Program for minimum number of jumps to reach the end

Here given code implementation process.

/*
    C program for
    Program for minimum number of jumps to reach the end
*/
#include <stdio.h>
#include <limits.h>

// This is finding the number of minimum steps 
// or jumps required to reach end of array elements
// When of array not contains negative values
int minimumJumpsToEnd(int arr[], int n)
{
    if (n <= 1)
    {
        // When less than 2 elements
        return 0;
    }
    if (arr[0] == 0)
    {
        // When first element is 0 then not possible to 
        // Visit next element
        // Therefor
        return -1;
    }
    // Use to collect number of jump process
    int jump[n];
    // Initial value
    jump[0] = 0;
    // Execute loop through by size of n
    for (int i = 1; i < n; ++i)
    {
        // Set initial steps jump
        jump[i] = INT_MAX;
        // Inner loop which is finding minimum jump between 0 to i
        for (int j = 0; j < i; ++j)
        {
            if (i <= j + arr[j] && jump[j] != INT_MAX)
            {
                if (jump[i] > (jump[j] + 1))
                {
                    // Update new minimum jump
                    jump[i] = jump[j] + 1;
                }
                // Stop inner loop execution
                j = i;
            }
        }
    }
    // When jump[ n-1] is INT_MAX 
    // That means we not reach the end of array elements
    return jump[n - 1];
}
void displayArray(int arr[], int n)
{
    for (int i = 0; i < n; ++i)
    {
        printf(" %d", arr[i]);
    }
    printf("\n");
}
int main(int argc, char const *argv[])
{
    int arr1[] = {
        2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
    };
    int arr2[] = {
        2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
    };
    // Test A
    int n = sizeof(arr1) / sizeof(arr1[0]);
    displayArray(arr1, n);
    // Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
    // Jump start with 2 -> 3 -> 1 -> 6 -> 9 
    // Jump between      ①    ②   ③   ④    
    // Ans : 4    
    printf(" Minimum Jump to end : %d\n", minimumJumpsToEnd(arr1, n));
    // Test B
    n = sizeof(arr2) / sizeof(arr2[0]);
    displayArray(arr2, n);
    // Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
    // Jump start with 2 -> 3 -> 2 -> 3 -> 2 -> 4 
    // Jump between      ①    ②   ③   ④   ⑤
    printf(" Minimum Jump to end : %d\n", minimumJumpsToEnd(arr2, n));
    return 0;
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
// Java Program 
// Program for minimum number of jumps to reach the end
public class Jumping
{
    // This is finding the number of minimum steps 
    // or jumps required to reach end of array elements
    // When of array not contains negative values
    public int minimumJumpsToEnd(int[] arr, int n)
    {
        if (n <= 1)
        {
            // When less than 2 elements
            return 0;
        }
        if (arr[0] == 0)
        {
            // When first element is 0 then not possible to 
            // Visit next element
            // Therefor
            return -1;
        }
        // Use to collect number of jump process
        int[] jump = new int[n];
        // Initial value
        jump[0] = 0;
        // Execute loop through by size of n
        for (int i = 1; i < n; ++i)
        {
            // Set initial steps jump
            jump[i] = Integer.MAX_VALUE;
            // Inner loop which is finding minimum jump between 0 to i
            for (int j = 0; j < i; ++j)
            {
                if (i <= j + arr[j] && jump[j] != Integer.MAX_VALUE)
                {
                    if (jump[i] > (jump[j] + 1))
                    {
                        // Update new minimum jump
                        jump[i] = jump[j] + 1;
                    }
                    // Stop inner loop execution
                    j = i;
                }
            }
        }
        // When jump[ n-1] is MAX Integer
        // That means we not reach the end of array elements
        return jump[n - 1];
    }
    public void displayArray(int[] arr, int n)
    {
        for (int i = 0; i < n; ++i)
        {
            System.out.print(" " + arr[i] );
        }
        System.out.print("\n");
    }
    public static void main(String args[])
    {
        Jumping task = new Jumping();
        int[] arr1 = {
            2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
        };
        int[] arr2 = {
            2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
        };
        // Test A
        int n = arr1.length;
        task.displayArray(arr1, n);
        // Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
        // Jump start with 2 → 3 → 1 → 6 → 9 
        // Jump between      ①  ②  ③   ④    
        // Ans : 4    
        System.out.println(" Minimum Jump to end : " + 
                         task.minimumJumpsToEnd(arr1, n) );
        // Test B
        n = arr2.length;
