find_dublicate_element_in_array_n+1_integers

A duplicate number within an array of size n+1 can be found efficiently using a few different approaches. The most common method involves leveraging the array’s indices to track element occurrences, often modifying the array in place. This allows for a solution with linear time complexity (O(n)) and constant extra space (O(1)).

Here’s a breakdown of the approach:

1. Initialization:
   • Assume the array contains numbers from 1 to n, with one number appearing twice.
   • Create a variable n to store the size of the array.
2. Iteration and Modification:
   • Iterate through the array elements using a loop.
   • For each element nums[i], calculate its corresponding index abs(nums[i]) - 1.
   • Check the sign of nums[abs(nums[i]) - 1].
     • If it’s negative, you’ve found the duplicate (it was already marked negative in a previous iteration).
     • If it’s positive, change the sign of nums[abs(nums[i]) - 1] to mark it as visited.
3. Duplicate Identification:
   • If a negative element is encountered, the absolute value of the corresponding index plus one ( abs(nums[i])) represents the duplicate number.

Here’s an example in Python:

def find_duplicate(nums):
    for num in nums:
        if nums[abs(num) - 1] < 0:
            return abs(num)
        nums[abs(num) - 1] *= -1
 
# Example usage:
nums = [1, 2, 3, 4, 2]
duplicate = find_duplicate(nums)
print(f"The duplicate number is: {duplicate}")

#include <iostream>
 
using namespace std;
 
int findDublicate(int *arr, int n) {
  for (int i = 0; i < n; i++) {
    if (arr[abs(arr[i]) - 1] < 0) {
      return abs(arr[i]);
    }
    arr[abs(arr[i]) - 1] *= -1;
  }
  return -1;
}
 
int main() {
  int arr[] = {1, 2, 3, 4, 5, 4};
  int n = sizeof(arr) / sizeof(arr[0]);
 
  cout << findDublicate(arr, n);
}

This approach modifies the original array, so if preserving the original array is necessary, a copy should be made first. [2]