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Find any one of the multiple repeating elements in read only array in C++
In this tutorial, we are going to write a program that finds the repeating element in the given array.
Let's see the steps to solve the problem.
Initialize the array.
Initialize a counter map to store the frequency of each element in the array.
Iterate over the array.
Count each element.
Print the element whose frequency is greater than 1.
Example
Let's see the code.
#include <bits/stdc++.h>
using namespace std;
int findRepeatingElement(int arr[], int n) {
map<int, int> frequencies;
for (int i = 0; i < n; i++) {
map<int, int>::iterator itr = frequencies.find(arr[i]);
if (itr != frequencies.end()) {
itr->second = itr->second + 1;
}
else {
frequencies.insert({arr[i], 1});
}
}
for (map<int, int>::iterator itr = frequencies.begin(); itr != frequencies.end(); ++itr) {
if (itr->second > 1) {
return itr->first;
}
}
}
int main() {
int arr[] = {1, 2, 3, 3, 4, 5, 5, 6};
cout << findRepeatingElement(arr, 8) << endl;
return 0;
}Output
If you run the above code, then you will get the following result
3
Conclusion
If you have any queries in the tutorial, mention them in the comment section.
Updated on: 01-Feb-2021
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