}
		while (start < last)
		{
			if (temp[start] == temp[last])
			{
				// When boundary element is similar
				// Change boundary position
				start += 1;
				last -= 1;
			}
			else if (temp[start] > temp[last])
			{
				// When left side boundary element is
				// greater than right side boundary element.
				// Change last boundary -1 element
				// by sum of last two boundary element.
				temp[last - 1] = temp[last - 1] + temp[last];
				// Reduce last boundary position
				last -= 1;
				// increase operation
				operation += 1;
			}
			else
			{
				// Change left side next element is
				// Sum of two starting boundary element.
				temp[start + 1] = temp[start + 1] + temp[start];
				// Increase start boundary position
				start += 1;
				// Increase operation
				operation += 1;
			}
		}
		print("\n Array : ");
		this.printArray(arr, n);
		print("\n Result : " + operation);
	}
}
fun main(args: Array < String > ): Unit
{
	val task: PalindromeOperation = PalindromeOperation();
	val arr1: Array < Int > = arrayOf(2, 1, 4, 3, 5, 2, 1, 2);
	val arr2: Array < Int > = arrayOf(4, 3, 2, 3, 4, 7);
	val arr3: Array < Int > = arrayOf(1, 2, 1);
	val arr4: Array < Int > = arrayOf(1, 4, 1, 5, 1);
	val arr5: Array < Int > = arrayOf(1, 2, 3, 1);
	// Test A
	var n: Int = arr1.count();
	/*
	    [2, 1 , 4 , 3 , 5 , 2, 1, 2]
	     -  -                  -  -
	    [Note boundary element 2, 1 is palindrome]
	    
	    2, 1 ,4 , 3 , [5 + 2] 1, 2
	    -  -             ➀    -  -
	    2, 1 [4 + 3] ,  7,   1, 2
	            ➁
	    2, 1,    7   , 7     1, 2  
	             -     - 
	    [ 2  1 7  7 1 2 ] is palindrome   
	    --------------------------------    
	    Result : 2 operation
	*/
	task.minMergePalindrome(arr1, n);
	// Test B
	n = arr2.count();
	/*
	    [4   3   2   3   4   7]
	    ---------------------
	    [4 + 3   2   3   4   7]
	       ➀ //operation
	       7     2 + 3   4   7
	             -   -
	               ➁  //operation    
	       7       5  +  4    7
	               -     -
	                  ➂ //operation
	       7          9       7
	    
	    [7 9 7] is palindrome
	   -------------------------   
	    Result = 3 operation
	*/
	task.minMergePalindrome(arr2, n);
	// Test C
	n = arr3.count();
	/*
	    [1, 2, 1]
	     -  -  - 
	    1 2 1  is palindrome
	    Result = 0
	*/
	task.minMergePalindrome(arr3, n);
	// Test D
	n = arr4.count();
	/*
	    [1, 4, 1 , 5, 1]
	     -  -  - 
	    
	    1, [4 + 1] , 5   1
	          ➀  
	    1,  5 , 5   1
	       
	    Result = 1
	*/
	task.minMergePalindrome(arr4, n);
	// Test E
	n = arr5.count();
	/*
	    [1 , [2 + 3], 1]
	            ➀ 
	    Result = 1
	*/
	task.minMergePalindrome(arr5, n);
}

Output

 Array :  2 1 4 3 5 2 1 2
 Result : 2
 Array :  4 3 2 3 4 7
 Result : 3
 Array :  1 2 1
 Result : 0
 Array :  1 4 1 5 1
 Result : 1
 Array :  1 2 3 1
 Result : 1


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