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Find the only repeating element in a sorted array of size n
• Difficulty Level : Easy
• Last Updated : 12 Apr, 2021

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 :  4

Input :  arr[] = { 1 , 1 , 2 , 3 , 4}
Output :  1```

A naive approach is to scan the whole array and check if an element occurs twice, then return. This approach takes O(n) time.
An efficient method is to use Binary Search
1- Check if the middle element is the repeating one.
2- If not then check if the middle element is at proper position if yes then start searching repeating element in right.
3- Otherwise the repeating one will be in left.

## C++

 `// C++ program to find the only repeating element in an``// array of size n and elements from range 1 to n-1.``#include ``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` `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 code``int` `main()``{``    ``int` `arr[] = {1, 2, 3, 3, 4, 5};``    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr);``    ``int` `index = findRepeatingElement(arr, 0, n-1);``    ``if` `(index != -1)``        ``cout << arr[index];``    ``return` `0;``}`

## Java

 `// Java program to find the only repeating element in an``// array of size n and elements from range 1 to n-1.` `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(String[] args)``    ``{``        ``int`  `arr[] = {``1``, ``2``, ``3``, ``3``, ``4``, ``5``};``        ``int` `index = findRepeatingElement(arr, ``0``, arr.length-``1``);``        ``if` `(index != -``1``)``            ``System.out.println(arr[index]);``    ``}``}`

## Python

 `# Python 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.``def` `findRepeatingElement(arr, low, high):` `    ``# low = 0 , high = n-1``    ``if` `low > high:``        ``return` `-``1` `    ``mid ``=` `(low ``+` `high) ``/` `2` `    ``# Check if the mid element is the repeating one``    ``if` `(arr[mid] !``=` `mid ``+` `1``):``    ` `        ``if` `(mid > ``0` `and` `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 ``=` `[``1``, ``2``, ``3``, ``3``, ``4``, ``5``]``n ``=` `len``(arr)``index ``=` `findRepeatingElement(arr, ``0``, n``-``1``)``if` `(index ``is` `not` `-``1``):``    ``print` `arr[index]` `#This code is contributed by Afzal Ansari`

## 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.`

## PHP

 ` ``\$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.``?>`

## Javascript

 ``

Output :

`3`

Time Complexity : O(log n)
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