Duplicates in an array in O(n) time and by using O(1) extra space | Set-3
Last Updated :
11 Jul, 2025
Given an array of n elements which contains 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. It is required that the order in which elements repeat should be maintained. If there is no repeating element present then print -1.
Examples:
Input : arr[] = {1, 2, 3, 1, 3, 6, 6}
Output : 1, 3, 6
Elements 1, 3 and 6 are repeating.
Second occurrence of 1 is found
first followed by repeated occurrence
of 3 and 6.
Input : arr[] = {0, 3, 1, 3, 0}
Output : 3, 0
Second occurrence of 3 is found
first followed by second occurrence
of 0.
We have discussed two approaches for this question in below posts:
Find duplicates in O(n) time and O(1) extra space | Set 1
Duplicates in an array in O(n) time and by using O(1) extra space | Set-2
There is a problem in first approach. It prints the repeated number more than once. For example: {1, 6, 3, 1, 3, 6, 6} it will give output as : 3 6 6. In second approach although each repeated item is printed only once, the order in which their repetition occurs is not maintained.
To print elements in order in which they are repeating, the second approach is modified. To mark the presence of an element size of the array, n is added to the index position arr[i] corresponding to array element arr[i]. Before adding n, check if value at index arr[i] is greater than or equal to n or not. If it is greater than or equal to, then this means that element arr[i] is repeating. To avoid printing repeating element multiple times, check if it is the first repetition of arr[i] or not. It is first repetition if value at index position arr[i] is less than 2*n.
This is because, if element arr[i] has occurred only once before then n is added to index arr[i] only once and thus value at index arr[i] is less than 2*n. Add n to index arr[i] so that value becomes greater than or equal to 2*n and it will prevent further printing of current repeating element. Also if value at index arr[i] is less than n then it is first occurrence (not repetition) of element arr[i]. So to mark this add n to element at index arr[i].
Below is the implementation of above approach:
C++
// C++ program to print all elements that
// appear more than once.
#include <bits/stdc++.h>
using namespace std;
// Function to find repeating elements
void printDuplicates(int arr[], int n)
{
int i;
// Flag variable used to
// represent whether repeating
// element is found or not.
int fl = 0;
for (i = 0; i < n; i++) {
// Check if current element is
// repeating or not. If it is
// repeating then value will
// be greater than or equal to n.
if (arr[arr[i] % n] >= n) {
// Check if it is first
// repetition or not. If it is
// first repetition then value
// at index arr[i] is less than
// 2*n. Print arr[i] if it is
// first repetition.
if (arr[arr[i] % n] < 2 * n) {
cout << arr[i] % n << " ";
fl = 1;
}
}
// Add n to index arr[i] to mark
// presence of arr[i] or to
// mark repetition of arr[i].
arr[arr[i] % n] += n;
}
// If flag variable is not set
// then no repeating element is
// found. So print -1.
if (!fl)
cout << "-1";
}
// Driver Function
int main()
{
int arr[] = { 1, 6, 3, 1, 3, 6, 6 };
int arr_size = sizeof(arr) / sizeof(arr[0]);
printDuplicates(arr, arr_size);
return 0;
}
// Java program to print all elements
// that appear more than once.
import java.io.*;
class GFG
{
// Function to find repeating elements
static void printDuplicates(int arr[], int n)
{
int i;
// Flag variable used to
// represent whether repeating
// element is found or not.
int fl = 0;
for (i = 0; i < n; i++)
{
// Check if current element is
// repeating or not. If it is
// repeating then value will
// be greater than or equal to n.
if (arr[arr[i] % n] >= n)
{
// Check if it is first
// repetition or not. If it is
// first repetition then value
// at index arr[i] is less than
// 2*n. Print arr[i] if it is
// first repetition.
if (arr[arr[i] % n] < 2 * n)
{
System.out.print( arr[i] % n + " ");
fl = 1;
}
}
// Add n to index arr[i] to mark
// presence of arr[i] or to
// mark repetition of arr[i].
arr[arr[i] % n] += n;
}
// If flag variable is not set
// then no repeating element is
// found. So print -1.
if (!(fl > 0))
System.out.println("-1");
}
// Driver Code
public static void main (String[] args)
{
int arr[] = { 1, 6, 3, 1, 3, 6, 6 };
int arr_size = arr.length;
printDuplicates(arr, arr_size);
}
}
// This code is contributed by anuj_67.
