# Find the only repetitive element between 1 to N-1

Given an array of size N filled with numbers from 1 to N-1 in random order. The array has only one repetitive element. The task is to find the repetitive element.

Examples:

Input: a[] = {1, 3, 2, 3, 4}
Output: 3
Explanation: The number 3 is the only repeating element.

Input: a[] = {1, 5, 1, 2, 3, 4}
Output: 1

Naive Approach:  To solve the problem follow the below idea:

Use two nested loops. The outer loop traverses through all elements and the inner loop check if the element picked by the outer loop appears anywhere else.

Below is the implementation of the above approach:

## C++

 `// C++ program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``#include ``using` `namespace` `std;` `int` `findRepeating(``int` `arr[], ``int` `N)``{``    ``for` `(``int` `i = 0; i < N; i++) {``        ``for` `(``int` `j = i + 1; j < N; j++) {``            ``if` `(arr[i] == arr[j])``                ``return` `arr[i];``        ``}``    ``}``}` `// Driver's code``int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function call``    ``cout << findRepeating(arr, N);``    ``return` `0;``}``// This code is added by Arpit Jain`

## Java

 `// Java program to find the only repeating``// element in an array where elements are``// from 1 to N-1.using System;``public` `class` `GFG {``    ``static` `int` `findRepeating(``int``[] arr)``    ``{``        ``for` `(``int` `i = ``0``; i < arr.length; i++) {``            ``for` `(``int` `j = i + ``1``; j < arr.length; j++) {``                ``if` `(arr[i] == arr[j])``                    ``return` `arr[i];``            ``}``        ``}``        ``return` `-``1``;``    ``}` `    ``// Driver's code``    ``static` `public` `void` `main(String[] args)``    ``{` `        ``// Code``        ``int``[] arr``            ``= ``new` `int``[] { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};``        ``int` `repeatingNum = findRepeating(arr);` `        ``// Function call``        ``System.out.println(repeatingNum);``    ``}``}` `// This code is contributed by phasing17`

## Python3

 `# Python3 program to find the only``# repeating element in an array where``# elements are from 1 to N-1.`  `def` `findRepeating(arr, N):``    ``for` `i ``in` `range``(N):``        ``for` `j ``in` `range``(i ``+` `1``, N):``            ``if` `(arr[i] ``=``=` `arr[j]):``                ``return` `arr[i]`  `# Driver's Code``if` `__name__ ``=``=` `"__main__"``:``  ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``  ``N ``=` `len``(arr)` `  ``# Function call``  ``print``(findRepeating(arr, N))` `# This code is contributed by Arpit Jain`

## C#

 `// C# program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``using` `System;` `public` `class` `GFG {``    ``static` `int` `findRepeating(``int``[] arr)``    ``{``        ``for` `(``int` `i = 0; i < arr.Length; i++) {``            ``for` `(``int` `j = i + 1; j < arr.Length; j++) {``                ``if` `(arr[i] == arr[j])``                    ``return` `arr[i];``            ``}``        ``}``        ``return` `-1;``    ``}` `    ``// Driver's code``    ``static` `public` `void` `Main()``    ``{``        ``int``[] arr``            ``= ``new` `int``[] { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``      ` `          ``// Function call``        ``int` `repeatingNum = findRepeating(arr);``        ``Console.WriteLine(repeatingNum);``    ``}``}` `// This code is contributed by Mohd Nizam`

## Javascript

 `        `

Output
```8

```

Time Complexity: O(N2)
Auxiliary Space: O(1)

## Find the only repetitive element using sorting:

Sort the given input array. Traverse the array and if value of the ith element is not equal to i+1, then the current element is repetitive as value of elements is between 1 and N-1 and every element appears only once except one element.

Follow the below steps to solve the problem:

• Sort the given array.
• Traverse the array and  compare the array elements with its index
• if arr[i] != i+1, it means that arr[i] is repetitive, So Just return arr[i]
• Otherwise, the array does not contain duplicates from 1 to n-1, In this case, return -1

Below is the implementation of the above approach:

## C++

 `// CPP program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``#include ``using` `namespace` `std;` `int` `findRepeating(``int` `arr[], ``int` `N)``{``    ``sort(arr, arr + N); ``// sort array` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``// compare array element with its index``        ``if` `(arr[i] != i + 1) {``            ``return` `arr[i];``        ``}``    ``}``    ``return` `-1;``}` `// driver's code``int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function call``    ``cout << findRepeating(arr, N);``    ``return` `0;``}``// this code is contributed by devendra solunke`