        task.displayArray(arr2, n);
        // Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
        // Jump start with 2 → 3 → 2 → 3 → 2 → 4 
        // Jump between      ①  ②  ③   ④  ⑤
        System.out.println(" Minimum Jump to end : " + 
                         task.minimumJumpsToEnd(arr2, n) );
    }
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
// Include header file
#include <iostream>
#include <limits.h>
using namespace std;
// C++ Program 
// Program for minimum number of jumps to reach the end
class Jumping
{
	public:
		// This is finding the number of minimum steps 
		// or jumps required to reach end of array elements
		// When of array not contains negative values
		int minimumJumpsToEnd(int arr[], int n)
		{
			if (n <= 1)
			{
				// When less than 2 elements
				return 0;
			}
			if (arr[0] == 0)
			{
				// When first element is 0 then not possible to 
				// Visit next element
				// Therefor
				return -1;
			}
			// Use to collect number of jump process
			int jump[n];
			// Initial value
			jump[0] = 0;
			// Execute loop through by size of n
			for (int i = 1; i < n; ++i)
			{
				// Set initial steps jump
				jump[i] = INT_MAX;
				// Inner loop which is finding minimum jump between 0 to i
				for (int j = 0; j < i; ++j)
				{
					if (i <= j + arr[j] && jump[j] != INT_MAX)
					{
						if (jump[i] > (jump[j] + 1))
						{
							// Update new minimum jump
							jump[i] = jump[j] + 1;
						}
						// Stop inner loop execution
						j = i;
					}
				}
			}
			// When jump[ n-1] is MAX Integer
			// That means we not reach the end of array elements
			return jump[n - 1];
		}
	void displayArray(int arr[], int n)
	{
		for (int i = 0; i < n; ++i)
		{
			cout << " " << arr[i];
		}
		cout << "\n";
	}
};
int main()
{
	Jumping *task = new Jumping();
	int arr1[] = {
		2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
	};
	int arr2[] = {
		2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
	};
	// Test A
	int n = sizeof(arr1) / sizeof(arr1[0]);
	task->displayArray(arr1, n);
	// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	// Jump start with 2 → 3 → 1 → 6 → 9 
	// Jump between      ①  ②  ③   ④    
	// Ans : 4    
	cout << " Minimum Jump to end : " << task->minimumJumpsToEnd(arr1, n) << endl;
	// Test B
	n = sizeof(arr2) / sizeof(arr2[0]);
	task->displayArray(arr2, n);
	// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	// Jump between      ①  ②  ③   ④  ⑤
	cout << " Minimum Jump to end : " << task->minimumJumpsToEnd(arr2, n) << endl;
	return 0;
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
// Include namespace system
using System;
// Csharp Program 
// Program for minimum number of jumps to reach the end
public class Jumping
{
	// This is finding the number of minimum steps 
	// or jumps required to reach end of array elements
	// When of array not contains negative values
	public int minimumJumpsToEnd(int[] arr, int n)
	{
		if (n <= 1)
		{
			// When less than 2 elements
			return 0;
		}
		if (arr[0] == 0)
		{
			// When first element is 0 then not possible to 
			// Visit next element
			// Therefor
			return -1;
		}
		// Use to collect number of jump process
		int[] jump = new int[n];
		// Initial value
		jump[0] = 0;
		// Execute loop through by size of n
		for (int i = 1; i < n; ++i)
		{
			// Set initial steps jump
			jump[i] = int.MaxValue;
			// Inner loop which is finding minimum jump between 0 to i
			for (int j = 0; j < i; ++j)
			{
				if (i <= j + arr[j] && jump[j] != int.MaxValue)
				{
					if (jump[i] > (jump[j] + 1))
					{
						// Update new minimum jump
						jump[i] = jump[j] + 1;
					}
					// Stop inner loop execution
					j = i;
				}
			}
		}
		// When jump[ n-1] is MAX Integer
		// That means we not reach the end of array elements
		return jump[n - 1];
	}
	public void displayArray(int[] arr, int n)
	{
		for (int i = 0; i < n; ++i)
		{
			Console.Write(" " + arr[i]);
		}
		Console.Write("\n");
	}
	public static void Main(String[] args)
	{
		Jumping task = new Jumping();
		int[] arr1 = {
			2 , 1 , 3 , 2 , 3 , 1 , 6 , 7 , 8 , 4 , 2 , 9
		};
		int[] arr2 = {
			2 , 3 , 0 , 1 , 2 , 3 , 0 , 1 , 2 , 6 , 4
		};
		// Test A
		int n = arr1.Length;
		task.displayArray(arr1, n);
		// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
		// Jump start with 2 → 3 → 1 → 6 → 9 
		// Jump between      ①  ②  ③   ④    
		// Ans : 4    
		Console.WriteLine(" Minimum Jump to end : " + 
                          task.minimumJumpsToEnd(arr1, n));
		// Test B
		n = arr2.Length;
		task.displayArray(arr2, n);
		// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
		// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
		// Jump between      ①  ②  ③   ④  ⑤
		Console.WriteLine(" Minimum Jump to end : " + 
                          task.minimumJumpsToEnd(arr2, n));
	}
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
package main
import "math"
import "fmt"
// Go Program 
// Program for minimum number of jumps to reach the end

