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FindDuplicate.cpp
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FindDuplicate.cpp
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// 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.
// Note : Duplicate number is always present in the given array/list.
// My Code
int duplicateNumber(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>
using namespace std;
#include "solution.h"
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 << duplicateNumber(input, size) << endl;
}
return 0;
}
// CN Code
int duplicateNumber(int *arr, int size)
{
for(int i=0; i<size; i++){
for(int j=i+1; j<size; j++){
if(arr[i] == arr[j]){
return arr[i];
}
}
}
}