## Java

 `// Java program to find the only repeating``// element in an array where elements are``// from 1 to N-1.` `import` `java.util.Arrays;` `public` `class` `GFG {``    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{``        ``Arrays.sort(arr); ``// sort array``      ` `        ``for` `(``int` `i = ``0``; i < N; i++) {``          ` `            ``// compare array element with its index``            ``if` `(arr[i] != i + ``1``) {``                ``return` `arr[i];``            ``}``        ``}``        ``return` `-``1``;``    ``}` `    ``// Driver's code``    ``static` `public` `void` `main(String[] args)``    ``{``        ``int``[] arr``            ``= ``new` `int``[] { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};``        ``int` `N = arr.length;``      ` `          ``// Function call``        ``int` `repeatingNum = findRepeating(arr, N);``        ``System.out.println(repeatingNum);``    ``}``}` `// This code is contributed by Abhijeet Kumar(abhijeet19403)`

## Python3

 `# Python3 program to find the only``# repeating element in an array where``# elements are from 1 to N-1.`  `def` `findRepeating(arr, N):``    ``arr.sort()``    ``for` `i ``in` `range``(``1``, N):``        ``if``(arr[i] !``=` `i``+``1``):``            ``return` `arr[i]`  `# Driver's Code``if` `__name__ ``=``=` `"__main__"``:``  ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``  ``N ``=` `len``(arr)` `  ``# Function call``  ``print``(findRepeating(arr, N))` `# This code is contributed by Arpit Jain`

## C#

 `// C# program to find the only repeating element in an array``// where elements are from 1 to N-1.` `using` `System;` `public` `class` `GFG {` `    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{``        ``Array.Sort(arr); ``// sort array``        ``for` `(``int` `i = 0; i < N; i++) {``            ``// compare array element with its index``            ``if` `(arr[i] != i + 1) {``                ``return` `arr[i];``            ``}``        ``}``        ``return` `-1;``    ``}` `      ``// Driver's code``    ``static` `public` `void` `Main()``    ``{``        ``int``[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``        ``int` `N = arr.Length;``      ` `          ``// Function call``        ``int` `repeatingNum = findRepeating(arr, N);``        ``Console.WriteLine(repeatingNum);``    ``}``}` `// This code is contributed by lokesh (lokeshmvs21).`

## Javascript

 ``

Output
```8

```

Time complexity: O(N * log N)
Auxiliary Space: O(1)

## Find the only repetitive element using a frequency array:

Use a array to store frequency of  elements appeared in array. If frequency of element is greater than one,then return it.

## C++

 `#include ``using` `namespace` `std;` `int` `findRepeating(``int` `arr[], ``int` `N)``{``    ``int` `freq[N+1];``// initializing array for all``       ``memset``(freq, 0, ``sizeof``(freq));``// elements with frequency 0.``    ``for` `(``int` `i = 0; i < N; i++) {``        ``freq[arr[i]]++;``    ``}``    ``for` `(``int` `i = 0; i < N; i++) {``        ``if` `(freq[arr[i]] > 1) {``            ``return` `arr[i];``        ``}``    ``}` `    ``return` `-1;``}` `int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function call``    ``cout << findRepeating(arr, N);``    ``return` `0;``}`

## Java

 `public` `class` `Main {``    ``static` `int` `findRepeating(``int` `arr[], ``int` `N) {``        ``int``[] freq = ``new` `int``[N + ``1``];``        ` `        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``freq[arr[i]]++;``        ``}``        ` `        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``if` `(freq[arr[i]] > ``1``) {``                ``return` `arr[i];``            ``}``        ``}``        ` `        ``return` `-``1``;``    ``}` `    ``public` `static` `void` `main(String[] args) {``        ``int` `arr[] = { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};``        ``int` `N = arr.length;` `        ``// Function call``        ``System.out.println(findRepeating(arr, N));``    ``}``}`

## Python3

 `def` `findRepeating(arr, N):``    ``freq ``=` `[``0``]``*``(N``+``1``) ``    ``for` `i ``in` `range``(N):``        ``freq[arr[i]] ``+``=` `1``    ``for` `i ``in` `range``(N):``        ``if` `freq[arr[i]] > ``1``:``            ``return` `arr[i]``    ``return` `-``1` `arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``N ``=` `len``(arr)` `# Function call``print``(findRepeating(arr, N))`

## C#

 `using` `System;``using` `System.Linq;` `class` `MainClass {``    ``public` `static` `int` `findRepeating(``int``[] arr, ``int` `N) {``        ``int``[] freq = ``new` `int``[N + 1];``        ``for` `(``int` `i = 0; i < N; i++) {``            ``freq[arr[i]]++;``        ``}``        ``for` `(``int` `i = 0; i < N; i++) {``            ``if` `(freq[arr[i]] > 1) {``                ``return` `arr[i];``            ``}``        ``}``        ``return` `-1;``    ``}` `    ``public` `static` `void` `Main (``string``[] args) {``        ``int``[] arr = {9, 8, 2, 6, 1, 8, 5, 3, 4, 7};``        ``int` `N = arr.Length;` `        ``// Function call``        ``Console.WriteLine(findRepeating(arr, N));``    ``}``}`

