Find the missing number in unordered arithmetic progression

Here given code implementation process.

import java.util.Arrays;
/*
    Java program
    Find the missing number in unordered arithmetic progression
*/
public class ArithmeticProgression
{
	public int missingNo(int[] series, 
  			int low, int high, int interval)
	{
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		int mid = low + (high - low) / 2;
		if (mid + 1 <= high && (series[mid + 1] - series[mid]) != interval)
		{
			return series[mid] + interval;
		}
		if (mid > 0 && (series[mid] - series[mid - 1]) != interval)
		{
			return series[mid - 1] + interval;
		}
		if ((mid * interval) + series[0] == series[mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return missingNo(series, low, mid - 1, interval);
	}
	public void missingArithmeticPro(int[] series, int n)
	{
		if (n > 1)
		{
			// Sort series
			Arrays.sort(series);
			// When n is more than 1
			// Find interval between sequence
			int interval = (series[n - 1] - series[0]) / n;
			// Find missing element
			int value = missingNo(series, 0, n - 1, interval);
			// Display missing element
			System.out.print("\n " + value);
		}
	}
	public static void main(String[] args)
	{
		ArithmeticProgression task = new ArithmeticProgression();
		// Arithmetic progression series with one missing element
		int[] a = {
			1 , 4 , 7 , 13 , 16 , 19
		};
		int[] b = {
			60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54
		};
		int[] c = {
			14 , 2 , 11 , 5
		};
		// Note : Here first and last elements are not considered as missing
		// Test A 
		int n = a.length;
		// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
		// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
		//                     -
		// -----------------------------------
		// Result = 10
		task.missingArithmeticPro(a, n);
		// Test B 
		n = b.length;
		// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
		// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
		//                        -
		// ---------------------------------------------------------
		// Result = 24
		task.missingArithmeticPro(b, n);
		// Test C 
		n = c.length;
		// arr[]  = [14, 2 , 11 , 5]
		// sort[] = ]2 , 5 , 11 , 14]
		// result = 8
		task.missingArithmeticPro(c, n);
	}
}

Output

 10
 24
 8
// Include header file
#include <iostream>
#include <algorithm>

using namespace std;
/*
    C++ program
    Find the missing number in unordered arithmetic progression
*/
class ArithmeticProgression
{
	public: int missingNo(int series[], int low, 
      			int high, int interval)
	{
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		int mid = low + (high - low) / 2;
		if (mid + 1 <= high && (series[mid + 1] - series[mid]) != interval)
		{
			return series[mid] + interval;
		}
		if (mid > 0 && (series[mid] - series[mid - 1]) != interval)
		{
			return series[mid - 1] + interval;
		}
		if ((mid *interval) + series[0] == series[mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return this->missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return this->missingNo(series, low, mid - 1, interval);
	}
	void missingArithmeticPro(int series[], int n)
	{
		if (n > 1)
		{
			// Sort series
			sort(series, series + n);
			// When n is more than 1
			// Find interval between sequence
			int interval = (series[n - 1] - series[0]) / n;
			// Find missing element
			int value = this->missingNo(series, 0, n - 1, interval);
			// Display missing element
			cout << "\n " << value;
		}
	}
};
int main()
{
	ArithmeticProgression *task = new ArithmeticProgression();
	// Arithmetic progression series with one missing element
	int a[] = {
		1 , 4 , 7 , 13 , 16 , 19
	};
	int b[] = {
		60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54
	};
	int c[] = {
		14 , 2 , 11 , 5
	};
	// Note : Here first and last elements are not considered as missing
	// Test A 
	int n = sizeof(a) / sizeof(a[0]);
	// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	//                     -
	// -----------------------------------
	// Result = 10
	task->missingArithmeticPro(a, n);
	// Test B 
	n = sizeof(b) / sizeof(b[0]);
	// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	//                        -
	// ---------------------------------------------------------
	// Result = 24
	task->missingArithmeticPro(b, n);
	// Test C 
	n = sizeof(c) / sizeof(c[0]);
	// arr[]  = [14, 2 , 11 , 5]
	// sort[] = ]2 , 5 , 11 , 14]
	// result = 8
	task->missingArithmeticPro(c, n);
	return 0;
}