// This is finding the number of minimum steps 
// or jumps required to reach end of array elements
// When of array not contains negative values
func minimumJumpsToEnd(arr[] int, n int) int {
	if n <= 1 {
		// When less than 2 elements
		return 0
	}
	if arr[0] == 0 {
		// When first element is 0 then not possible to 
		// Visit next element
		// Therefor
		return -1
	}
	// Use to collect number of jump process
	var jump = make([]int,n)
	// Execute loop through by size of n
	for i := 1 ; i < n ; i++ {
		// Set initial steps jump
		jump[i] = math.MaxInt64
		// Inner loop which is finding minimum jump between 0 to i
		for j := 0 ; j < i ; j++ {
			if i <= j + arr[j] && jump[j] != math.MaxInt64 {
				if jump[i] > (jump[j] + 1) {
					// Update new minimum jump
					jump[i] = jump[j] + 1
				}
				// Stop inner loop execution
				j = i
			}
		}
	}
	// When jump[ n-1] is MAX Integer
	// That means we not reach the end of array elements
	return jump[n - 1]
}
func displayArray(arr[] int, n int) {
	for i := 0 ; i < n ; i++ {
		fmt.Print(" ", arr[i])
	}
	fmt.Print("\n")
}
func main() {
	var arr1 = [] int {2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 }
	var arr2 = [] int { 2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4 }
	// Test A
	var n int = len(arr1)
	displayArray(arr1, n)
	// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	// Jump start with 2 → 3 → 1 → 6 → 9 
	// Jump between      ①  ②  ③   ④    
	// Ans : 4    
	fmt.Println(" Minimum Jump to end : ", minimumJumpsToEnd(arr1, n))
	// Test B
	n = len(arr2)
	displayArray(arr2, n)
	// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	// Jump between      ①  ②  ③   ④  ⑤
	fmt.Println(" Minimum Jump to end : ", minimumJumpsToEnd(arr2, n))
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
<?php
// Php Program 
// Program for minimum number of jumps to reach the end
class Jumping
{
	// This is finding the number of minimum steps 
	// or jumps required to reach end of array elements
	// When of array not contains negative values
	public	function minimumJumpsToEnd($arr, $n)
	{
		if ($n <= 1)
		{
			// When less than 2 elements
			return 0;
		}
		if ($arr[0] == 0)
		{
			// When first element is 0 then not possible to 
			// Visit next element
			// Therefor
			return -1;
		}
		// Use to collect number of jump process
		$jump = array_fill(0, $n, 0);
		// Execute loop through by size of n
		for ($i = 1; $i < $n; ++$i)
		{
			// Set initial steps jump
			$jump[$i] = PHP_INT_MAX;
			// Inner loop which is finding minimum jump between 0 to i
			for ($j = 0; $j < $i; ++$j)
			{
				if ($i <= $j + $arr[$j] && $jump[$j] != PHP_INT_MAX)
				{
					if ($jump[$i] > ($jump[$j] + 1))
					{
						// Update new minimum jump
						$jump[$i] = $jump[$j] + 1;
					}
					// Stop inner loop execution
					$j = $i;
				}
			}
		}
		// When jump[ n-1] is MAX Integer
		// That means we not reach the end of array elements
		return $jump[$n - 1];
	}
	public	function displayArray($arr, $n)
	{
		for ($i = 0; $i < $n; ++$i)
		{
			echo(" ".$arr[$i]);
		}
		echo("\n");
	}
}