## Javascript

 `function` `findRepeating(arr, N) {``    ``var` `freq = ``new` `Array(N + 1).fill(0);``    ``for` `(``var` `i = 0; i < N; i++) {``        ``freq[arr[i]]++;``    ``}``    ``for` `(``var` `i = 0; i < N; i++) {``        ``if` `(freq[arr[i]] > 1) {``            ``return` `arr[i];``        ``}``    ``}``    ``return` `-1;``}` `var` `arr = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7];``var` `N = arr.length;` `// Function call``console.log(findRepeating(arr, N));`

Output
```8

```

Time Complexity: O(N)
Auxiliary Space: O(N)

## Find the only repetitive element using the hash set:

Use a hash table to store elements visited. If an already visited element appears again, return it.

Follow the below steps to solve the problem:

• Create a hash set to store the visited elements
• Traverse the array
• If the given element is already present in the hash set then, return this element
• else insert this element into the hash set
• Return -1, if no repeating is found

Below is the implementation of the above approach:

## C++

 `// C++ program to find the only repeating``// element in an array where elements are``// from 1 to N-1.` `#include ``using` `namespace` `std;` `int` `findRepeating(``int` `arr[], ``int` `N)``{``    ``unordered_set<``int``> s;``    ``for` `(``int` `i = 0; i < N; i++) {``        ``if` `(s.find(arr[i]) != s.end())``            ``return` `arr[i];``        ``s.insert(arr[i]);``    ``}` `    ``// If input is correct, we should``    ``// never reach here``    ``return` `-1;``}` `// Driver's code``int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``  ` `      ``// Function call``    ``cout << findRepeating(arr, N);``    ``return` `0;``}`

## Java

 `import` `java.util.*;``// Java program to find the only repeating``// element in an array where elements are``// from 1 to N-1.` `class` `GFG {` `    ``static` `int` `findRepeating(``int` `arr[], ``int` `N)``    ``{``        ``HashSet s = ``new` `HashSet();``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``if` `(s.contains(arr[i]))``                ``return` `arr[i];``            ``s.add(arr[i]);``        ``}` `        ``// If input is correct, we should``        ``// never reach here``        ``return` `-``1``;``    ``}` `    ``// Driver's code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};``        ``int` `N = arr.length;``      ` `          ``// Function call``        ``System.out.println(findRepeating(arr, N));``    ``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 program to find the only``# repeating element in an array``# where elements are from 1 to n-1.`  `def` `findRepeating(arr, N):``    ``s ``=` `set``()``    ``for` `i ``in` `range``(N):``        ``if` `arr[i] ``in` `s:``            ``return` `arr[i]``        ``s.add(arr[i])` `    ``# If input is correct, we should``    ``# never reach here``    ``return` `-``1`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``]``  ``N ``=` `len``(arr)` `  ``# Function call``  ``print``(findRepeating(arr, N))` `# This code is contributed``# by Shrikant13`

## C#

 `// C# program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``using` `System;``using` `System.Collections.Generic;` `class` `GFG {` `    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{``        ``HashSet<``int``> s = ``new` `HashSet<``int``>();``        ``for` `(``int` `i = 0; i < N; i++) {``            ``if` `(s.Contains(arr[i]))``                ``return` `arr[i];``            ``s.Add(arr[i]);``        ``}` `        ``// If input is correct, we should``        ``// never reach here``        ``return` `-1;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``        ``int` `N = arr.Length;``      ` `          ``// Function call``        ``Console.WriteLine(findRepeating(arr, N));``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Javascript

 `   ` `// JavaScript program to find the only repeating``// element in an array where elements are``// from 1 to n-1.` `function` `findRepeating(arr,n)``{``  ``s = ``new` `Set();``   ``for` `(let i = 0; i < n; i++)``   ``{``      ``if` `(s.has(arr[i]))``         ``return` `arr[i];``      ``s.add(arr[i]);``   ``}``   ` `   ``// If input is correct, we should ``   ``// never reach here``   ``return` `-1;``}` `// driver code``let arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];``let n = arr.length;``document.write(findRepeating(arr, n));` `// This code is contributed by shinjanpatra.`

Output
```8

```

Time Complexity: O(N)
Auxiliary Space: O(N)

## Find the only repetitive element using the Sum of first N elements:

We know sum of first n-1 natural numbers is (N – 1)*N/2. We compute sum of array elements and subtract natural number sum from it to find the only missing element.

