Check if array contains duplicates
Last Updated :
23 Jul, 2025
Given an integer array arr[], check if the array contains any duplicate value.
Examples:
Input: arr[] = {4, 5, 6, 4}
Output: true
Explanation: 4 is the duplicate value.
Input: arr[] = {1, 2, 3, 4}
Output: false
Explanation: All values are distinct.
[Naive Approach] By using two nested loops – O(n ^ 2) Time and O(1) Space
The simple idea is to use a nested loop to compare each element in the array with every other element. If any two elements are found to be the same, return true, indicating the array has a duplicate element. If no duplicates are found after all comparisons, return false.
C++
// C++ Program check if there are any duplicates
// in the array using nested loops approach
#include <bits/stdc++.h>
using namespace std;
bool checkDuplicates(vector<int> &arr) {
int n = arr.size();
// Outer loop to pick each element one by one
for(int i = 0; i < n - 1; i++) {
// Inner loop to compare the current element with
// the rest of the elements
for(int j = i + 1; j < n; j++) {
// If a duplicate is found, return true
if(arr[i] == arr[j])
return true;
}
}
// If no duplicates are found, return false
return false;
}
int main () {
vector<int> arr = {4, 5, 6, 4};
cout << (checkDuplicates(arr) ? "true" : "false") << endl;
return 0;
}
// C Program check if there are any duplicates
// in the array using nested loops approach
#include <stdio.h>
#include <stdbool.h>
bool checkDuplicates(int *arr, int n) {
// Outer loop to pick each element one by one
for(int i = 0; i < n - 1; i++) {
// Inner loop to compare the current element
// with the rest of the elements
for(int j = i + 1; j < n; j++) {
// If a duplicate is found, return true
if(arr[i] == arr[j])
return true;
}
}
// If no duplicates are found, return false
return false;
}
int main () {
int arr[] = {4, 5, 6, 4};
int n = sizeof(arr) / sizeof(arr[0]);
printf(checkDuplicates(arr, n) ? "true" : "false");
return 0;
}
// Java Program check if there are any duplicates
// in the array using nested loops approach
import java.util.*;
class GfG {
static boolean checkDuplicates(int[] arr) {
int n = arr.length;
// Outer loop to pick each element one by one
for (int i = 0; i < n - 1; i++) {
// Inner loop to compare the current element
// with the rest of the elements
for (int j = i + 1; j < n; j++) {
// If a duplicate is found Return true
if (arr[i] == arr[j])
return true;
}
}
// If no duplicates are found, return false
return false;
}
public static void main(String[] args) {
int[] arr = {4, 5, 6, 4};
System.out.println(checkDuplicates(arr));
}
}
# Python Program check if there are any duplicates
# in the array using nested loops approach
def check_duplicates(arr):
n = len(arr)
# Outer loop to pick each element one by one
for i in range(n - 1):
# Inner loop to compare the current element with the
# rest of the elements
for j in range(i + 1, n):
# If a duplicate is found return True
if arr[i] == arr[j]:
return True
# If no duplicates are found, return False
return False
if __name__ == "__main__":
arr = [4, 5, 6, 4]
print(check_duplicates(arr))
// C# Program check if there are any duplicates
// in the array using nested loops approach
using System;
using System.Collections.Generic;
class GfG {
static bool CheckDuplicates(int[] arr) {
int n = arr.Length;
// Outer loop to pick each element one by one
for (int i = 0; i < n - 1; i++) {
// Inner loop to compare the current element with
// the rest of the elements
for (int j = i + 1; j < n; j++) {
// If a duplicate is found return true
if (arr[i] == arr[j]) {
return true;
}
}
}
// If no duplicates are found, return false
return false;
}
static void Main() {
int[] arr = { 4, 5, 6, 4 };
Console.WriteLine(CheckDuplicates(arr));
}
}
// JavaScript Program check if there are any duplicates
// in the array using nested loops approach
function checkDuplicates(arr) {
let n = arr.length;
// Outer loop to pick each element one by one
for (let i = 0; i < n - 1; i++) {
// Inner loop to compare the current element with
// the rest of the elements
for (let j = i + 1; j < n; j++) {
// If a duplicate is found return true
if (arr[i] === arr[j]) {
return true;
}
}
}
// If no duplicates are found, return false
return false;
}
let arr = [4, 5, 6, 4];
console.log(checkDuplicates(arr));
Time Complexity: O(N2), As we are using nested loop over the given array, where N is the number of elements in the given array.