function main()
{
	$task = new Jumping();
	$arr1 = array(2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9);
	$arr2 = array(2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4);
	// Test A
	$n = count($arr1);
	$task->displayArray($arr1, $n);
	// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	// Jump start with 2 → 3 → 1 → 6 → 9 
	// Jump between      ①  ②  ③   ④    
	// Ans : 4    
	echo(" Minimum Jump to end : ".
         $task->minimumJumpsToEnd($arr1, $n).
		"\n");
	// Test B
	$n = count($arr2);
	$task->displayArray($arr2, $n);
	// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	// Jump between      ①  ②  ③   ④  ⑤
	echo(" Minimum Jump to end : ".
         $task->minimumJumpsToEnd($arr2, $n).
		"\n");
}
main();

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
// Node JS Program 
// Program for minimum number of jumps to reach the end
class Jumping
{
	// This is finding the number of minimum steps 
	// or jumps required to reach end of array elements
	// When of array not contains negative values
	minimumJumpsToEnd(arr, n)
	{
		if (n <= 1)
		{
			// When less than 2 elements
			return 0;
		}
		if (arr[0] == 0)
		{
			// When first element is 0 then not possible to 
			// Visit next element
			// Therefor
			return -1;
		}
		// Use to collect number of jump process
		var jump = Array(n).fill(0);
		// Execute loop through by size of n
		for (var i = 1; i < n; ++i)
		{
			// Set initial steps jump
			jump[i] = Number.MAX_VALUE;
			// Inner loop which is finding minimum jump between 0 to i
			for (var j = 0; j < i; ++j)
			{
				if (i <= j + arr[j] && jump[j] != Number.MAX_VALUE)
				{
					if (jump[i] > (jump[j] + 1))
					{
						// Update new minimum jump
						jump[i] = jump[j] + 1;
					}
					// Stop inner loop execution
					j = i;
				}
			}
		}
		// When jump[ n-1] is MAX Integer
		// That means we not reach the end of array elements
		return jump[n - 1];
	}
	displayArray(arr, n)
	{
		for (var i = 0; i < n; ++i)
		{
			process.stdout.write(" " + arr[i]);
		}
		process.stdout.write("\n");
	}
}

function main()
{
	var task = new Jumping();
	var arr1 = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9];
	var arr2 = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4];
	// Test A
	var n = arr1.length;
	task.displayArray(arr1, n);
	// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	// Jump start with 2 → 3 → 1 → 6 → 9 
	// Jump between      ①  ②  ③   ④    
	// Ans : 4    
	console.log(" Minimum Jump to end : " + 
                task.minimumJumpsToEnd(arr1, n));
	// Test B
	n = arr2.length;
	task.displayArray(arr2, n);
	// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	// Jump between      ①  ②  ③   ④  ⑤
	console.log(" Minimum Jump to end : " + 
                task.minimumJumpsToEnd(arr2, n));
}
main();

