Find the only repeating element in a sorted array of size n Last Updated : 19 Apr, 2024 Comments Improve Suggest changes Like Article Like Report Given a sorted array of n elements containing elements in range from 1 to n-1 i.e. one element occurs twice, the task is to find the repeating element in an array. Examples : Input : arr[] = { 1, 2 , 3 , 4 , 4}Output : 4Input : arr[] = { 1 , 1 , 2 , 3 , 4}Output : 1Brute Force: Traverse the input array using a for a loop.For each element in the array, traverse the remaining part of the array using another for loop.For each subsequent element, check if it is equal to the current element.If a match is found, return the current element.If no match is found, continue with the next element in the outer loop.If the outer loop completes without finding a match, return -1 to indicate that there is no repeating element in the array.Below is the implementation of the above approach: C++ // C++ program to find the only repeating element in an // array of size n and elements from range 1 to n-1. #include <bits/stdc++.h> using namespace std; // Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. int FindRepeatingElement(int arr[], int size){ for(int i=0; i<size; i++){ for(int j=i+1; j<size; j++){ if(arr[i] == arr[j]) return i; } } return -1; } // Driver code int main() { int arr[] = {1, 2 , 3 , 4 , 4}; int n = sizeof(arr) / sizeof(arr[0]); int index = FindRepeatingElement(arr, n); if (index != -1) cout << arr[index]; return 0; } Java import java.util.*; public class Main { // Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. public static int findRepeatingElement(int arr[], int size){ for(int i=0; i<size; i++){ for(int j=i+1; j<size; j++){ if(arr[i] == arr[j]) return i; } } return -1; } // Driver code public static void main(String[] args) { int arr[] = {1, 2 , 3 , 4 , 4}; int n = arr.length; int index = findRepeatingElement(arr, n); if (index != -1) System.out.println(arr[index]); } } Python3 # Python3 program to find the only repeating element in an # array of size n and elements from range 1 to n-1. # Returns second appearance of a repeating element # The function assumes that array elements are in range from # 1 to n-1. def FindRepeatingElement(arr, size): for i in range(size): for j in range(i+1, size): if arr[i] == arr[j]: return arr[i] return -1 # Driver code if __name__ == '__main__': arr = [1, 2, 3, 4, 4] n = len(arr) element = FindRepeatingElement(arr, n) if element != -1: print(element) C# using System; public class Program { // Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. public static int FindRepeatingElement(int[] arr, int size) { for (int i = 0; i < size; i++) { for (int j = i + 1; j < size; j++) { if (arr[i] == arr[j]) { return i; } } } return -1; } // Driver code public static void Main() { int[] arr = { 1, 2, 3, 4, 4 }; int n = arr.Length; int index = FindRepeatingElement(arr, n); if (index != -1) { Console.WriteLine(arr[index]); } } } JavaScript // JavaScript program to find the only repeating element in an // array of size n and elements from range 1 to n-1. // Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. function FindRepeatingElement(arr, size) { for(let i=0; i<size; i++) { for(let j=i+1; j<size; j++) { if(arr[i] == arr[j]) return i; } } return -1; } // Driver code let arr = [1, 2 , 3 , 4 , 4]; let n = arr.length; let index = FindRepeatingElement(arr, n); if (index != -1) console.log(arr[index]); // This code is contributed by akashish__ Output4Time Complexity: O(N^2)Auxiliary Space: O(1) An efficient method is to use Floyd’s Tortoise and Hare algorithm. C++ // C++ program to find the only repeating element in an // array of size n and elements from range 1 to n-1. #include <bits/stdc++.h> using namespace std; // Returns the repeating element in the array // The function assumes that array elements are in range from // 1 to n-1. int FindRepeatingElement(int arr[], int size){ for(int i = 0; i < size; i++){ if(arr[abs(arr[i])] >= 0) arr[abs(arr[i])] = -arr[abs(arr[i])]; else return abs(arr[i]); } return -1; } // Driver code int main() { int arr[] = {1,3,3,4,5,6,7,8}; int n = sizeof(arr) / sizeof(arr[0]); int repeatingElement = FindRepeatingElement(arr, n); cout << repeatingElement; return 0; } Java class Test { // Returns the repeating element in the array // The function assumes that array elements are in range from // 1 to n-1. static int findRepeatingElement(int arr[]){ int tortoise = arr[0]; int hare = arr[0]; // Phase 1: Finding intersection point of Tortoise and Hare do { tortoise = arr[tortoise]; hare = arr[arr[hare]]; } while(tortoise != hare); // Phase 2: Finding the entrance to the cycle tortoise = arr[0]; while(tortoise != hare){ tortoise = arr[tortoise]; hare = arr[hare]; } return hare; } // Driver method public static void main(String[] args) { int arr[] = {1, 2, 3, 3, 4, 5}; int repeatingElement = findRepeatingElement(arr); System.out.println(repeatingElement); } } Python3 # Python program to find the only repeating element in an # array of size n and elements from range 1 to n-1 def findRepeatingElement(arr): tortoise = arr[0] hare = arr[0] # Phase 1: Finding intersection point of Tortoise and Hare while True: tortoise = arr[tortoise] hare = arr[arr[hare]] if tortoise == hare: break # Phase 2: Finding the entrance to the cycle tortoise = arr[0] while tortoise != hare: tortoise = arr[tortoise] hare = arr[hare] return hare # Driver code arr = [1, 2, 3, 3, 4, 5] repeatingElement = findRepeatingElement(arr) print(repeatingElement) C# // C# program to find the only repeating // element in an array of size n and // elements from range 1 to n-1. using System; class Test { // Returns index of second appearance of a // repeating element. The function assumes that // array elements are in range from 1 to n-1. static int findRepeatingElement(int []arr, int low, int high) { // low = 0 , high = n-1; if (low > high) return -1; int mid = (low + high) / 2; // Check if the mid element // is the repeating one if (arr[mid] != mid + 1) { if (mid > 0 && arr[mid]==arr[mid-1]) return mid; // If mid element is not at its position // that means the repeated element is in left return findRepeatingElement(arr, low, mid-1); } // If mid is at proper position // then repeated one is in right. return findRepeatingElement(arr, mid+1, high); } // Driver method public static void Main() { int []arr = {1, 2, 3, 3, 4, 5}; int index = findRepeatingElement(arr, 0, arr.Length-1); if (index != -1) Console.Write(arr[index]); } } // This code is contributed by Nitin Mittal. JavaScript <script> // JavaScript program to find the only repeating element in an // array of size n and elements from range 1 to n-1. // Returns index of second appearance of a repeating element // The function assumes that array elements are in range from // 1 to n-1. function findRepeatingElement(arr, low, high) { // low = 0 , high = n-1; if (low > high) return -1; var mid = parseInt((low + high) / 2); // Check if the mid element is the repeating one if (arr[mid] != mid + 1) { if (mid > 0 && arr[mid] == arr[mid - 1]) return mid; // If mid element is not at its position that means // the repeated element is in left return findRepeatingElement(arr, low, mid - 1); } // If mid is at proper position then repeated one is in // right. return findRepeatingElement(arr, mid + 1, high); } // Driver code var arr = [1, 2, 3, 3, 4, 5]; var n = arr.length; var index = findRepeatingElement(arr, 0, n - 1); if (index != -1) document.write(arr[index]); // This code is contributed by rdtank. </script> PHP <?php // PHP program to find the only // repeating element in an array // of size n and elements from // range 1 to n-1. // Returns index of second // appearance of a repeating // element. The function assumes // that array elements are in // range from 1 to n-1. function findRepeatingElement($arr, $low, $high) { // low = 0 , high = n-1; if ($low > $high) return -1; $mid = floor(($low + $high) / 2); // Check if the mid element // is the repeating one if ($arr[$mid] != $mid + 1) { if ($mid > 0 && $arr[$mid] == $arr[$mid - 1]) return $mid; // If mid element is not at // its position that means // the repeated element is in left return findRepeatingElement($arr, $low, $mid - 1); } // If mid is at proper position // then repeated one is in right. return findRepeatingElement($arr, $mid + 1, $high); } // Driver code $arr = array(1, 2, 3, 3, 4, 5); $n = sizeof($arr); $index = findRepeatingElement($arr, 0, $n - 1); if ($index != -1) echo $arr[$index]; // This code is contributed // by nitin mittal. ?> Output3Time Complexity : O(log n) Space Complexity: O(1) Comment More infoAdvertise with us Next Article Asymptotic Notations for Analysis of Algorithms K kartik Improve Article Tags : Divide and Conquer DSA Arrays Binary Search Practice Tags : ArraysBinary SearchDivide and Conquer Similar Reads Basics & PrerequisitesTime and Space ComplexityMany times there are more than one ways to solve a problem with different algorithms and we need a way to compare multiple ways. 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