Given a constant array of N elements which contain elements from 1 to N – 1, with any of these numbers appearing any number of times.
Examples:
Input: N = 5, arr[] = {1, 3, 4, 2, 2}
Output: 2
Explanation:
2 is the number occurring more than once.Input: N = 5, arr[] = {3, 1, 3, 4, 2}
Output: 3
Explanation:
3 is the number occurring more than once.
Naive Approach: The naive method is to first sort the given array and then look for adjacent positions of the array to find the duplicate number.
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above method the idea is to use the concept of Counting Sort. Since the range of elements in the array is known, hence we could use this sorting technique to improvise the time complexity.
The idea is to initialize another array(say count[]) with the same size N and initialize all the elements as 0. Then count the occurrences of each element of the array and update the count in the count[]. Print all the element whose count is greater than 1.
Below is the implementation of above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to find the duplicate // number using countig sort method int findDuplicate(int arr[], int n) { int countArr[n + 1], i; // Initialize all the elements // of the countArr to 0 for (i = 0; i <= n; i++) countArr[i] = 0; // Count the occurences of each // element of the array for (i = 0; i <= n; i++) countArr[arr[i]]++; bool a = false; // Find the element with more // than one count for (i = 1; i <= n; i++) { if (countArr[i] > 1) { a = true; cout << i << ' '; } } // If unique elements are ther // print "-1" if (!a) cout << "-1"; } // Driver Code int main() { // Given N int n = 4; // Given array arr[] int arr[] = { 1, 3, 4, 2, 2 }; // Function Call findDuplicate(arr, n); return 0; } |
Java
// Java program for the above approach import java.util.*; class GFG { // Function to find the duplicate number // using countig sort method public static int findDuplicate(int arr[], int n) { int countArr[] = new int[n + 1], i; // Initialize all the elements of the // countArr to 0 for (i = 0; i <= n; i++) countArr[i] = 0; // Count the occurences of each // element of the array for (i = 0; i <= n; i++) countArr[arr[i]]++; bool a = false; // Find the element with more // than one count for (i = 1; i <= n; i++) { if (countArr[i] > 1) { a = true; cout << i << ' '; } } if (!a) System.out.println("-1"); } // Driver Code public static void main(String[] args) { int n = 4; int arr[] = { 1, 3, 4, 2, 2 }; // Function Call findDuplicate(arr, n); } } |
Python3
# Python3 program for the above approach # Function to find the duplicate # number using countig sort method def findDuplicate(arr, n): # Initialize all the elements # of the countArr to 0 countArr = [0] * (n + 1) # Count the occurences of each # element of the array for i in range(n + 1): countArr[arr[i]] += 1 a = False # Find the element with more # than one count for i in range(1, n + 1): if(countArr[i] > 1): a = True print(i, end = " ") # If unique elements are there # print "-1" if(not a): print(-1) # Driver code if __name__ == '__main__': # Given N n = 4 # Given array arr[] arr = [ 1, 3, 4, 2, 2 ] # Function Call findDuplicate(arr, n) # This code is contributed by Shivam Singh |
C#
// C# program for the above approach using System; class GFG{ // Function to find the duplicate number // using countig sort method static void findDuplicate(int []arr, int n) { int []countArr = new int[n + 1]; int i; // Initialize all the elements of the // countArr to 0 for(i = 0; i <= n; i++) countArr[i] = 0; // Count the occurences of each // element of the array for(i = 0; i <= n; i++) countArr[arr[i]]++; bool a = false; // Find the element with more // than one count for(i = 1; i <= n; i++) { if (countArr[i] > 1) { a = true; Console.Write(i + " "); } } if (!a) Console.WriteLine("-1"); } // Driver Code public static void Main(String[] args) { int n = 4; int []arr = { 1, 3, 4, 2, 2 }; // Function Call findDuplicate(arr, n); } } // This code is contributed by Amit Katiyar |
2
Time Complexity: O(N)
Auxiliary Space: O(N)
Related articles:
- Find duplicates in O(n) time and O(1) extra space | Set 1
- Duplicates in an array in O(n) and by using O(1) extra space | Set-2
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