missingandrepeatingnumber.cpp

Given an unsorted array of size n. Array elements are in the range of 1 to n. One number from set {1, 2, …n} is missing and one number occurs twice in the array. Find these two numbers.

Examples: 

Input: arr[] = {3, 1, 3}
Output: Missing = 2, Repeating = 3
Explanation: In the array, 2 is missing and 3 occurs twice 

// C++ program to Find the repeating
// and missing elements

#include <bits/stdc++.h>
using namespace std;

void printTwoElements(int arr[], int size)
{
	int i;
	cout << "The repeating element is ";

	for (i = 0; i < size; i++) {
		if (arr[abs(arr[i]) - 1] > 0)
			arr[abs(arr[i]) - 1] = -arr[abs(arr[i]) - 1];
		else
			cout << abs(arr[i]) << "\n";
	}

	cout << "and the missing element is ";
	for (i = 0; i < size; i++) {
		if (arr[i] > 0)
			cout << (i + 1);
	}
}

/* Driver code */
int main()
{
	int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
	int n = sizeof(arr) / sizeof(arr[0]);
	printTwoElements(arr, n);
}

// This code is contributed by Shivi_Aggarwal
// Java program to Find the repeating
// and missing elements

import java.io.*;

class GFG {

	static void printTwoElements(int arr[], int size)
	{
		int i;
		System.out.print("The repeating element is ");

		for (i = 0; i < size; i++) {
			int abs_val = Math.abs(arr[i]);
			if (arr[abs_val - 1] > 0)
				arr[abs_val - 1] = -arr[abs_val - 1];
			else
				System.out.println(abs_val);
		}

		System.out.print("and the missing element is ");
		for (i = 0; i < size; i++) {
			if (arr[i] > 0)
				System.out.println(i + 1);
		}
	}

	// Driver code
	public static void main(String[] args)
	{
		int arr[] = { 7, 3, 4, 5, 5, 6, 2 };
		int n = arr.length;
		printTwoElements(arr, n);
	}
}

// This code is contributed by Gitanjali