Auxiliary Space: O(1), As we are not using any extra space.
[Expected Approach] By using HashSet Data Structure – O(n) Time and O(n) Space
The main idea is to insert each value into a HashSet, which only stores unique elements. If any insertion fails or if any elements are already exits in HashSet, it means a duplicate exists, so return true. If all insertions succeed, it indicates that all elements are unique, so return false.
C++
// C++ Program check if there are any duplicates
// in the array using Hashing
#include <iostream>
#include <unordered_set>
#include <vector>
using namespace std;
bool checkDuplicates(vector<int>& arr) {
int n = arr.size();
// Create an unordered_set to store the unique elements
unordered_set<int> st;
// Iterate through each element
for (int i = 0; i < n; i++) {
// If the element is already present, return true
// Else insert the element into the set
if (st.find(arr[i]) != st.end())
return true;
else
st.insert(arr[i]);
}
// If no duplicates are found, return false
return false;
}
int main() {
vector<int> arr = { 4, 5, 6, 4 };
cout << (checkDuplicates(arr) ? "true" : "false") << endl;
return 0;
}
// Java Program check if there are any duplicates
// in the array using Hashing
import java.util.HashSet;
import java.util.Set;
class GfG {
// Function to check if there are any duplicates
static boolean checkDuplicates(int[] arr) {
int n = arr.length;
// Create a HashSet to store the unique elements
Set<Integer> st = new HashSet<>();
// Iterate through each element
for (int i = 0; i < n; i++) {
// If the element is already present, return true
// Else insert the element into the set
if (st.contains(arr[i]))
return true;
else
st.add(arr[i]);
}
// If no duplicates are found, return false
return false;
}
public static void main(String[] args) {
int[] arr = { 4, 5, 6, 4 };
System.out.println(checkDuplicates(arr));
}
}
# Python Program check if there are any duplicates
# in the array using Hashing
def checkDuplicates(arr):
n = len(arr)
# Create a set to store the unique elements
st = set()
# Iterate through each element
for i in range(n):
# If the element is already present, return true
# Else insert the element into the set
if arr[i] in st:
return True
else:
st.add(arr[i])
# If no duplicates are found, return false
return False
if __name__ == "__main__":
arr = [4, 5, 6, 4]
print(checkDuplicates(arr))
// C# Program check if there are any duplicates
// in the array using Hashing
using System;
using System.Collections.Generic;
class GfG {
// Function to check if there are any duplicates
static bool CheckDuplicates(int[] arr) {
int n = arr.Length;
// Create a HashSet to store the unique elements
HashSet<int> st = new HashSet<int>();
// Iterate through each element
for (int i = 0; i < n; i++) {
// If the element is already present, return true
// Else insert the element into the set
if (st.Contains(arr[i]))
return true;
else
st.Add(arr[i]);
}
// If no duplicates are found, return false
return false;
}
public static void Main(string[] args) {
int[] arr = { 4, 5, 6, 4 };
Console.WriteLine(CheckDuplicates(arr));
}
}
// JavaScript Program check if there are any duplicates
// in the array using Hashing
function checkDuplicates(arr) {
const n = arr.length;
// Create a Set to store the unique elements
const st = new Set();
// Iterate through each element
for (let i = 0; i < n; i++) {
// If the element is already present, return true
// Else insert the element into the set
if (st.has(arr[i]))
return true;
else
st.add(arr[i]);
}
// If no duplicates are found, return false
return false;
}
const arr = [4, 5, 6, 4];
console.log(checkDuplicates(arr));
Time Complexity: O(n), As we are traversing over the array once.