Output

 10
 24
 8
// Include namespace system
using System;
/*
    Csharp program
    Find the missing number in unordered arithmetic progression
*/
public class ArithmeticProgression
{
	public int missingNo(int[] series, int low, int high, int interval)
	{
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		int mid = low + (high - low) / 2;
		if (mid + 1 <= high && (series[mid + 1] - series[mid]) != interval)
		{
			return series[mid] + interval;
		}
		if (mid > 0 && (series[mid] - series[mid - 1]) != interval)
		{
			return series[mid - 1] + interval;
		}
		if ((mid * interval) + series[0] == series[mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return this.missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return this.missingNo(series, low, mid - 1, interval);
	}
	public void missingArithmeticPro(int[] series, int n)
	{
		if (n > 1)
		{
			// Sort series
			Array.Sort(series);
			// When n is more than 1
			// Find interval between sequence
			int interval = (series[n - 1] - series[0]) / n;
			// Find missing element
			int value = this.missingNo(series, 0, n - 1, interval);
			// Display missing element
			Console.Write("\n " + value);
		}
	}
	public static void Main(String[] args)
	{
		ArithmeticProgression task = new ArithmeticProgression();
		// Arithmetic progression series with one missing element
		int[] a = {
			1 , 4 , 7 , 13 , 16 , 19
		};
		int[] b = {
			60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54
		};
		int[] c = {
			14 , 2 , 11 , 5
		};
		// Note : Here first and last elements are not considered as missing
		// Test A 
		int n = a.Length;
		// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
		// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
		//                     -
		// -----------------------------------
		// Result = 10
		task.missingArithmeticPro(a, n);
		// Test B 
		n = b.Length;
		// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
		// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
		//                        -
		// ---------------------------------------------------------
		// Result = 24
		task.missingArithmeticPro(b, n);
		// Test C 
		n = c.Length;
		// arr[]  = [14, 2 , 11 , 5]
		// sort[] = ]2 , 5 , 11 , 14]
		// result = 8
		task.missingArithmeticPro(c, n);
	}
}

Output

 10
 24
 8
package main
import "sort"
import "fmt"
/*
    Go program
    Find the missing number in unordered arithmetic progression
*/
type ArithmeticProgression struct {}
func getArithmeticProgression() * ArithmeticProgression {
	var me *ArithmeticProgression = &ArithmeticProgression {}
	return me
}
func(this ArithmeticProgression) missingNo(series[] int, 
	low int, high int, interval int) int {
	if low >= high {
		// When no result
		return -1
	}
	// Find middle element
	var mid int = low + (high - low) / 2
	if mid + 1 <= high && (series[mid + 1] - series[mid]) != interval {
		return series[mid] + interval
	}
	if mid > 0 && (series[mid] - series[mid - 1]) != interval {
		return series[mid - 1] + interval
	}
	if (mid * interval) + series[0] == series[mid] {
		// When the middle position element is a valid element 
		// of the given arithmetic progression.
		// Check sequence of right-sided elements from middle.
		return this.missingNo(series, mid + 1, high, interval)
	}
	// Check sequence of left-sided elements from middle.
	return this.missingNo(series, low, mid - 1, interval)
}
func(this ArithmeticProgression) missingArithmeticPro(series[] int, n int) {
	if n > 1 {
		// Sort series
		sort.Ints(series)
		// When n is more than 1
		// Find interval between sequence
		var interval int = (series[n - 1] - series[0]) / n
		// Find missing element
		var value int = this.missingNo(series, 0, n - 1, interval)
		// Display missing element
		fmt.Print("\n ", value)
	}
}
func main() {
	var task * ArithmeticProgression = getArithmeticProgression()
	// Arithmetic progression series with one missing element
	var a = [] int {
		1,
		4,
		7,
		13,
		16,
		19,
	}
	var b = [] int {
		60,
		6,
		66,
		12,
		30,
		48,
		36,
		42,
		18,
		54,
	}
	var c = [] int {
		14,
		2,
		11,
		5,
	}
	// Note : Here first and last elements are not considered as missing
	// Test A 
	var n int = len(a)
	// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	//                     -
	// -----------------------------------
	// Result = 10
	task.missingArithmeticPro(a, n)
	// Test B 
	n = len(b)
	// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	//                        -
	// ---------------------------------------------------------
	// Result = 24
	task.missingArithmeticPro(b, n)
	// Test C 
	n = len(c)
	// arr[]  = [14, 2 , 11 , 5]
	// sort[] = ]2 , 5 , 11 , 14]
	// result = 8
	task.missingArithmeticPro(c, n)
}

