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

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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 6

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

We have discussed an approach for this question in the 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:  

  1. Traverse the given array from start to end.
  2. For every element in the array increment the arr[i]%n‘th element by n.
  3. 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;
}

                    

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.print(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);
    }
}

                    

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.

                    

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);
    }
}

                    

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.
?>

                    

Javascript

<script>
  
// Javascript 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 (let i = 0; i < n; i++)
        {
            let index = arr[i] % n;
            arr[index] += n;
        }
   
        // Now check which value exists more
        // than once by dividing with the size
        // of array
        for (let i = 0; i < n; i++)
        {
            if ((arr[i] / n) >= 2)
                document.write(i + " ");
        }
    }
      
    // Driver code
    let arr=[1, 6, 3, 1, 3, 6, 6];
    let arr_size = arr.length;
    document.write("The repeating elements are: <br>");
      
    // Function call
    printRepeating(arr, arr_size);
      
    //  This code is contributed by avanitrachhadiya2155
      
</script>

                    

Output
The repeating elements are: 
1 3 6 

Complexity Analysis: 

  • Time Complexity: O(n). 
    Only two traversals are needed. So the time complexity is O(n)
  • Auxiliary Space: O(1). 
    As no extra space is needed, so the space complexity is constant



Last Updated : 12 Sep, 2023
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