# Find the only repeating element in a sorted array of size n

• Difficulty Level : Easy
• Last Updated : 06 Jul, 2022

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.

Observation: If an element ‘X’ is repeating, then it must be at index ‘X’ in the array. So the problem reduces to find any element whose value is same as its index.

## 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` `size){` `    ``int` `lo = 0;` `    ``int` `hi = size - 1;` `    ``int` `mid;` `    `  `    ``while``(lo <= hi){` `        ``mid = (lo+hi)/2;` `        `  `        ``if``(arr[mid] <= mid){` `            ``hi = mid-1;` `        ``}` `        ``else``{` `            ``lo = mid + 1;` `        ``}` `    ``}` `    `  `    ``return` `lo;` `}`   `// Driver code ` `int` `main() ` `{ ` `    ``int` `arr[] = {1, 2, 3, 3, 4, 5}; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `index = FindRepeatingElement(arr, n); ` `    ``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]);` `    ``}` `}`

## Python3

 `# 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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