Follow the below steps to solve the problem:

• Calculate the sum of array elements and the sum of first (N-1) natural numbers
• Return (array sum) – ((N-1) natural numbers sum)

Below is the implementation of the above approach:

## C++

 `// CPP program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``#include ``using` `namespace` `std;` `int` `findRepeating(``int` `arr[], ``int` `N)``{``    ``// Find array sum and subtract sum``    ``// first N-1 natural numbers from it``    ``// to find the result.``    ``return` `accumulate(arr, arr + N, 0) - ((N - 1) * N / 2);``}` `// Driver's code``int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``  ` `      ``// Function call``    ``cout << findRepeating(arr, N);``    ``return` `0;``}`

## Java

 `// Java program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{``        ``// Find array sum and subtract sum``        ``// first n-1 natural numbers from it``        ``// to find the result.` `        ``int` `sum = ``0``;``        ``for` `(``int` `i = ``0``; i < N; i++)``            ``sum += arr[i];``        ``return` `sum - (((N - ``1``) * N) / ``2``);``    ``}` `    ``// Driver's code``    ``public` `static` `void` `main(String args[])``    ``{``        ``int``[] arr = { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};``        ``int` `N = arr.length;``          ` `          ``// Function call``        ``System.out.println(findRepeating(arr, N));``    ``}``}` `// This code is contributed by rachana soma`

## Python3

 `# Python3 program to find the only``# repeating element in an array where``# elements are from 1 to N-1.`  `def` `findRepeating(arr, N):` `    ``# Find array sum and subtract sum``    ``# first n-1 natural numbers from it``    ``# to find the result.``    ``return` `sum``(arr) ``-` `(((N ``-` `1``) ``*` `N) ``/``/` `2``)`  `# Driver's Code``if` `__name__ ``=``=` `"__main__"``:``  ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``  ``N ``=` `len``(arr)` `  ``# Function call``  ``print``(findRepeating(arr, N))` `# This code is contributed``# by mohit kumar`

## C#

 `// C# program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``using` `System;` `class` `GFG {` `    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{``        ``// Find array sum and subtract sum``        ``// first N-1 natural numbers from it``        ``// to find the result.` `        ``int` `sum = 0;``        ``for` `(``int` `i = 0; i < N; i++)``            ``sum += arr[i];``        ``return` `sum - (((N - 1) * N) / 2);``    ``}` `    ``// Driver's code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``        ``int` `N = arr.Length;``      ` `          ``// Function call``        ``Console.WriteLine(findRepeating(arr, N));``    ``}``}` `/* This code contributed by PrinciRaj1992 */`

## Javascript

 `// javascript program to find the only repeating ``// element in an array where elements are ``// from 1 to n-1.    ``function` `findRepeating(arr , n)``{` `        ``// Find array sum and subtract sum``        ``// first n-1 natural numbers from it``        ``// to find the result.``        ``var` `sum = 0;``        ``for` `(i = 0; i < n; i++)``            ``sum += arr[i];``        ``return` `sum - (((n - 1) * n) / 2);``    ``}` `    ``// Driver code``        ``var` `arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];``        ``var` `n = arr.length;``        ``document.write(findRepeating(arr, n));` `// This code is contributed by Rajput-Ji`

Output
```8

```

Time Complexity: O(N)
Auxiliary Space: O(1)

Note: This approach Causes overflow for large arrays.

## Find the only repetitive element using XOR:

The idea is based on the fact that x ^ x = 0 and if x ^ y = z then x ^ z = y

Follow the below steps to solve the problem:

• Compute XOR of elements from 1 to n-1.
• Compute XOR of array elements.
• XOR of the above two would be our result.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``#include ``using` `namespace` `std;` `int` `findRepeating(``int` `arr[], ``int` `N)``{` `    ``// res is going to store value of``    ``// 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^``    ``// arr[1] ^ .... arr[N-1]``    ``int` `res = 0;``    ``for` `(``int` `i = 0; i < N - 1; i++)``        ``res = res ^ (i + 1) ^ arr[i];``    ``res = res ^ arr[N - 1];``    ``return` `res;``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);``  ` `      ``// Function call``    ``cout << findRepeating(arr, N);``    ``return` `0;``}`

## Java

 `// Java program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``import` `java.io.*;``class` `GFG {` `    ``static` `int` `findRepeating(``int` `arr[], ``int` `N)``    ``{` `        ``// res is going to store value of``        ``// 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^``        ``// arr[1] ^ .... arr[n-1]``        ``int` `res = ``0``;``        ``for` `(``int` `i = ``0``; i < N - ``1``; i++)``            ``res = res ^ (i + ``1``) ^ arr[i];``        ``res = res ^ arr[N - ``1``];` `        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};``        ``int` `N = arr.length;``      ` `          ``// Function call``        ``System.out.println(findRepeating(arr, N));``    ``}``}` `// This code is contributed by``// Smitha Dinesh Semwal.`

