Given an integer array of size 2N + 1. In this given array, N numbers are present twice and one number is present only once in the array.

TITLE:-

Find the Unique Element

PROBLEM:-

Given an integer array of size 2N + 1. In this given array, N numbers are present twice and one number is present only once in the array.

You need to find and return that number which is unique in the array.

Note : Given array will always contain odd number of elements.

Input format :

Line 1 : Array size i.e. 2N+1
Line 2 : Array elements (separated by space)

Output Format :

Unique element present in the array

Constraints :

1 <= N <= 10^6

Sample Input :

7
2 3 1 6 3 6 2

Sample Output :

1

SOLUTION:-

#include<iostream>
#include <algorithm>
using namespace std;

// arr - input array
// size - size of array

int FindUnique(int arr[], int size){
    /* Don't write main().
     * Don't read input, it is passed as function argument.
     * Return output and don't print it.
     * Taking input and printing output is handled automatically.
     */
   int counter=1;
   int find=arr[0];
    for(int i=1;i<size;i++)
    {
        if(arr[i]==find)
        {
            counter++;
        }
    }
    if(counter==1)
        return find;
    
    int smallout=FindUnique(arr+1,size-1);
    return smallout;
}

int main() {

	int size;

	cin>>size;
	int *input=new int[1+size];	
	
	for(int i=0;i<size;i++)
		cin>>input[i];
	
	cout<<FindUnique(input,size)<<endl;
		
	return 0;
}




• Tweet
• Share
• Share
• Share
• Share

About Admin MC3

This is dummy text. It is not meant to be read. Accordingly, it is difficult to figure out when to end it. But then, this is dummy text. It is not meant to be read. Period.

Related Post