Duplicates in an array in O(n) and by using O(1) extra space
Last Updated :
24 Dec, 2024
Given an array arr[] of n elements that contains elements from 0 to n-1, with any of these numbers appearing any number of times. The task is to find the repeating numbers.
Note: The repeating element should be printed only once.
Example:
Input: n = 7, arr[] = [1, 2, 3, 6, 3, 6, 1]
Output: 1, 3, 6
Explanation: The numbers 1 , 3 and 6 appears more than once in the array.
Input : n = 5, arr[] = [1, 2, 3, 4 ,3]
Output: 3
Explanation: The number 3 appears more than once in the array.
Please refer to Duplicates in an array in O(n) and by using O(n) extra space for implementation of naive approach.
Approach:
The basic idea is based on the hash map to solve the problem. But, the numbers in the array are from 0 to n-1, and the input array has length n. So, the input array itself can be used as a hash map. 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.
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.
C++
// C++ code to find duplicates in an array
// in O(1) space
#include <bits/stdc++.h>
using namespace std;
vector<int> findDuplicates(vector<int> &arr) {
int n = arr.size();
// 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
vector<int> result;
for (int i = 0; i < n; i++) {
if ((arr[i] / n) >= 2)
result.push_back(i);
}
return result;
}
int main() {
vector<int> arr = {2, 3, 1, 2, 3, 1, 6};
vector<int> duplicates = findDuplicates(arr);
for (int element : duplicates) {
cout << element << " ";
}
return 0;
}
Java
// Java code to find duplicates in an array
// in O(1) space
import java.util.*;
class GfG {
static int[] findDuplicates(int[] arr) {
int n = arr.length;
// 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
List<Integer> result = new ArrayList<>();
for (int i = 0; i < n; i++) {
if ((arr[i] / n) >= 2) {
result.add(i);
}
}
// If no duplicates found, return array with -1
if (result.isEmpty()) {
result.add(-1);
}
// Convert list to array and return
return result.stream().mapToInt(i -> i).toArray();
}
public static void main(String[] args) {
int[] arr = { 2, 3, 1, 2, 3, 1, 6 };
int[] duplicates = findDuplicates(arr);
for (int element : duplicates) {
System.out.print(element + " ");
}
}
}
Python
# Python code to find duplicates in an array
# in O(1) space
def findDuplicates(arr):
n = len(arr)
# 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(n):
index = arr[i] % n
arr[index] += n
# Now check which value exists more
# than once by dividing with the size
# of array
result = []
for i in range(n):
if (arr[i] // n) >= 2:
result.append(i)
return result
if __name__ == "__main__":
arr = [2, 3, 1, 2, 3, 1, 6]
duplicates = findDuplicates(arr)
print(" ".join(map(str, duplicates)))
C#
// C# code to find duplicates in an array
// in O(1) space
using System;
using System.Collections.Generic;
class GfG {
static int[] FindDuplicates(int[] arr) {
int n = arr.Length;
// 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
List<int> result = new List<int>();
for (int i = 0; i < n; i++) {
if ((arr[i] / n) >= 2) {
result.Add(i);
}
}
// If no duplicates found, return an array with -1
if (result.Count == 0) {
result.Add(-1);
}
return result.ToArray();
}
static void Main(string[] args) {
int[] arr = { 2, 3, 1, 2, 3, 1, 6 };
int[] duplicates = FindDuplicates(arr);
foreach (int element in duplicates) {
Console.Write(element + " ");
}
}
}
JavaScript
// JavaScript code to find duplicates in an array
// in O(1) space
function findDuplicates(arr) {
let n = arr.length;
// 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
let result = [];
for (let i = 0; i < n; i++) {
if (Math.floor(arr[i] / n) >= 2) {
result.push(i);
}
}
return result;
}
// Driver code
let arr = [2, 3, 1, 2, 3, 1, 6];
let duplicates = findDuplicates(arr);
console.log(duplicates.join(" "));
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
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