# Python 3 program to print all elements that
# appear more than once.
# Function to find repeating elements
def printDuplicates(arr, n):
# Flag variable used to
# represent whether repeating
# element is found or not.
fl = 0;
for i in range (0, n):
# Check if current element is
# repeating or not. If it is
# repeating then value will
# be greater than or equal to n.
if (arr[arr[i] % n] >= n):
# Check if it is first
# repetition or not. If it is
# first repetition then value
# at index arr[i] is less than
# 2*n. Print arr[i] if it is
# first repetition.
if (arr[arr[i] % n] < 2 * n):
print(arr[i] % n, end = " ")
fl = 1;
# Add n to index arr[i] to mark
# presence of arr[i] or to
# mark repetition of arr[i].
arr[arr[i] % n] += n;
# If flag variable is not set
# then no repeating element is
# found. So print -1.
if (fl == 0):
print("-1")
# Driver Function
arr = [ 1, 6, 3, 1, 3, 6, 6 ];
arr_size = len(arr);
printDuplicates(arr, arr_size);
# This code is contributed
# by Akanksha Rai
// C# program to print all elements
// that appear more than once.
using System;
class GFG
{
// Function to find repeating elements
static void printDuplicates(int []arr, int n)
{
int i;
// Flag variable used to
// represent whether repeating
// element is found or not.
int fl = 0;
for (i = 0; i < n; i++)
{
// Check if current element is
// repeating or not. If it is
// repeating then value will
// be greater than or equal to n.
if (arr[arr[i] % n] >= n)
{
// Check if it is first
// repetition or not. If it is
// first repetition then value
// at index arr[i] is less than
// 2*n. Print arr[i] if it is
// first repetition.
if (arr[arr[i] % n] < 2 * n)
{
Console.Write( arr[i] % n + " ");
fl = 1;
}
}
// Add n to index arr[i] to mark
// presence of arr[i] or to
// mark repetition of arr[i].
arr[arr[i] % n] += n;
}
// If flag variable is not set
// then no repeating element is
// found. So print -1.
if (!(fl > 0))
Console.Write("-1");
}
// Driver Code
public static void Main ()
{
int []arr = { 1, 6, 3, 1, 3, 6, 6 };
int arr_size = arr.Length;
printDuplicates(arr, arr_size);
}
}
// This code is contributed
// by 29AjayKumar
<?php
// PHP program to print all elements that
// appear more than once.
// Function to find repeating elements
function printDuplicates($arr, $n)
{
$i;
// Flag variable used to
// represent whether repeating
// element is found or not.
$fl = 0;
for ($i = 0; $i < $n; $i++)
{
// Check if current element is
// repeating or not. If it is
// repeating then value will
// be greater than or equal to n.
if ($arr[$arr[$i] % $n] >= $n)
{
// Check if it is first
// repetition or not. If it is
// first repetition then value
// at index arr[i] is less than
// 2*n. Print arr[i] if it is
// first repetition.
if ($arr[$arr[$i] % $n] < 2 * $n)
{
echo $arr[$i] % $n . " ";
$fl = 1;
}
}
// Add n to index arr[i] to mark
// presence of arr[i] or to
// mark repetition of arr[i].
$arr[$arr[$i] % $n] += $n;
}
// If flag variable is not set
// then no repeating element is
// found. So print -1.
if (!$fl)
echo "-1";
}
// Driver Function
$arr = array(1, 6, 3, 1, 3, 6, 6);
$arr_size = sizeof($arr);
printDuplicates($arr, $arr_size);
// This code is contributed
// by Mukul Singh
<script>
// JavaScript program to print all elements that
// appear more than once.
// Function to find repeating elements
function printDuplicates(arr, n)
{
let i;
// Flag variable used to
// represent whether repeating
// element is found or not.
let fl = 0;
for (i = 0; i < n; i++) {
// Check if current element is
// repeating or not. If it is
// repeating then value will
// be greater than or equal to n.
if (arr[arr[i] % n] >= n) {
// Check if it is first
// repetition or not. If it is
// first repetition then value
// at index arr[i] is less than
// 2*n. Print arr[i] if it is
// first repetition.
if (arr[arr[i] % n] < 2 * n) {
document.write(arr[i] % n + " ");
fl = 1;
}
}
// Add n to index arr[i] to mark
// presence of arr[i] or to
// mark repetition of arr[i].
arr[arr[i] % n] += n;
}
// If flag variable is not set
// then no repeating element is
// found. So print -1.
if (!fl)
document.write("-1");
}
// Driver Function
let arr = [ 1, 6, 3, 1, 3, 6, 6 ];
let arr_size = arr.length;
printDuplicates(arr, arr_size);
// This code is contributed by Surbhi Tyagi.
</script>
Complexity Analysis:
- Time Complexity: O(N), as we are using a loop to traverse N times.
- Auxiliary Space: O(1), as we are not using any extra space.
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