Question

You have been given an integer array/list(ARR) of size N which contains numbers from 0 to (N - 2). Each number is present at least once. That is, if N = 5, the array/list constitutes values ranging from 0 to 3 and among these, there is a single integer value that is present twice. You need to find and return that duplicate number present in the array.

Answer 1

Answer:

#include <iostream>

using namespace std;

int findUnique(int *arr, int size)

{

   int res = arr[0];

       for (int i = 1; i < size; i++)

           res = res ^ arr[i];

       return res;

}

int main()

{

int t;

cin >> t;

while (t--)

{

 int size;

 cin >> size;

 int *input = new int[size];

 for (int i = 0; i < size; ++i)

 {

  cin >> input[i];

 }

 cout << findUnique(input, size) << endl;

}

return 0;

}

Explanation:

Answer 2

Answer:

public static int findDuplicate(int[] arr) {

 

       int sum=0;

       for(int i=0;i<=arr.length-2;i++)

       {

          sum=sum+i;

       }

       int n=0;

       for(int i=0;i<arr.length;i++)

       {

        n=n+arr[i];  

       }

       int x=n-sum;

        return x;

}

Explanation:

subtract the total sum of elements in the array -the sum of the numbers ranging from 0 to n-2