Finding duplicate elements in an array is a common programming task. This article explains a simple approach using two nested loops to identify all repeating elements in an array.
#include <stdio.h>
void findDuplicates(int arr[], int n) {
printf("Repeating elements are:\n");
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
printf("%d\n", arr[i]);
break;
}
}
}
}
int main() {
int arr[] = {1, 2, 3, 4, 2, 3, 5};
int n = sizeof(arr) / sizeof(arr[0]);
findDuplicates(arr, n);
return 0;
}
Input: [1, 2, 3, 4, 2, 3, 5]
Output:
Repeating elements are:
2
3
#include <iostream>
using namespace std;
void findDuplicates(int arr[], int n) {
cout << "Repeating elements are:\n";
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (arr[i] == arr[j]) {
cout << arr[i] << endl;
break;
}
}
}
}
int main() {
int arr[] = {1, 2, 3, 4, 2, 3, 5};
int n = sizeof(arr) / sizeof(arr[0]);
findDuplicates(arr, n);
return 0;
}
Input: [1, 2, 3, 4, 2, 3, 5]
Output:
Repeating elements are:
2
3
By mastering this basic approach, you can further explore more optimized methods like hashing or sorting to handle larger datasets efficiently. Experiment with the code examples and modify the inputs to deepen your understanding.