Output

 10
 24
 8
<?php
/*
    Php program
    Find the missing number in unordered arithmetic progression
*/
class ArithmeticProgression
{
	public	function missingNo($series, $low, $high, $interval)
	{
		if ($low >= $high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		$mid = $low + (int)(($high - $low) / 2);
		if ($mid + 1 <= $high && 
            ($series[$mid + 1] - $series[$mid]) != $interval)
		{
			return $series[$mid] + $interval;
		}
		if ($mid > 0 && ($series[$mid] - $series[$mid - 1]) != $interval)
		{
			return $series[$mid - 1] + $interval;
		}
		if (($mid * $interval) + $series[0] == $series[$mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return $this->missingNo($series, $mid + 1, $high, $interval);
		}
		// Check sequence of left-sided elements from middle.
		return $this->missingNo($series, $low, $mid - 1, $interval);
	}
	public	function missingArithmeticPro($series, $n)
	{
		if ($n > 1)
		{
			// Sort series
			sort($series);
			// When n is more than 1
			// Find interval between sequence
			$interval = (int)(($series[$n - 1] - $series[0]) / $n);
			// Find missing element
			$value = $this->missingNo($series, 0, $n - 1, $interval);
			// Display missing element
			echo("\n ".$value);
		}
	}
}

function main()
{
	$task = new ArithmeticProgression();
	// Arithmetic progression series with one missing element
	$a = array(1, 4, 7, 13, 16, 19);
	$b = array(60, 6, 66, 12, 30, 48, 36, 42, 18, 54);
	$c = array(14, 2, 11, 5);
	// Note : Here first and last elements are not considered as missing
	// Test A 
	$n = count($a);
	// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	//                     -
	// -----------------------------------
	// Result = 10
	$task->missingArithmeticPro($a, $n);
	// Test B 
	$n = count($b);
	// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	//                        -
	// ---------------------------------------------------------
	// Result = 24
	$task->missingArithmeticPro($b, $n);
	// Test C 
	$n = count($c);
	// arr[]  = [14, 2 , 11 , 5]
	// sort[] = ]2 , 5 , 11 , 14]
	// result = 8
	$task->missingArithmeticPro($c, $n);
}
main();

Output

 10
 24
 8
/*
    Node JS program
    Find the missing number in unordered arithmetic progression
*/
class ArithmeticProgression
{
	missingNo(series, low, high, interval)
	{
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		var mid = low + parseInt((high - low) / 2);
		if (mid + 1 <= high && 
            (series[mid + 1] - series[mid]) != interval)
		{
			return series[mid] + interval;
		}
		if (mid > 0 && (series[mid] - series[mid - 1]) != interval)
		{
			return series[mid - 1] + interval;
		}
		if ((mid * interval) + series[0] == series[mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return this.missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return this.missingNo(series, low, mid - 1, interval);
	}
	missingArithmeticPro(series, n)
	{
		if (n > 1)
		{
			// Sort series
			series.sort(function(a, b)
			{
				return a - b;
			});
			// When n is more than 1
			// Find interval between sequence
			var interval = parseInt((series[n - 1] - series[0]) / n);
			// Find missing element
			var value = this.missingNo(series, 0, n - 1, interval);
			// Display missing element
			process.stdout.write("\n " + value);
		}
	}
}