Auxiliary space: O(n), As we are using unordered set which takes O(n) space in worst case, where n is the size of the array.
[Other Approach] By Sorting the array - O(n * log(n)) Time and O(1) Space
The main idea is to first sort the array arr[] and then iterate through it to check if any adjacent elements are equal. If a pair of adjacent elements is equal, the function returns true, indicating the presence of duplicates. If no duplicates are found after traversing the entire array, the function returns false.
C++
// C++ Program check if there are any duplicates
// in the array using Sorting
#include <bits/stdc++.h>
using namespace std;
bool checkDuplicates(vector<int>& arr) {
int n = arr.size();
// Sort the input array
sort(arr.begin(), arr.end());
// Iterate through the sorted array
for (int i = 1; i < n; ++i) {
// Check if adjacent elements are equal
if (arr[i] == arr[i - 1])
return true;
}
// If no duplicates are found, return false
return false;
}
int main() {
vector<int> arr = { 4, 5, 6, 4 };
cout << (checkDuplicates(arr) ? "true" : "false") << endl;
return 0;
}
// C Program check if there are any duplicates
// in the array using Sorting
#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
// Function to compare two integers for qsort
int compare(const void* a, const void* b) {
return (*(int*)a - *(int*)b);
}
bool checkDuplicates(int* arr, int n) {
// Sort the array
qsort(arr, n, sizeof(int), compare);
// Iterate through the sorted array
for (int i = 1; i < n; i++) {
// Check if adjacent elements are equal
if (arr[i] == arr[i - 1])
return true;
}
// If no duplicates are found, return false
return false;
}
int main() {
int arr[] = { 4, 5, 6, 4 };
int n = sizeof(arr) / sizeof(arr[0]);
printf("%s\n", checkDuplicates(arr, n) ? "true" : "false");
return 0;
}
// Java Program check if there are any duplicates
// in the array using Sorting
import java.util.Arrays;
class GFG {
static boolean checkDuplicates(int[] arr) {
int n = arr.length;
// Sort the array
Arrays.sort(arr);
// Iterate through the sorted array
for (int i = 1; i < n; i++) {
// Check if adjacent elements are equal
if (arr[i] == arr[i - 1])
return true;
}
// If no duplicates are found, return false
return false;
}
public static void main(String[] args) {
int[] arr = { 4, 5, 6, 4 };
System.out.println(checkDuplicates(arr));
}
}
# Python Program check if there are any duplicates
# in the array using Sorting
def check_duplicates(arr):
n = len(arr)
# Sort the array
arr.sort()
# Iterate through the sorted array
for i in range(1, n):
# Check if adjacent elements are equal
if arr[i] == arr[i - 1]:
return True
# If no duplicates are found, return False
return False
if __name__ == "__main__":
arr = [4, 5, 6, 4]
print(check_duplicates(arr))
// C# Program check if there are any duplicates
// in the array using Sorting
using System;
class GFG {
static bool CheckDuplicates(int[] arr) {
int n = arr.Length;
// Sort the array
Array.Sort(arr);
// Iterate through the sorted array
for (int i = 1; i < n; i++) {
// Check if adjacent elements are equal
if (arr[i] == arr[i - 1])
return true;
}
// If no duplicates are found, return false
return false;
}
static void Main() {
int[] arr = { 4, 5, 6, 4 };
Console.WriteLine(CheckDuplicates(arr));
}
}
// JavaScript Program check if there are any duplicates
// in the array using Sorting
function checkDuplicates(arr) {
let n = arr.length;
// Sort the array
arr.sort((a, b) => a - b);
// Iterate through the sorted array
for (let i = 1; i < n; i++) {
// Check if adjacent elements are equal
if (arr[i] === arr[i - 1])
return true;
}
// If no duplicates are found, return false
return false;
}
const arr = [4, 5, 6, 4];
console.log(checkDuplicates(arr));
Time Complexity: O(n * logn), As we are using sorting function which takes nlogn time.
Auxiliary space: O(1), As we are not using extra space.
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