Find duplicates in an Array with values 1 to N using counting sort
Given a constant array of N elements which contain elements from 1 to N – 1, with any of these numbers appearing any number of times.
Examples:
Input: N = 5, arr[] = {1, 3, 4, 2, 2}
Output: 2
Explanation:
2 is the number occurring more than once.
Input: N = 5, arr[] = {3, 1, 3, 4, 2}
Output: 3
Explanation:
3 is the number occurring more than once.
Naive Approach: The naive method is to first sort the given array and then look for adjacent positions of the array to find the duplicate number.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findDuplicates( int arr[], int n) {
sort(arr, arr + n);
bool flag = false ;
for ( int i = 0; i < n - 1; i++) {
if (arr[i] == arr[i + 1]) {
flag = true ;
return arr[i];
}
}
if (!flag) {
return -1;
}
}
int main() {
int arr[] = { 1, 3, 4, 2, 2 };
int n = sizeof (arr) / sizeof (arr[0]);
cout << findDuplicates(arr, n);
return 0;
}
|
Java
import java.util.Arrays;
public class GFG {
public static int findDuplicates( int [] arr, int n)
{
Arrays.sort(arr);
boolean flag = false ;
for ( int i = 0 ; i < n - 1 ; i++) {
if (arr[i] == arr[i + 1 ]) {
flag = true ;
return arr[i];
}
}
if (!flag) {
return - 1 ;
}
return - 1 ;
}
public static void main(String[] args)
{
int [] arr = { 1 , 3 , 4 , 2 , 2 };
int n = arr.length;
System.out.println(findDuplicates(arr, n));
}
}
|
Python3
def find_duplicates(arr):
arr.sort()
for i in range ( len (arr) - 1 ):
if arr[i] = = arr[i + 1 ]:
return arr[i]
return - 1
arr = [ 1 , 3 , 4 , 2 , 2 ]
n = len (arr)
print (find_duplicates(arr))
|
C#
using System;
public class GFG
{
public static int FindDuplicates( int [] arr, int n)
{
Array.Sort(arr);
bool flag = false ;
for ( int i = 0; i < n - 1; i++)
{
if (arr[i] == arr[i + 1])
{
flag = true ;
return arr[i];
}
}
if (!flag)
{
return -1;
}
return -1;
}
public static void Main( string [] args)
{
int [] arr = { 1, 3, 4, 2, 2 };
int n = arr.Length;
Console.WriteLine(FindDuplicates(arr, n));
}
}
|
Javascript
function findDuplicates(arr) {
arr.sort();
for (let i = 0; i < arr.length - 1; i++) {
if (arr[i] === arr[i + 1]) {
return arr[i];
}
}
return -1;
}
let arr = [1, 3, 4, 2, 2];
let n = arr.length;
console.log(findDuplicates(arr));
|
Time Complexity: O(N*log N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above method the idea is to use the concept of Counting Sort. Since the range of elements in the array is known, hence we could use this sorting technique to improvise the time complexity.
The idea is to initialize another array(say count[]) with the same size N and initialize all the elements as 0. Then count the occurrences of each element of the array and update the count in the count[]. Print all the element whose count is greater than 1.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int findDuplicate( int arr[], int n)
{
int countArr[n + 1], i;
for (i = 0; i <= n; i++)
countArr[i] = 0;
for (i = 0; i <= n; i++)
countArr[arr[i]]++;
bool a = false ;
for (i = 1; i <= n; i++) {
if (countArr[i] > 1) {
a = true ;
cout << i << ' ' ;
}
}
if (!a)
cout << "-1" ;
}
int main()
{
int n = 4;
int arr[] = { 1, 3, 4, 2, 2 };
findDuplicate(arr, n);
return 0;
}
|
Java
import java.util.*;
class GFG {
public static int
findDuplicate( int arr[], int n)
{
int countArr[] = new int [n + 1 ], i;
for (i = 0 ; i <= n; i++)
countArr[i] = 0 ;
for (i = 0 ; i <= n; i++)
countArr[arr[i]]++;
bool a = false ;
for (i = 1 ; i <= n; i++) {
if (countArr[i] > 1 ) {
a = true ;
System.out.print(i + " " );
}
}
if (!a)
System.out.println( "-1" );
}
public static void main(String[] args)
{
int n = 4 ;
int arr[] = { 1 , 3 , 4 , 2 , 2 };
findDuplicate(arr, n);
}
}
|
Python3
def findDuplicate(arr, n):
countArr = [ 0 ] * (n + 1 )
for i in range (n + 1 ):
countArr[arr[i]] + = 1
a = False
for i in range ( 1 , n + 1 ):
if (countArr[i] > 1 ):
a = True
print (i, end = " " )
if ( not a):
print ( - 1 )