function main()
{
	var task = new ArithmeticProgression();
	// Arithmetic progression series with one missing element
	var a = [1, 4, 7, 13, 16, 19];
	var b = [60, 6, 66, 12, 30, 48, 36, 42, 18, 54];
	var c = [14, 2, 11, 5];
	// Note : Here first and last elements are not considered as missing
	// Test A 
	var n = a.length;
	// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	//                     -
	// -----------------------------------
	// Result = 10
	task.missingArithmeticPro(a, n);
	// Test B 
	n = b.length;
	// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	//                        -
	// ---------------------------------------------------------
	// Result = 24
	task.missingArithmeticPro(b, n);
	// Test C 
	n = c.length;
	// arr[]  = [14, 2 , 11 , 5]
	// sort[] = ]2 , 5 , 11 , 14]
	// result = 8
	task.missingArithmeticPro(c, n);
}
main();

Output

 10
 24
 8
#    Python 3 program
#    Find the missing number in unordered arithmetic progression
class ArithmeticProgression :
	def missingNo(self, series, low, high, interval) :
		if (low >= high) :
			#  When no result
			return -1
		
		#  Find middle element
		mid = low + int((high - low) / 2)
		if (mid + 1 <= high and(series[mid + 1] - series[mid]) != interval) :
			return series[mid] + interval
		
		if (mid > 0 and(series[mid] - series[mid - 1]) != interval) :
			return series[mid - 1] + interval
		
		if ((mid * interval) + series[0] == series[mid]) :
			#  When the middle position element is a valid element 
			#  of the given arithmetic progression.
			#  Check sequence of right-sided elements from middle.
			return self.missingNo(series, mid + 1, high, interval)
		
		#  Check sequence of left-sided elements from middle.
		return self.missingNo(series, low, mid - 1, interval)
	
	def missingArithmeticPro(self, series, n) :
		if (n > 1) :
			#  Sort series
			series.sort()
			#  When n is more than 1
			#  Find interval between sequence
			interval = int((series[n - 1] - series[0]) / n)
			#  Find missing element
			value = self.missingNo(series, 0, n - 1, interval)
			#  Display missing element
			print("\n ", value, end = "")
		
	

def main() :
	task = ArithmeticProgression()
	#  Arithmetic progression series with one missing element
	a = [1, 4, 7, 13, 16, 19]
	b = [60, 6, 66, 12, 30, 48, 36, 42, 18, 54]
	c = [14, 2, 11, 5]
	#  Note : Here first and last elements are not considered as missing
	#  Test A 
	n = len(a)
	#  arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	#  sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	#                      -
	#  -----------------------------------
	#  Result = 10
	task.missingArithmeticPro(a, n)
	#  Test B 
	n = len(b)
	#  arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	#  sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	#                         -
	#  ---------------------------------------------------------
	#  Result = 24
	task.missingArithmeticPro(b, n)
	#  Test C 
	n = len(c)
	#  arr[]  = [14, 2 , 11 , 5]
	#  sort[] = ]2 , 5 , 11 , 14]
	#  result = 8
	task.missingArithmeticPro(c, n)

if __name__ == "__main__": main()

Output

  10
  24
  8
#    Ruby program
#    Find the missing number in unordered arithmetic progression
class ArithmeticProgression 
	def missingNo(series, low, high, interval) 
		if (low >= high) 
			#  When no result
			return -1
		end

		#  Find middle element
		mid = low + (high - low) / 2
		if (mid + 1 <= high && (series[mid + 1] - series[mid]) != interval) 
			return series[mid] + interval
		end

		if (mid > 0 && (series[mid] - series[mid - 1]) != interval) 
			return series[mid - 1] + interval
		end

		if ((mid * interval) + series[0] == series[mid]) 
			#  When the middle position element is a valid element 
			#  of the given arithmetic progression.
			#  Check sequence of right-sided elements from middle.
			return self.missingNo(series, mid + 1, high, interval)
		end

