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.

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