Given an array arr[] of size N, which contains a permutation of numbers from 1 to M, as well as an element that is repeated(one or more times), the task is to find the repeating element.
Examples:
Input: arr[]={2, 6, 4, 3, 1, 5, 2}, N=7
Output:
2
Explanation: In arr[], all elements from 0 to 6 occurs once, except 2 which is repeated once.Input: arr[]={2, 1, 3, 1, 1, 1}, N=6
Output:
1
Naive Approach: The naive approach would be to sort the array and check for adjacent elements that are equal.
C++
// C++ program to find the repeating element// in an array using naive approach#include <bits/stdc++.h>using namespace std;// Function to find the repeating elementint findRepeatingElement(int arr[], int n, int m) { // Sort the given array sort(arr, arr + n); // Traverse the sorted array to find the repeating // element for (int i = 0; i < n-1; i++) { if (arr[i] == arr[i + 1]) { return arr[i]; } } // If no repeating element found return -1;}// Driver codeint main() { // Input array int arr[] = { 2, 6, 4, 3, 1, 5, 2 }; int n = sizeof(arr) / sizeof(arr[0]); int m = 4; // Function call to find the repeating element int repeating_element = findRepeatingElement(arr, n, m); // Print the repeating element cout<< repeating_element << endl; return 0;} |
Java
// Java program to find the repeating element// in an array using naive approachimport java.util.Arrays;public class GFG { // Function to find the repeating element public static int findRepeatingElement(int[] arr, int n, int m) { // Sort the given array Arrays.sort(arr); // Traverse the sorted array to find the repeating element for (int i = 0; i < n - 1; i++) { if (arr[i] == arr[i + 1]) { return arr[i]; } } // If no repeating element found return -1; } // Driver code public static void main(String[] args) { // Input array int[] arr = { 2, 1, 3, 1, 1, 1}; int n = arr.length; int m = 4; // Function call to find the repeating element int repeating_element = findRepeatingElement(arr, n, m); // Print the repeating element System.out.println(repeating_element); }} |
Python3
def find_repeating_element(arr, n, m): arr.sort() for i in range(n - 1): if arr[i] == arr[i + 1]: return arr[i] return -1# Input arrayarr = [2, 6, 4, 3, 1, 5, 2]n = len(arr)m = 4# Function call to find the repeating elementrepeating_element = find_repeating_element(arr, n, m)# Print the repeating elementprint(repeating_element)# This code is contributed by shivhack999 |
C#
using System;class Program{ // Function to find the repeating element static int findRepeatingElement(int[] arr, int n, int m) { // Sort the given array Array.Sort(arr); // Traverse the sorted array to find the repeating // element for (int i = 0; i < n-1; i++) { if (arr[i] == arr[i + 1]) { return arr[i]; } } // If no repeating element found return -1; } // Driver code static void Main(string[] args) { // Input array int[] arr = { 2, 6, 4, 3, 1, 5, 2 }; int n = arr.Length; int m = 4; // Function call to find the repeating element int repeating_element = findRepeatingElement(arr, n, m); // Print the repeating element Console.WriteLine(repeating_element); }} |
Javascript
// Function to find the repeating elementfunction findRepeatingElement(arr, n, m) { // Sort the given array arr.sort(); // Traverse the sorted array to find the repeating // element for (let i = 0; i < n-1; i++) { if (arr[i] == arr[i + 1]) { return arr[i]; } } // If no repeating element found return -1;}// Input arraylet arr = [2, 6, 4, 3, 1, 5, 2];let n = arr.length;let m = 4; // Function call to find the repeating elementlet repeating_element = findRepeatingElement(arr, n, m);// Print the repeating elementconsole.log(repeating_element); |
Output
2
Time Complexity: O(NlogN)
Auxiliary Space: O(1)
Approach: Follow the steps below to solve the problem:
- Initialize two variables M and sum to store the maximum element and the sum of the array respectively.
- Traverse array arr and do the following:
- Add the current element to sum
- Compare the current element to M to calculate the maximum element.