## Python3

 `# Python3 program to find the only``# repeating element in an array where``# elements are from 1 to N-1.`  `def` `findRepeating(arr, N):` `    ``# res is going to store value of``    ``# 1 ^ 2 ^ 3 .. ^ (N-1) ^ arr[0] ^``    ``# arr[1] ^ .... arr[n-1]``    ``res ``=` `0``    ``for` `i ``in` `range``(``0``, N``-``1``):``        ``res ``=` `res ^ (i``+``1``) ^ arr[i]``    ``res ``=` `res ^ arr[N``-``1``]` `    ``return` `res`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``  ``N ``=` `len``(arr)` `  ``# Function call``  ``print``(findRepeating(arr, N))` `# This code is contributed by Smitha Dinesh Semwal.`

## C#

 `// C# program to find the only repeating``// element in an array where elements are``// from 1 to N-1.``using` `System;``public` `class` `GFG {` `    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{` `        ``// res is going to store value of``        ``// 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^``        ``// arr[1] ^ .... arr[N-1]``        ``int` `res = 0;``        ``for` `(``int` `i = 0; i < N - 1; i++)``            ``res = res ^ (i + 1) ^ arr[i];``        ``res = res ^ arr[N - 1];` `        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main(String[] args)``    ``{``        ``int``[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``        ``int` `N = arr.Length;``      ` `          ``// Function call``        ``Console.WriteLine(findRepeating(arr, N));``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 `// JavaScript program to find the only repeating``// element in an array where elements are``// from 1 to n-1.`  `function` `findRepeating(arr,n)``{` `  ``// res is going to store value of``  ``// 1 ^ 2 ^ 3 .. ^ (n-1) ^ arr[0] ^ ``  ``// arr[1] ^ .... arr[n-1]``  ``let res = 0;``  ``for` `(let i=0; i

## PHP

 ``

Output
```8

```

Time Complexity: O(N)
Auxiliary Space: O(1)

## Find the only repetitive element using indexing:

As there are only positive numbers, so visit the index equal to the current element and make it negative. If an index value is already negative, then it means that current element is repeated

Follow the below steps to solve the problem:

• Iterate through the array.
• For every index visit arr[arr[i]], if it is positive change the sign of elements at that index, else print the element.

Below is the implementation of the above approach:

## C++

 `// CPP program to find the only``// repeating element in an array``// where elements are from 1 to N-1.` `#include ``using` `namespace` `std;` `// Function to find repeated element``int` `findRepeating(``int` `arr[], ``int` `N)``{``    ``int` `missingElement = 0;` `    ``// indexing based``    ``for` `(``int` `i = 0; i < N; i++) {` `        ``int` `element = arr[``abs``(arr[i])];` `        ``if` `(element < 0) {``            ``missingElement = arr[i];``            ``break``;``        ``}` `        ``arr[``abs``(arr[i])] = -arr[``abs``(arr[i])];``    ``}` `    ``return` `abs``(missingElement);``}` `// Driver code``int` `main()``{``    ``int` `arr[] = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``// Function call``    ``cout << findRepeating(arr, N);` `    ``return` `0;``}`

## Java

 `// Java program to find the only``// repeating element in an array``// where elements are from 1 to N-1.``import` `java.lang.Math.*;` `class` `GFG {``  ` `    ``// Function to find repeated element``    ``static` `int` `findRepeating(``int` `arr[], ``int` `N)``    ``{``        ``int` `missingElement = ``0``;` `        ``// indexing based``        ``for` `(``int` `i = ``0``; i < N; i++) {` `            ``int` `absVal = Math.abs(arr[i]);``            ``int` `element = arr[absVal];` `            ``if` `(element < ``0``) {``                ``missingElement = arr[i];``                ``break``;``            ``}` `            ``absVal = Math.abs(arr[i]);``            ``arr[absVal] = -arr[absVal];``        ``}` `        ``return` `Math.abs(missingElement);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `};` `        ``int` `N = arr.length;``    ` `          ``// Function call``        ``System.out.println(findRepeating(arr, N));``    ``}``}``// This code is contributed by``// Smitha Dinesh Semwal.`

## Python3

 `# Python3 program to find the only``# repeating element in an array``# where elements are from 1 to N-1.` `# Function to find repeated element`  `def` `findRepeating(arr, N):` `    ``missingElement ``=` `0` `    ``# indexing based``    ``for` `i ``in` `range``(``0``, N):` `        ``element ``=` `arr[``abs``(arr[i])]` `        ``if``(element < ``0``):``            ``missingElement ``=` `arr[i]``            ``break` `        ``arr[``abs``(arr[i])] ``=` `-``arr[``abs``(arr[i])]` `    ``return` `abs``(missingElement)`  `# Driver code``if` `__name__ ``=``=` `"__main__"``:``  ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``  ``N ``=` `len``(arr)``  ` `  ``# Function call``  ``print``(findRepeating(arr, N))` `# This code is contributed by Smitha Dinesh Semwal.`