		#  Check sequence of left-sided elements from middle.
		return self.missingNo(series, low, mid - 1, interval)
	end

	def missingArithmeticPro(series, n) 
		if (n > 1) 
			#  Sort series
			series = series.sort
			#  When n is more than 1
			#  Find interval between sequence
			interval = (series[n - 1] - series[0]) / n
			#  Find missing element
			value = self.missingNo(series, 0, n - 1, interval)
			#  Display missing element
			print("\n ", value)
		end

	end

end

def main() 
	task = ArithmeticProgression.new()
	#  Arithmetic progression series with one missing element
	a = [1, 4, 7, 13, 16, 19]
	b = [60, 6, 66, 12, 30, 48, 36, 42, 18, 54]
	c = [14, 2, 11, 5]
	#  Note : Here first and last elements are not considered as missing
	#  Test A 
	n = a.length
	#  arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	#  sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	#                      -
	#  -----------------------------------
	#  Result = 10
	task.missingArithmeticPro(a, n)
	#  Test B 
	n = b.length
	#  arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	#  sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	#                         -
	#  ---------------------------------------------------------
	#  Result = 24
	task.missingArithmeticPro(b, n)
	#  Test C 
	n = c.length
	#  arr[]  = [14, 2 , 11 , 5]
	#  sort[] = ]2 , 5 , 11 , 14]
	#  result = 8
	task.missingArithmeticPro(c, n)
end

main()

Output

 10
 24
 8
import scala.collection.mutable._;
/*
    Scala program
    Find the missing number in unordered arithmetic progression
*/
class ArithmeticProgression()
{
	def missingNo(series: Array[Int], low: Int, 
      			high: Int, interval: Int): Int = {
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		var mid: Int = low + (high - low) / 2;
		if (mid + 1 <= high && (series(mid + 1) - series(mid)) != interval)
		{
			return series(mid) + interval;
		}
		if (mid > 0 && (series(mid) - series(mid - 1)) != interval)
		{
			return series(mid - 1) + interval;
		}
		if ((mid * interval) + series(0) == series(mid))
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return missingNo(series, low, mid - 1, interval);
	}
	def missingArithmeticPro(series: Array[Int], n: Int): Unit = {
		if (n > 1)
		{
			// Sort series
			var v = series.sorted;
			// When n is more than 1
			// Find interval between sequence
			var interval: Int = (v(n - 1) - v(0)) / n;
			// Find missing element
			var value: Int = missingNo(v, 0, n - 1, interval);
			// Display missing element
			print("\n " + value);
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: ArithmeticProgression = new ArithmeticProgression();
		// Arithmetic progression series with one missing element
		var a: Array[Int] = Array(1, 4, 7, 13, 16, 19);
		var b: Array[Int] = Array(60, 6, 66, 12, 30, 48, 36, 42, 18, 54);
		var c: Array[Int] = Array(14, 2, 11, 5);
		// Note : Here first and last elements are not considered as missing
		// Test A 
		var n: Int = a.length;
		// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
		// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
		//                     -
		// -----------------------------------
		// Result = 10
		task.missingArithmeticPro(a, n);
		// Test B 
		n = b.length;
		// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
		// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
		//                        -
		// ---------------------------------------------------------
		// Result = 24
		task.missingArithmeticPro(b, n);
		// Test C 
		n = c.length;
		// arr[]  = [14, 2 , 11 , 5]
		// sort[] = ]2 , 5 , 11 , 14]
		// result = 8
		task.missingArithmeticPro(c, n);
	}
}