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
import sys
#  Python 3 Program 
#  Program for minimum number of jumps to reach the end
class Jumping :
	#  This is finding the number of minimum steps 
	#  or jumps required to reach end of list elements
	#  When of list not contains negative values
	def minimumJumpsToEnd(self, arr, n) :
		if (n <= 1) :
			#  When less than 2 elements
			return 0
		
		if (arr[0] == 0) :
			#  When first element is 0 then not possible to 
			#  Visit next element
			#  Therefor
			return -1
		
		#  Use to collect number of jump process
		jump = [0] * (n)
		i = 1
		#  Execute loop through by size of n
		while (i < n) :
			#  Set initial steps jump
			jump[i] = sys.maxsize
			j = 0
			#  Inner loop which is finding minimum jump between 0 to i
			while (j < i) :
				if (i <= j + arr[j] and jump[j] != sys.maxsize) :
					if (jump[i] > (jump[j] + 1)) :
						#  Update new minimum jump
						jump[i] = jump[j] + 1
					
					#  Stop inner loop execution
					j = i
				
				j += 1
			
			i += 1
		
		#  When jump[ n-1] is MAX Integer
		#  That means we not reach the end of list elements
		return jump[n - 1]
	
	def displayArray(self, arr, n) :
		i = 0
		while (i < n) :
			print(" ", arr[i], end = "")
			i += 1
		
		print(end = "\n")
	

def main() :
	task = Jumping()
	arr1 = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9]
	arr2 = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	#  Test A
	n = len(arr1)
	task.displayArray(arr1, n)
	#  Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	#  Jump start with 2 → 3 → 1 → 6 → 9 
	#  Jump between      ①  ②  ③   ④    
	#  Ans : 4    
	print(" Minimum Jump to end : ", 
          task.minimumJumpsToEnd(arr1, n))
	#  Test B
	n = len(arr2)
	task.displayArray(arr2, n)
	#  Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	#  Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	#  Jump between      ①  ②  ③   ④  ⑤
	print(" Minimum Jump to end : ", 
          task.minimumJumpsToEnd(arr2, n))

if __name__ == "__main__": main()

Output

  2  1  3  2  3  1  6  7  8  4  2  9
 Minimum Jump to end :  4
  2  3  0  1  2  3  0  1  2  6  4
 Minimum Jump to end :  5
#  Ruby Program 
#  Program for minimum number of jumps to reach the end
class Jumping 
	#  This is finding the number of minimum steps 
	#  or jumps required to reach end of array elements
	#  When of array not contains negative values
	def minimumJumpsToEnd(arr, n) 
		if (n <= 1) 
			#  When less than 2 elements
			return 0
		end

		if (arr[0] == 0) 
			#  When first element is 0 then not possible to 
			#  Visit next element
			#  Therefor
			return -1
		end

		#  Use to collect number of jump process
		jump = Array.new(n) {0}
		i = 1
		#  Execute loop through by size of n
		while (i < n) 
			#  Set initial steps jump
			jump[i] = (2 ** (0. size * 8 - 2))
			j = 0
			#  Inner loop which is finding minimum jump between 0 to i
			while (j < i) 
				if (i <= j + arr[j] && jump[j] != (2 ** (0. size * 8 - 2))) 
					if (jump[i] > (jump[j] + 1)) 
						#  Update new minimum jump
						jump[i] = jump[j] + 1
					end

					#  Stop inner loop execution
					j = i
				end

				j += 1
			end

			i += 1
		end

		#  When jump[ n-1] is MAX Integer
		#  That means we not reach the end of array elements
		return jump[n - 1]
	end

	def displayArray(arr, n) 
		i = 0
		while (i < n) 
			print(" ", arr[i])
			i += 1
		end

		print("\n")
	end

end

def main() 
	task = Jumping.new()
	arr1 = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9]
	arr2 = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	#  Test A
	n = arr1.length
	task.displayArray(arr1, n)
	#  Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	#  Jump start with 2 → 3 → 1 → 6 → 9 
	#  Jump between      ①  ②  ③   ④    
	#  Ans : 4    
	print(" Minimum Jump to end : ", 
          task.minimumJumpsToEnd(arr1, n), "\n")
	#  Test B
	n = arr2.length
	task.displayArray(arr2, n)
	#  Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	#  Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	#  Jump between      ①  ②  ③   ④  ⑤
	print(" Minimum Jump to end : ", 
          task.minimumJumpsToEnd(arr2, n), "\n")
end

main()