## C#

 `// C# program to find the only``// repeating element in an array``// where elements are from 1 to N-1.``using` `System;` `public` `class` `GFG {` `    ``// Function to find repeated element``    ``static` `int` `findRepeating(``int``[] arr, ``int` `N)``    ``{``        ``int` `missingElement = 0;` `        ``// indexing based``        ``for` `(``int` `i = 0; i < N; i++) {` `            ``int` `absVal = Math.Abs(arr[i]);``            ``int` `element = arr[absVal];` `            ``if` `(element < 0) {``                ``missingElement = arr[i];``                ``break``;``            ``}` `            ``absVal = Math.Abs(arr[i]);``            ``arr[absVal] = -arr[absVal];``        ``}` `        ``return` `Math.Abs(missingElement);``    ``}` `    ``// Driver Code``    ``public` `static` `void` `Main(``string``[] args)``    ``{``        ``int``[] arr = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };` `        ``int` `N = arr.Length;``        ` `          ``// Function call``        ``Console.WriteLine(findRepeating(arr, N));``    ``}``}` `// this code is contributed by phasing17`

## Javascript

 `     ``// JavaScript program for the above approach;` `     ``// Function to find repeated element``     ``function` `findRepeating(arr, n) {``         ``let missingElement = 0;` `         ``// indexing based``         ``for` `(let i = 0; i < n; i++) {` `             ``let absVal = Math.abs(arr[i]);``             ``let element = arr[absVal];` `             ``if` `(element < 0) {``                 ``missingElement = arr[i];``                 ``break``;``             ``}` `             ``let absVal = Math.abs(arr[i]);``             ``arr[absVal] = -arr[absVal];``         ``}` `         ``return` `Math.abs(missingElement);` `     ``}` `     ``// driver code` `     ``let arr = [5, 4, 3, 9, 8,``         ``9, 1, 6, 2, 5];` `     ``let n = arr.length;` `     ``document.write(findRepeating(arr, n));``// This code is contributed by Potta Lokesh`

## PHP

 ``

Output
```8

```

Time Complexity: O(N)
Auxiliary Space: O(1)

## Find the only repetitive element using Linked-List cycle method:

Use two pointers the fast and the slow. The fast one goes forward two steps each time, while the slow one goes only step each time. They must meet the same item when slow==fast.

In fact, they meet in a circle, the duplicate number must be the entry point of the circle when visiting the array from array[0].

Next we just need to find the entry point. We use a point(we can use the fast one before) to visit from the beginning with one step each time, do the same job to slow. When fast==slow, they meet at the entry point of the circle.

Follow the below steps to solve the problem:

• Declare two integer pointers as slow and fast
• Move the slow pointer one time and fast pointer two times, until slow is not equal to fast
• Once they are equal then again start the fast pointer from the start of the array
• Move both the pointers, one step at a time until both of them are equal
• Return slow or fast pointer as the answer

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `int` `findDuplicate(vector<``int``>& nums)``{``    ``int` `slow = nums[0];``//slow pointer``    ``int` `fast = nums[0];``//fast pointer` `    ``do` `{``        ``slow = nums[slow];``//moves one step-->if speed is x``        ``fast = nums[nums[fast]];``//it moves fast than slow pointer-->speed is 2x``    ``} ``while` `(slow != fast);` `    ``fast = nums[0];``//initialize fast to the starting point``  ``//loop untill  both pointer will meet at the same point``    ``while` `(slow != fast) {``        ``slow = nums[slow];``        ``fast = nums[fast];``    ``}``//return duplicate number``    ``return` `slow;``}` `// Driver code``int` `main()``{``    ``vector<``int``> v{ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``  ` `      ``// Function call``    ``int` `ans = findDuplicate(v);``    ``cout << ans << endl;``    ``return` `0;``}`

## Java

 `import` `java.io.*;``import` `java.util.*;` `class` `GFG {` `  ``public` `static` `int` `findDuplicate(ArrayList nums)``  ``{``    ``int` `slow = nums.get(``0``);``    ``int` `fast = nums.get(``0``);` `    ``do` `{``      ``slow = nums.get(slow);``      ``fast = nums.get(nums.get(fast));``    ``} ``    ``while` `(slow != fast);` `    ``fast = nums.get(``0``);``    ``while` `(slow != fast) {``      ``slow = nums.get(slow);``      ``fast = nums.get(fast);``    ``}` `    ``return` `slow;``  ``}` `  ``public` `static` `void` `main(String[] args)``  ``{``    ``ArrayList arr = ``new` `ArrayList( Arrays. asList( ``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7` `) );` `    ``// Function call``    ``int` `ans = findDuplicate(arr);``    ``System.out.println(ans);``  ``}``}` `// This code is contributed by adityapatil12`

