121314

# Function to find repeating elements in the array
def findRepeatingElements(arr):
    arr.sort()  # Sort the array to easily find duplicates
    
    print("The repeating elements are:", end=" ")
    for i in range(len(arr) - 1):
        # If current element is equal to next element, it's a repeating element
        if arr[i] == arr[i + 1]:
            print(arr[i], end=" ")

# Driver code
arr = [1, 1, 2, 3, 4, 4, 5, 2]  # Example input
findRepeatingElements(arr)  # Call the function to find repeating elements

1234567891011121314151617

// Function to find repeating elements in the array
function findRepeatingElements(arr) {
    arr.sort((a, b) => a - b);  // Sort the array to easily find duplicates
    
    process.stdout.write("The repeating elements are: ");
    for (let i = 0; i < arr.length - 1; i++) {
        // If current element is equal to next element, it's a repeating element
        if (arr[i] === arr[i + 1]) {
            process.stdout.write(arr[i] + " ");
        }
    }
}

// Driver code
const arr = [1, 1, 2, 3, 4, 4, 5, 2];  // Example input
findRepeatingElements(arr);  // Call the function to find repeating elements

Complexity Analysis

Time Complexity: O(N log N) + O(N), O(N log N) is required for sorting the array, and O(N) for iterating through the array to find a loop or perform additional checks.

Space Complexity: O(1), No extra space is used; the operations are performed in-place.

Optimal Approach

Algorithm

Store the elements in the hashmap with its frequency of occurrence.
Iterate through the hashmap. If occurrence is more than 1, return the element.