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
// Scala Program 
// Program for minimum number of jumps to reach the end
class Jumping()
{
	// This is finding the number of minimum steps 
	// or jumps required to reach end of array elements
	// When of array not contains negative values
	def minimumJumpsToEnd(arr: Array[Int], n: Int): Int = {
		if (n <= 1)
		{
			// When less than 2 elements
			return 0;
		}
		if (arr(0) == 0)
		{
			// When first element is 0 then not possible to 
			// Visit next element
			// Therefor
			return -1;
		}
		// Use to collect number of jump process
		var jump: Array[Int] = Array.fill[Int](n)(0);
		var i: Int = 1;
		// Execute loop through by size of n
		while (i < n)
		{
			// Set initial steps jump
			jump(i) = Int.MaxValue;
			var j: Int = 0;
			// Inner loop which is finding minimum jump between 0 to i
			while (j < i)
			{
				if (i <= j + arr(j) && jump(j) != Int.MaxValue)
				{
					if (jump(i) > (jump(j) + 1))
					{
						// Update new minimum jump
						jump(i) = jump(j) + 1;
					}
					// Stop inner loop execution
					j = i;
				}
				j += 1;
			}
			i += 1;
		}
		// When jump[ n-1] is MAX Integer
		// That means we not reach the end of array elements
		return jump(n - 1);
	}
	def displayArray(arr: Array[Int], n: Int): Unit = {
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr(i));
			i += 1;
		}
		print("\n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Jumping = new Jumping();
		var arr1: Array[Int] = Array(2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9);
		var arr2: Array[Int] = Array(2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4);
		// Test A
		var n: Int = arr1.length;
		task.displayArray(arr1, n);
		// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
		// Jump start with 2 → 3 → 1 → 6 → 9 
		// Jump between      ①  ②  ③   ④    
		// Ans : 4    
		println(" Minimum Jump to end : " + 
                task.minimumJumpsToEnd(arr1, n));
		// Test B
		n = arr2.length;
		task.displayArray(arr2, n);
		// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
		// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
		// Jump between      ①  ②  ③   ④  ⑤
		println(" Minimum Jump to end : " + 
                task.minimumJumpsToEnd(arr2, n));
	}
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5
import Foundation;
// Swift 4 Program 
// Program for minimum number of jumps to reach the end
class Jumping
{
	// This is finding the number of minimum steps 
	// or jumps required to reach end of array elements
	// When of array not contains negative values
	func minimumJumpsToEnd(_ arr: [Int], _ n: Int) -> Int
	{
		if (n <= 1)
		{
			// When less than 2 elements
			return 0;
		}
		if (arr[0] == 0)
		{
			// When first element is 0 then not possible to 
			// Visit next element
			// Therefor
			return -1;
		}
		// Use to collect number of jump process
		var jump: [Int] = Array(repeating: 0, count: n);
		var i: Int = 1;
		// Execute loop through by size of n
		while (i < n)
		{
			// Set initial steps jump
			jump[i] = Int.max;
			var j: Int = 0;
			// Inner loop which is finding minimum jump between 0 to i
			while (j < i)
			{
				if (i <= j + arr[j] && jump[j]  != Int.max)
				{
					if (jump[i] > (jump[j] + 1))
					{
						// Update new minimum jump
						jump[i] = jump[j] + 1;
					}
					// Stop inner loop execution
					j = i;
				}
				j += 1;
			}
			i += 1;
		}
		// When jump[ n-1] is MAX Integer
		// That means we not reach the end of array elements
		return jump[n - 1];
	}
	func displayArray(_ arr: [Int], _ n: Int)
	{
		var i: Int = 0;
		while (i < n)
		{
			print(" ", arr[i], terminator: "");
			i += 1;
		}
		print(terminator: "\n");
	}
}
func main()
{
	let task: Jumping = Jumping();
	let arr1: [Int] = [2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9];
	let arr2: [Int] = [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4];
	// Test A
	var n: Int = arr1.count;
	task.displayArray(arr1, n);
	// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	// Jump start with 2 → 3 → 1 → 6 → 9 
	// Jump between      ①  ②  ③   ④    
	// Ans : 4    
	print(" Minimum Jump to end : ", task.minimumJumpsToEnd(arr1, n));
	// Test B
	n = arr2.count;
	task.displayArray(arr2, n);
	// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	// Jump between      ①  ②  ③   ④  ⑤
	print(" Minimum Jump to end : ", task.minimumJumpsToEnd(arr2, n));
}
main();