Output

 10
 24
 8
import Foundation;
/*
    Swift 4 program
    Find the missing number in unordered arithmetic progression
*/
class ArithmeticProgression
{
	func missingNo(_ series: [Int], _ low: Int, 
      _ high: Int, _ interval: Int) -> Int
	{
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		let mid: Int = low + (high - low) / 2;
		if (mid + 1 <= high && 
            (series[mid + 1] - series[mid])  != interval)
		{
			return series[mid] + interval;
		}
		if (mid > 0 && (series[mid] - series[mid - 1])  != interval)
		{
			return series[mid - 1] + interval;
		}
		if ((mid * interval) + series[0] == series[mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return self.missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return self.missingNo(series, low, mid - 1, interval);
	}
	func missingArithmeticPro(_ v: [Int], _ n: Int)
	{
		if (n > 1)
		{
			// Sort series
			let series = v.sorted();
			// When n is more than 1
			// Find interval between sequence
			let interval: Int = (series[n - 1] - series[0]) / n;
			// Find missing element
			let value: Int = self.missingNo(series, 0, n - 1, interval);
			// Display missing element
			print("\n ", value, terminator: "");
		}
	}
}
func main()
{
	let task: ArithmeticProgression = ArithmeticProgression();
	// Arithmetic progression series with one missing element
	let a: [Int] = [1, 4, 7, 13, 16, 19];
	let b: [Int] = [60, 6, 66, 12, 30, 48, 36, 42, 18, 54];
	let c: [Int] = [14, 2, 11, 5];
	// Note : Here first and last elements are not considered as missing
	// Test A 
	var n: Int = a.count;
	// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	//                     -
	// -----------------------------------
	// Result = 10
	task.missingArithmeticPro(a, n);
	// Test B 
	n = b.count;
	// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	//                        -
	// ---------------------------------------------------------
	// Result = 24
	task.missingArithmeticPro(b, n);
	// Test C 
	n = c.count;
	// arr[]  = [14, 2 , 11 , 5]
	// sort[] = ]2 , 5 , 11 , 14]
	// result = 8
	task.missingArithmeticPro(c, n);
}
main();

Output

  10
  24
  8
/*
    Kotlin program
    Find the missing number in unordered arithmetic progression
*/
class ArithmeticProgression
{
	fun missingNo(series: Array < Int > , 
                   low: Int, high: Int, interval: Int): Int
	{
		if (low >= high)
		{
			// When no result
			return -1;
		}
		// Find middle element
		val mid: Int = low + (high - low) / 2;
		if (mid + 1 <= high && (series[mid + 1] - series[mid]) != interval)
		{
			return series[mid] + interval;
		}
		if (mid > 0 && (series[mid] - series[mid - 1]) != interval)
		{
			return series[mid - 1] + interval;
		}
		if ((mid * interval) + series[0] == series[mid])
		{
			// When the middle position element is a valid element 
			// of the given arithmetic progression.
			// Check sequence of right-sided elements from middle.
			return this.missingNo(series, mid + 1, high, interval);
		}
		// Check sequence of left-sided elements from middle.
		return this.missingNo(series, low, mid - 1, interval);
	}
	fun missingArithmeticPro(series: Array < Int > , n: Int): Unit
	{
		if (n > 1)
		{
			// Sort series
			series.sort();
			// When n is more than 1
			// Find interval between sequence
			val interval: Int = (series[n - 1] - series[0]) / n;
			// Find missing element
			val value: Int = this.missingNo(series, 0, n - 1, interval);
			// Display missing element
			print("\n " + value);
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: ArithmeticProgression = ArithmeticProgression();
	// Arithmetic progression series with one missing element
	val a: Array < Int > = arrayOf(1, 4, 7, 13, 16, 19);
	val b: Array < Int > = arrayOf(60, 6, 66, 12, 30, 48, 36, 42, 18, 54);
	val c: Array < Int > = arrayOf(14, 2, 11, 5);
	// Note : Here first and last elements are not considered as missing
	// Test A 
	var n: Int = a.count();
	// arr[]  = [1 , 4 , 7 , 13 , 16 , 19]
	// sort[] = ]1 , 4 , 7 , 13 , 16 , 19]
	//                     -
	// -----------------------------------
	// Result = 10
	task.missingArithmeticPro(a, n);
	// Test B 
	n = b.count();
	// arr[]  = [ 60 , 6 , 66 , 12 , 30 , 48 , 36 , 42 , 18 , 54 ]
	// sort[] = ] 6 , 12 , 18 , 30 , 36 , 42 , 48 , 54 , 60 , 66 ]
	//                        -
	// ---------------------------------------------------------
	// Result = 24
	task.missingArithmeticPro(b, n);
	// Test C 
	n = c.count();
	// arr[]  = [14, 2 , 11 , 5]
	// sort[] = ]2 , 5 , 11 , 14]
	// result = 8
	task.missingArithmeticPro(c, n);
}

Output

 10
 24
 8



Comment

Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.

New Comment




© 2022, kalkicode.com, All rights reserved