if __name__ = = '__main__' :
n = 4
arr = [ 1 , 3 , 4 , 2 , 2 ]
findDuplicate(arr, n)
|
C#
using System;
class GFG{
static void findDuplicate( int []arr, int n)
{
int []countArr = new int [n + 1];
int i;
for (i = 0; i <= n; i++)
countArr[i] = 0;
for (i = 0; i <= n; i++)
countArr[arr[i]]++;
bool a = false ;
for (i = 1; i <= n; i++)
{
if (countArr[i] > 1)
{
a = true ;
Console.Write(i + " " );
}
}
if (!a)
Console.WriteLine( "-1" );
}
public static void Main(String[] args)
{
int n = 4;
int []arr = { 1, 3, 4, 2, 2 };
findDuplicate(arr, n);
}
}
|
Javascript
<script>
function
findDuplicate(arr, n)
{
let countArr = Array.from({length: n+1}, (_, i) => 0), i;
for (i = 0; i <= n; i++)
countArr[i] = 0;
for (i = 0; i <= n; i++)
countArr[arr[i]]++;
let a = false ;
for (i = 1; i <= n; i++) {
if (countArr[i] > 1) {
a = true ;
document.write(i + " " );
}
}
if (!a)
document.write( "-1" );
}
let n = 4;
let arr = [ 1, 3, 4, 2, 2 ];
findDuplicate(arr, n);
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N)
Related articles:
Approch 3 : The given code use a hashing technique to locate duplicates in an array. To be more precise, the frequency of each element in the array is stored using an unordered map. An element is considered to be duplicated if its frequency is more than 1, in which case the method returns that element.
Algorithm Steps of given approch :
- Create an unordered map called freq to store the frequency of each element in the array.
- Iterate over the input array using a for loop.
- For each element in the array, increment its frequency in the freq map using the ++freq[arr[i]] notation.
- If the frequency of the current element is greater than 1, return the element as it is a duplicate.
- If no duplicates are found, return -1.
- End of the program.
C++
#include <bits/stdc++.h>
using namespace std;
int findDuplicates( int arr[], int n) {
unordered_map< int , int > freq;
for ( int i=0; i<n; i++) {
if (++freq[arr[i]] > 1) {
return arr[i];
}
}
return -1;
}
int main() {
int arr[] = {1, 3, 4, 2, 2};
int n = sizeof (arr) / sizeof (arr[0]);
cout << findDuplicates(arr, n);
return 0;
}
|
Java
import java.util.HashMap;
import java.util.Map;
public class Main {
public static int findDuplicates( int [] arr) {
Map<Integer, Integer> freq = new HashMap<>();
for ( int i = 0 ; i < arr.length; i++) {
if (freq.containsKey(arr[i])) {
freq.put(arr[i], freq.get(arr[i]) + 1 );
if (freq.get(arr[i]) > 1 ) {
return arr[i];
}
} else {
freq.put(arr[i], 1 );
}
}
return - 1 ;
}
public static void main(String[] args) {
int [] arr = { 1 , 3 , 4 , 2 , 2 };
System.out.println(findDuplicates(arr));
}
}
|
Python
def find_duplicates(arr):
freq = {}
for num in arr:
if num in freq:
return num
else :
freq[num] = 1
return - 1
def main():
arr = [ 1 , 3 , 4 , 2 , 2 ]
result = find_duplicates(arr)
print (result)
if __name__ = = "__main__" :
main()
|
C#
using System;
using System.Collections.Generic;
class Program
{
static int FindDuplicates( int [] arr)
{
Dictionary< int , int > freq = new Dictionary< int , int >();
foreach ( int num in arr)
{
if (freq.ContainsKey(num))
{
freq[num]++;
if (freq[num] > 1)
{
return num;
}
}
else
{
freq[num] = 1;
}
}
return -1;
}
static void Main()
{
int [] arr = { 1, 3, 4, 2, 2 };
int result = FindDuplicates(arr);
Console.WriteLine(result);
}
}
|
Javascript
function findDuplicates(arr) {
const freq = {};
for (let i = 0; i < arr.length; i++) {
if (freq.hasOwnProperty(arr[i])) {
freq[arr[i]]++;
if (freq[arr[i]] > 1) {
return arr[i];
}
} else {
freq[arr[i]] = 1;
}
}
return -1;
}
const arr = [1, 3, 4, 2, 2];
console.log(findDuplicates(arr));
|
Time Complexity : O(n), where n is the size of the input array.
Auxiliary Space : O(n).
- Find duplicates in O(n) time and O(1) extra space | Set 1
- Duplicates in an array in O(n) and by using O(1) extra space | Set-2
Last Updated :
07 Nov, 2023
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