Sort the array in ascending order. This ensures that any duplicate numbers are adjacent to each other.
Iterate through the sorted array and check if the current element is equal to the next element (i.e., arr[i] == arr[i+1]).
If the condition arr[i] == arr[i+1] is true, then arr[i] is a duplicate number.
Return or store the duplicate element as needed.

Code

#include <bits/stdc++.h>
using namespace std;

// find the duplicate by sorting and checking adjacent elements
int findDuplicate(vector<int>& arr) {
  // get size
  int n = arr.size();
  // sort array in-place
  sort(arr.begin(), arr.end());
  // scan adjacent pairs
  for (int i = 0; i < n - 1; i++) {
    // return when a duplicate is found
    if (arr[i] == arr[i + 1]) {
      return arr[i];
    }
  }
  // fallback if no duplicate found
  return -1;
}

// program entry
int main() {
  // declare and initialize array
  vector<int> arr = {1, 3, 4, 2, 2};
  // print result
  cout << "The duplicate element is " << findDuplicate(arr) << endl;
  // exit
  return 0;
}
import java.util.*;

// utility class holding the logic
class Solution {
    // find the duplicate by sorting and checking adjacent elements
    static int findDuplicate(int[] arr) {
        // get size
        int n = arr.length;
        // sort array in-place
        Arrays.sort(arr);
        // scan adjacent pairs
        for (