Given an array of n elements containing elements from 0 to n-1, with any of these numbers appearing any number of times, find these repeating numbers in O(n) and using only constant memory space.

**Example:**

Input:n = 7 , array = {1, 2, 3, 1, 3, 6, 6}Output:1, 3 and 6.Explanation:Duplicate element in the array are 1 , 3 and 6Input:n = 6, array = {5, 3, 1, 3, 5, 5}Output:3 and 5.Explanation:Duplicate element in the array are 3 and 6

We have discussed an approach for this question in below post:

Duplicates in an array in O(n) and by using O(1) extra space | Set-2.

But there is a problem in the above approach. It prints the repeated number more than once.

## We strongly recommend that you click here and practice it, before moving on to the solution.

** Approach:** The basic idea is to use a HashMap to solve the problem. But there is a catch, the numbers in the array are from 0 to n-1, and the input array has length n. So, the input array can be used as a HashMap. While traversing the array, if an element

*a*is encountered then increase the value of

*a%n*‘th element by n. The frequency can be retrieved by dividing the

*a%n*‘th element by n.

__Algorithm__:

- Traverse the given array from start to end.
- For every element in the array increment the
*arr[i]%n*‘th element by n. - Now traverse the array again and print all those indices i for which
*arr[i]/n*is greater than 1. Which guarantees that the number*n*has been added to that index.

**Note: **This approach works because all elements are in the range from 0 to n-1 and arr[i]/n would be greater than 1 only if a value “i” has appeared more than once.

Below is the implementation of the above approach:

## CPP

`// C++ program to print all elements that` `// appear more than once.` `#include <iostream>` `using` `namespace` `std;` `// function to find repeating elements` `void` `printRepeating(` `int` `arr[], ` `int` `n)` `{` ` ` `// First check all the values that are` ` ` `// present in an array then go to that` ` ` `// values as indexes and increment by` ` ` `// the size of array` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{` ` ` `int` `index = arr[i] % n;` ` ` `arr[index] += n;` ` ` `}` ` ` `// Now check which value exists more` ` ` `// than once by dividing with the size` ` ` `// of array` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `if` `((arr[i] / n) >= 2)` ` ` `cout << i << ` `" "` `;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 1, 6, 3, 1, 3, 6, 6 };` ` ` `int` `arr_size = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]);` ` ` `cout << ` `"The repeating elements are: \n"` `;` ` ` `// Function call` ` ` `printRepeating(arr, arr_size);` ` ` `return` `0;` `}` |

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## Java

`// Java program to print all elements that` `// appear more than once.` `import` `java.util.*;` `class` `GFG {` ` ` `// function to find repeating elements` ` ` `static` `void` `printRepeating(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` `// First check all the values that are` ` ` `// present in an array then go to that` ` ` `// values as indexes and increment by` ` ` `// the size of array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `{` ` ` `int` `index = arr[i] % n;` ` ` `arr[index] += n;` ` ` `}` ` ` `// Now check which value exists more` ` ` `// than once by dividing with the size` ` ` `// of array` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++) ` ` ` `{` ` ` `if` `((arr[i] / n) >= ` `2` `)` ` ` `System.out.println(i + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `arr[] = { ` `1` `, ` `6` `, ` `3` `, ` `1` `, ` `3` `, ` `6` `, ` `6` `};` ` ` `int` `arr_size = arr.length;` ` ` `System.out.println(` `"The repeating elements are: "` `);` ` ` `// Function call` ` ` `printRepeating(arr, arr_size);` ` ` `}` `}` |

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## Python3

`# Python3 program to` `# print all elements that` `# appear more than once.` `# function to find` `# repeating elements` `def` `printRepeating(arr, n):` ` ` `# First check all the` ` ` `# values that are` ` ` `# present in an array` ` ` `# then go to that` ` ` `# values as indexes` ` ` `# and increment by` ` ` `# the size of array` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `index ` `=` `arr[i] ` `%` `n` ` ` `arr[index] ` `+` `=` `n` ` ` `# Now check which value` ` ` `# exists more` ` ` `# than once by dividing` ` ` `# with the size` ` ` `# of array` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `if` `(arr[i]` `/` `n) >` `=` `2` `:` ` ` `print` `(i, end` `=` `" "` `)` `# Driver code` `arr ` `=` `[` `1` `, ` `6` `, ` `3` `, ` `1` `, ` `3` `, ` `6` `, ` `6` `]` `arr_size ` `=` `len` `(arr)` `print` `(` `"The repeating elements are:"` `)` `# Function call` `printRepeating(arr, arr_size)` `# This code is contributed` `# by Shreyanshi Arun.` |

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## C#

`// C# program to print all elements that` `// appear more than once.` `using` `System;` `class` `GFG {` ` ` `// function to find repeating elements` ` ` `static` `void` `printRepeating(` `int` `[] arr, ` `int` `n)` ` ` `{` ` ` `// First check all the values that are` ` ` `// present in an array then go to that` ` ` `// values as indexes and increment by` ` ` `// the size of array` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `{` ` ` `int` `index = arr[i] % n;` ` ` `arr[index] += n;` ` ` `}` ` ` `// Now check which value exists more` ` ` `// than once by dividing with the size` ` ` `// of array` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `if` `((arr[i] / n) >= 2)` ` ` `Console.Write(i + ` `" "` `);` ` ` `}` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[] arr = { 1, 6, 3, 1, 3, 6, 6 };` ` ` `int` `arr_size = arr.Length;` ` ` `Console.Write(` `"The repeating elements are: "` ` ` `+ ` `"\n"` `);` ` ` `// Function call` ` ` `printRepeating(arr, arr_size);` ` ` `}` `}` |

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## PHP

`<?php` `// PHP program to print all ` `// elements that appear more` `// than once.` `// function to find ` `// repeating elements` `function` `printRepeating( ` `$arr` `, ` `$n` `)` `{` ` ` `// First check all the values ` ` ` `// that are present in an array ` ` ` `// then go to that values as indexes ` ` ` `// and increment by the size of array` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `$index` `= ` `$arr` `[` `$i` `] % ` `$n` `;` ` ` `$arr` `[` `$index` `] += ` `$n` `;` ` ` `}` ` ` `// Now check which value ` ` ` `// exists more than once ` ` ` `// by dividing with the` ` ` `// size of array` ` ` `for` `(` `$i` `= 0; ` `$i` `< ` `$n` `; ` `$i` `++)` ` ` `{` ` ` `if` `((` `$arr` `[` `$i` `] / ` `$n` `) >= 2)` ` ` `echo` `$i` `, ` `" "` `;` ` ` `}` `}` `// Driver code` `$arr` `= ` `array` `(1, 6, 3, 1, 3, 6, 6);` `$arr_size` `= sizeof(` `$arr` `) / ` ` ` `sizeof(` `$arr` `[0]);` `echo` `"The repeating elements are: \n"` `;` `// Function call` `printRepeating( ` `$arr` `, ` `$arr_size` `);` `// This code is contributed by nitin mittal.` `?>` |

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**Output**

The repeating elements are: 1 3 6

__Complexity Analysis__:

**Time Complexity:**O(n).

Only two traversal is needed. So the time complexity is O(n)**Auxiliary Space:**O(1).

As no extra space is needed, so the space complexity is constant

This article is contributed by **Sahil Chhabra (akku)**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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