Output

  2  1  3  2  3  1  6  7  8  4  2  9
 Minimum Jump to end :  4
  2  3  0  1  2  3  0  1  2  6  4
 Minimum Jump to end :  5
// Kotlin Program 
// Program for minimum number of jumps to reach the end
class Jumping
{
	// This is finding the number of minimum steps 
	// or jumps required to reach end of array elements
	// When of array not contains negative values
	fun minimumJumpsToEnd(arr: Array < Int > , n: Int): Int
	{
		if (n <= 1)
		{
			// When less than 2 elements
			return 0;
		}
		if (arr[0] == 0)
		{
			// When first element is 0 then not possible to 
			// Visit next element
			// Therefor
			return -1;
		}
		// Use to collect number of jump process
		var jump: Array < Int > = Array(n)
		{
			0
		};
		var i: Int = 1;
		// Execute loop through by size of n
		while (i < n)
		{
			// Set initial steps jump
			jump[i] = Int.MAX_VALUE;
			var j: Int = 0;
			// Inner loop which is finding minimum jump between 0 to i
			while (j < i)
			{
				if (i <= j + arr[j] && jump[j] != Int.MAX_VALUE)
				{
					if (jump[i] > (jump[j] + 1))
					{
						// Update new minimum jump
						jump[i] = jump[j] + 1;
					}
					// Stop inner loop execution
					j = i;
				}
				j += 1;
			}
			i += 1;
		}
		// When jump[ n-1] is MAX Integer
		// That means we not reach the end of array elements
		return jump[n - 1];
	}
	fun displayArray(arr: Array < Int > , n: Int): Unit
	{
		var i: Int = 0;
		while (i < n)
		{
			print(" " + arr[i]);
			i += 1;
		}
		print("\n");
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Jumping = Jumping();
	val arr1: Array < Int > = arrayOf(2, 1, 3, 2, 3, 1, 6, 7, 8, 4, 2, 9);
	val arr2: Array < Int > = arrayOf(2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4);
	// Test A
	var n: Int = arr1.count();
	task.displayArray(arr1, n);
	// Given [2, 1, 3,  2, 3,  1 , 6, 7, 8, 4, 2, 9 ]
	// Jump start with 2 → 3 → 1 → 6 → 9 
	// Jump between      ①  ②  ③   ④    
	// Ans : 4    
	println(" Minimum Jump to end : " + 
            task.minimumJumpsToEnd(arr1, n));
	// Test B
	n = arr2.count();
	task.displayArray(arr2, n);
	// Given [2, 3, 0, 1, 2, 3, 0, 1, 2, 6, 4]
	// Jump start with 2 → 3 → 2 → 3 → 2 → 4 
	// Jump between      ①  ②  ③   ④  ⑤
	println(" Minimum Jump to end : " + 
            task.minimumJumpsToEnd(arr2, n));
}

Output

 2 1 3 2 3 1 6 7 8 4 2 9
 Minimum Jump to end : 4
 2 3 0 1 2 3 0 1 2 6 4
 Minimum Jump to end : 5


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