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Find the only repetitive element between 1 to n-1 using C++
In this problem, we are given an unordered array arr[] of size N containing values from 1 to N-1 with one value occuring twice in the array. Our task is to find the only repetitive element between 1 to n-1.
Let’s take an example to understand the problem,
Input
arr[] = {3, 5, 4, 1, 2, 1}Output
1
Solution Approach
A simple solution to the problem is traversing the array and for each value find whether the element exists somewhere else in the array. Return the value with double occurrence.
Example 1
Program to illustrate the working of our solution
#include <iostream>
using namespace std;
int findRepValArr(int arr[], int n){
for(int i = 0; i < n; i++)
for(int j = i+1; j < n; j++)
if(arr[i] == arr[j])
return arr[i];
}
int main(){
int arr[] = { 5, 3, 2, 6, 6, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"The repetitive value in the array is "<<findRepValArr(arr, n);
return 0;
}Output
The repetitive value in the array is 6
Another approach to solve the problem is by using the fact that the repetitive value in the array is found by subtracting the sum of all integers from 1 to (N-1) from the array sum.
Sum of 1st N-1 natural number (Sn) = n*(n-1)/2
doubleVal = arrSum - (Sn)
Example 2
Program to illustrate the working of our solution
#include <iostream>
using namespace std;
int findRepValArr(int arr[], int n){
int arrSum = 0;
for(int i = 0; i < n; i++)
arrSum += arr[i];
int sn = (((n)*(n-1))/2);
return arrSum - sn;
}
int main(){
int arr[] = { 5, 3, 2, 6, 6, 1, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
cout<<"The repetitive value in the array is "<<findRepValArr(arr, n);
return 0;
}Output
The repetitive value in the array is 6
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