## Python3

 `class` `GFG :``    ``@staticmethod``    ``def`  `findDuplicate( nums) :``        ``slow ``=` `nums[``0``]``        ``fast ``=` `nums[``0``]``        ``while` `True` `:``            ``slow ``=` `nums[slow]``            ``fast ``=` `nums[nums[fast]]``            ``if``((slow !``=` `fast) ``=``=` `False``) :``                    ``break``        ``fast ``=` `nums[``0``]``        ``while` `(slow !``=` `fast) :``            ``slow ``=` `nums[slow]``            ``fast ``=` `nums[fast]``        ``return` `slow``    ``@staticmethod``    ``def` `main( args) :``        ``arr ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``        ` `        ``# Function call``        ``ans ``=` `GFG.findDuplicate(arr)``        ``print``(ans)``    ` `if` `__name__``=``=``"__main__"``:``    ``GFG.main([])``    ` `    ``# This code is contributed by aadityaburujwale.`

## C#

 `// C# code for the above approach``using` `System;` `public` `class` `GFG {` `  ``static` `int` `findDuplicate(``int``[] nums)``  ``{``    ``int` `slow = nums[0];``    ``int` `fast = nums[0];` `    ``do` `{``      ``slow = nums[slow];``      ``fast = nums[nums[fast]];``    ``} ``while` `(slow != fast);` `    ``fast = nums[0];``    ``while` `(slow != fast) {``      ``slow = nums[slow];``      ``fast = nums[fast];``    ``}` `    ``return` `slow;``  ``}` `  ``static` `public` `void` `Main()``  ``{` `    ``// Code``    ``int``[] v = { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };` `    ``// Function call``    ``int` `ans = findDuplicate(v);``    ``Console.Write(ans);``  ``}``}` `// This code is contributed by lokesh.`

## Javascript

 `function` `findDuplicate(nums){``    ``var` `slow = nums[0];``    ``var` `fast = nums[0];` `    ``do` `{``      ``slow = nums[slow];``      ``fast = nums[nums[fast]];``    ``} ``    ``while` `(slow != fast);` `    ``fast = nums[0];``    ``while` `(slow != fast) {``      ``slow = nums[slow];``      ``fast = nums[fast];``    ``}` `    ``return` `slow;``}` `    ``var` `arr = [ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 ];` `    ``// Function call``    ``var` `ans = findDuplicate(arr);``    ``console.log(ans);``    ` `    ``// This code is contributed by aadityaburujwale.`

Output
```8

```

Time Complexity: O(N)
Auxiliary Space: O(1)

## Find the only repetitive element Using Binary Search:

 Initialize low to 1 (since the array has values in the range of 1 to n, where n is the size of the array) and high to nums.size() – 1.Enter a while loop that iteratively reduces the search range by half:Calculate mid as the average of low and high.Iterate over the nums array and count the number of elements that are less than or equal to mid.Based on the count of elements less than or equal to mid, perform a binary search on the left half of the array if the count is less than or equal to mid, or on the right half of the array otherwise.Return the low variable, which is the duplicate number in the array.The reason for this is that the code performs a binary search on the array, which has a time complexity of O(log n), and it iterates over the array to count the number of elements less than or equal to the midpoint in each iteration. The worst-case time complexity of this iteration is O(n log n) because the total number of iterations required is proportional to the height of the binary search tree, which is log n, and at each iteration, we are visiting all elements in the array, which has a time complexity of O(n). Therefore, the overall time complexity is O(n log n).implementation is given below:-

## C++

 `#include ``using` `namespace` `std;`` ` `int` `findDuplicate(vector<``int``>& nums)``{``    ``int` `low = 1, high = nums.size() - 1, cnt;``        ` `        ``while``(low <=  high)``        ``{``            ``int` `mid = low + (high - low) / 2;``            ``cnt = 0;``            ``// cnt number less than equal to mid``            ``for``(``int` `n : nums)``            ``{``                ``if``(n <= mid)``                    ``++cnt;``            ``}``            ``// binary search on left``            ``if``(cnt <= mid)``                ``low = mid + 1;``            ``else``            ``// binary search on right``                ``high = mid - 1;``            ` `        ``}``        ``return` `low;``}`` ` `// Driver code``int` `main()``{``    ``vector<``int``> v{ 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };``   ` `      ``// Function call``    ``int` `ans = findDuplicate(v);``    ``cout << ans << endl;``//8``    ``return` `0;``}` `//priyapramanick17`

