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FindUnique.cpp
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FindUnique.cpp
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// You have been given an integer array/list(ARR) of size N. Where N is equal to [2M + 1].
// Now, in the given array/list, 'M' numbers are present twice and one number is present only once.
// You need to find and return that number which is unique in the array/list.
// Note: Unique element is always present in the array/list according to the given condition.
//My Code
int findUnique(int *arr, int size)
{
int count, res;
for(int i=0; i<size; i++){
int j= arr[i];
count = 0;
for(int k=0; k<size; k++){
if (arr[k] == j){
count++;
}
}
if(count == 1){
res = arr[i];
}
}
return res;
}
// Main Code
#include <iostream>
#include "solution.h"
using namespace std;
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;
}
// CN Code
int findUnique(int *arr, int size)
{
for(int i=0; i<size; i++){
int j=0;
for(; j<size; j++){
if(i != j && arr[i] == arr[j]){
break;
}
}
if(j == size){
return arr[i];
}
}
}