- Store the sum of permutation from 1 to M in a variable say, sum1, using the formula:
Sum of elements from 1 to X= X*(X+1)/2
- Calculate the answer as the difference between sum and sum1 divided by the number of extra characters i.e. (sum-sum1)/(N-M).
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate the repeating character in a given// permutationint repeatingElement(int arr[], int N){ // variables to store maximum element and sum of the // array respectively. int M = 0, sum = 0; for (int i = 0; i < N; i++) { // calculate sum of array sum += arr[i]; // calculate maximum element in the array M = max(M, arr[i]); } // calculating sum of permutation int sum1 = M * (M + 1) / 2; // calculate required answer int ans = (sum - sum1) / (N - M); return ans;}// Driver codeint main(){ // Input int arr[] = { 2, 6, 4, 3, 1, 5, 2 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call cout << repeatingElement(arr, N) << endl; return 0;} |
Java
// Java Program for the above approachimport java.io.*;class GFG{ // Function to calculate the repeating character in a given // permutation public static int repeatingElement(int arr[], int N) { // variables to store maximum element and sum of the // array respectively. int M = 0, sum = 0; for (int i = 0; i < N; i++) { // calculate sum of array sum += arr[i]; // calculate maximum element in the array M = Math.max(M, arr[i]); } // calculating sum of permutation int sum1 = M * (M + 1) / 2; // calculate required answer int ans = (sum - sum1) / (N - M); return ans; } // Driver code public static void main (String[] args) { // Input int arr[] = { 2, 6, 4, 3, 1, 5, 2 }; int N = arr.length; // Function call System.out.println(repeatingElement(arr, N)); }}// This code is contributed by lokeshpotta20 |
Python3
# Python 3 program for the above approach# Function to calculate the repeating character in a given# permutationdef repeatingElement(arr, N): # variables to store maximum element and sum of the # array respectively. M = 0 sum = 0 for i in range(N): # calculate sum of array sum += arr[i] # calculate maximum element in the array M = max(M, arr[i]) # calculating sum of permutation sum1 = M * (M + 1) // 2 # calculate required answer ans = (sum - sum1) // (N - M) return ans# Driver codeif __name__ == '__main__': # Input arr = [2, 6, 4, 3, 1, 5, 2] N = len(arr) # Function call print(repeatingElement(arr, N)) # This code is contributed by SURENDRA_GANGWAR. |
C#
// C++ program for the above approachusing System;// Function to calculate the repeating character in a given// permutationpublic class GFG{ public static int repeatingElement(int[] arr, int N) { // variables to store maximum element and sum of the // array respectively. int M = 0, sum = 0; for (int i = 0; i < N; i++) { // calculate sum of array sum += arr[i]; // calculate maximum element in the array M = Math.Max(M, arr[i]); } // calculating sum of permutation int sum1 = M * (M + 1) / 2; // calculate required answer int ans = (sum - sum1) / (N - M); return ans; } // Driver code public static void Main() { // Input int[] arr = { 2, 6, 4, 3, 1, 5, 2 }; int N = 7; // Function call Console.WriteLine(repeatingElement(arr, N)); }}// This code is contributed by Sohom Das |
Javascript
// JavaScript program for the above approach // Function to calculate the repeating character in a given // permutation function repeatingElement(arr, N) { // variables to store maximum element and sum of the // array respectively. let M = 0, sum = 0; for (let i = 0; i < N; i++) { // calculate sum of array sum += arr[i]; // calculate maximum element in the array M = Math.max(M, arr[i]); } // calculating sum of permutation let sum1 = parseInt(M * (M + 1) / 2); // calculate required answer let ans = parseInt((sum - sum1) / (N - M)); return ans; } // Driver code // Input let arr = [2, 6, 4, 3, 1, 5, 2]; let N = arr.length; // Function call document.write(repeatingElement(arr, N)); // This code is contributed by Potta Lokesh </script> |
Output
2
Time Complexity: O(N)
Auxiliary Space: O(1)
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