## Java

 `import` `java.util.ArrayList;``import` `java.util.List;` `public` `class` `FindDuplicate {``    ``public` `static` `void` `main(String[] args) {``        ``List nums = ``new` `ArrayList<>();``        ``nums.add(``9``);``        ``nums.add(``8``);``        ``nums.add(``2``);``        ``nums.add(``6``);``        ``nums.add(``1``);``        ``nums.add(``8``);``        ``nums.add(``5``);``        ``nums.add(``3``);``        ``nums.add(``4``);``        ``nums.add(``7``);` `        ``// Function call``        ``int` `ans = findDuplicate(nums);``        ``System.out.println(ans); ``// 8``    ``}` `    ``public` `static` `int` `findDuplicate(List nums) {``        ``int` `low = ``1``;``        ``int` `high = nums.size() - ``1``;``        ``int` `cnt;` `        ``while` `(low <= high) {``            ``int` `mid = low + (high - low) / ``2``;``            ``cnt = ``0``;` `            ``// Count the number of elements less than or equal to mid``            ``for` `(``int` `n : nums) {``                ``if` `(n <= mid)``                    ``cnt++;``            ``}` `            ``// Binary search on the left``            ``if` `(cnt <= mid)``                ``low = mid + ``1``;``            ``else``                ``// Binary search on the right``                ``high = mid - ``1``;``        ``}``        ``return` `low;``    ``}``}`

## Python

 `def` `find_duplicate(nums):``    ``# Initialize the low and high values for binary search``    ``low ``=` `1``    ``high ``=` `len``(nums) ``-` `1` `    ``while` `low <``=` `high:``        ``mid ``=` `low ``+` `(high ``-` `low) ``/``/` `2`  `# Calculate the middle index``        ``cnt ``=` `0` `        ``# Count the numbers less than or equal to mid``        ``for` `n ``in` `nums:``            ``if` `n <``=` `mid:``                ``cnt ``+``=` `1` `        ``# Perform binary search on the left half``        ``if` `cnt <``=` `mid:``            ``low ``=` `mid ``+` `1``        ``else``:``            ``# Perform binary search on the right half``            ``high ``=` `mid ``-` `1` `    ``return` `low` `# Driver code``if` `__name__ ``=``=` `"__main__"``:``    ``v ``=` `[``9``, ``8``, ``2``, ``6``, ``1``, ``8``, ``5``, ``3``, ``4``, ``7``]``   ` `    ``# Function call``    ``ans ``=` `find_duplicate(v)``    ``print``(ans)  ``# Output should be 8`

## C#

 `using` `System;``using` `System.Collections.Generic;` `class` `Program``{``    ``// Function to find the duplicate element in a list of integers``    ``public` `static` `int` `FindDuplicate(List<``int``> nums)``    ``{``        ``int` `low = 1;``        ``int` `high = nums.Count - 1;``        ``int` `cnt;` `        ``while` `(low <= high)``        ``{``            ``int` `mid = low + (high - low) / 2;``            ``cnt = 0;` `            ``// Count numbers less than or equal to mid in the list``            ``foreach` `(``int` `n ``in` `nums)``            ``{``                ``if` `(n <= mid)``                ``{``                    ``cnt++;``                ``}``            ``}` `            ``// If the count of numbers less than or equal to mid is less than or equal to mid,``            ``// it means the duplicate element is on the right side of mid.``            ``if` `(cnt <= mid)``            ``{``                ``low = mid + 1; ``// Binary search on the right side``            ``}``            ``else``            ``{``                ``high = mid - 1; ``// Binary search on the left side``            ``}``        ``}` `        ``return` `low; ``// The duplicate element is found at 'low'``    ``}` `    ``static` `void` `Main(``string``[] args)``    ``{``        ``List<``int``> nums = ``new` `List<``int``> { 9, 8, 2, 6, 1, 8, 5, 3, 4, 7 };` `        ``// Function call``        ``int` `ans = FindDuplicate(nums);``        ``Console.WriteLine(ans); ``// Print the duplicate element``    ``}``}`

## Javascript

 `function` `findDuplicate(nums) {``    ``let low = 1;``    ``let high = nums.length - 1;``    ``let cnt;` `    ``while` `(low <= high) {``        ``let mid = low + Math.floor((high - low) / 2);``        ``cnt = 0;``        ``// cnt number less than equal to mid``        ``for` `(let n of nums) {``            ``if` `(n <= mid)``                ``++cnt;``        ``}``        ``// binary search on left``        ``if` `(cnt <= mid)``            ``low = mid + 1;``        ``else``            ``// binary search on right``            ``high = mid - 1;` `    ``}``    ``return` `low;``}` `// Driver code``let v = [9, 8, 2, 6, 1, 8, 5, 3, 4, 7];` `// Function call``let ans = findDuplicate(v);``console.log(ans); ``//8`

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