Find duplicates in constant array with elements 0 to N-1 in O(1) space
Last Updated :
28 Feb, 2023
Given a constant array of n elements which contains elements from 1 to n-1, with any of these numbers appearing any number of times. Find any one of these repeating numbers in O(n) and using only constant memory space.
Examples:
Input : arr[] = {1, 2, 3, 4, 5, 6, 3}
Output : 3
As the given array is constant methods given in below articles will not work.
- 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
- We are taking two variables i & j starting from 0
- We will run loop until i reached last elem or found repeated elem
- We will pre-increment the j value so that we can compare elem with next elem
- If we don't find elem, we will increase i as j will be pointing last elem and then reposition j with
Implementation:
C++
#include <iostream>
using namespace std;
// function to find one duplicate
int findduplicate(int a[], int n)
{
int i = 0, j = 0;
while (i < n) {
if (a[i] == a[++j])
return a[j];
if (j == n - 1) {
i++;
j = i;
}
}
return -1;
}
int main()
{
int arr[] = { 1, 2, 4, 3, 4, 5, 6, 3 };
int n = sizeof(arr) / sizeof(arr[0]);
cout << findduplicate(arr, n) << endl;
return 0;
}
// This code is contributed by akashish__
// Java program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
import java.io.*;
import java.util.*;
public class GFG {
// function to find one duplicate
static int findduplicate(int []a, int n)
{
int i=0,j=0;
while(i<n){
if(a[i]==a[++j]) return a[j];
if(j==n-1) {
i++;
j=i;
}
}
return -1;
}
public static void main(String args[])
{
int []arr = {1, 2, 4, 3, 4, 5, 6, 3};
int n = arr.length;
System.out.print(findduplicate(arr, n));
}
}
// This code is contributed by
// Love Raj
// Java program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
import java.io.*;
import java.util.*;
public class GFG {
// function to find one duplicate
static int findduplicate(int []arr, int n)
{
// return -1 because in these cases
// there can not be any repeated element
if (n <= 1)
return -1;
// initialize fast and slow
int slow = arr[0];
int fast = arr[arr[0]];
// loop to enter in the cycle
while (fast != slow)
{
// move one step for slow
slow = arr[slow];
// move two step for fast
fast = arr[arr[fast]];
}
// loop to find entry
// point of the cycle
fast = 0;
while (slow != fast)
{
slow = arr[slow];
fast = arr[fast];
}
return slow;
}
// Driver Code
public static void main(String args[])
{
int []arr = {1, 2, 3, 4, 5, 6, 3};
int n = arr.length;
System.out.print(findduplicate(arr, n));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)
# Python 3 program to find a duplicate
# element in an array with values in
# range from 0 to n-1.
# function to find one duplicate
def findduplicate(arr, n):
# return -1 because in these cases
# there can not be any repeated element
if (n <= 1):
return -1
# initialize fast and slow
slow = arr[0]
fast = arr[arr[0]]
# loop to enter in the cycle
while (fast != slow) :
# move one step for slow
slow = arr[slow]
# move two step for fast
fast = arr[arr[fast]]
# loop to find entry point of the cycle
fast = 0
while (slow != fast):
slow = arr[slow]
fast = arr[fast]
return slow
# Driver Code
if __name__ == "__main__":
arr = [1, 2, 3, 4, 5, 6, 3 ]
n = len(arr)
print(findduplicate(arr, n))
# This code is contributed by ita_c
// C# program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
using System;
using System.Collections.Generic;
class GFG {
// function to find one duplicate
static int findduplicate(int []arr, int n)
{
// return -1 because in these cases
// there can not be any repeated element
if (n <= 1)
return -1;
// initialize fast and slow
int slow = arr[0];
int fast = arr[arr[0]];
// loop to enter in the cycle
while (fast != slow)
{
// move one step for slow
slow = arr[slow];
// move two step for fast
fast = arr[arr[fast]];
}
// loop to find entry
// point of the cycle
fast = 0;
while (slow != fast)
{
slow = arr[slow];
fast = arr[fast];
}
return slow;
}
// Driver Code
public static void Main()
{
int []arr = {1, 2, 3, 4, 5, 6, 3};
int n = arr.Length;
Console.Write(findduplicate(arr, n));
}
}
// This code is contributed by
// Manish Shaw (manishshaw1)
<?php
// PHP program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
// function to find one duplicate
function findduplicate($arr, $n)
{
// return -1 because in these cases
// there can not be any repeated element
if ($n <= 1)
return -1;
// initialize fast and slow
$slow = $arr[0];
$fast = $arr[$arr[0]];
// loop to enter in the cycle
while ($fast != $slow)
{
// move one step for slow
$slow = $arr[$slow];
// move two step for fast
$fast = $arr[$arr[$fast]];
}
// loop to find entry point of the cycle
$fast = 0;
while ($slow != $fast)
{
$slow = $arr[$slow];
$fast = $arr[$fast];
}
return $slow;
}
// Driver Code
$arr = array( 1, 2, 3, 4, 5, 6, 3 );
$n = sizeof($arr);
echo findduplicate($arr, $n);
// This code is contributed by Tushil
?>
<script>
// Javascript program to find a duplicate
// element in an array with values in
// range from 0 to n-1.
// function to find one duplicate
function findduplicate(arr, n)
{
// return -1 because in these cases
// there can not be any repeated element
if (n <= 1)
return -1;
// initialize fast and slow
let slow = arr[0];
let fast = arr[arr[0]];
// loop to enter in the cycle
while (fast != slow)
{
// move one step for slow
slow = arr[slow];
// move two step for fast
fast = arr[arr[fast]];
}
// loop to find entry
// point of the cycle
fast = 0;
while (slow != fast)
{
slow = arr[slow];
fast = arr[fast];
}
return slow;
}
let arr = [1, 2, 3, 4, 5, 6, 3];
let n = arr.length;
document.write(findduplicate(arr, n));
</script>
// function to find one duplicate
function findduplicate( a, n)
{
let i = 0, j = 0;
while (i < n) {
if (a[i] == a[++j])
return a[j];
if (j == n - 1) {
i++;
j = i;
}
}
return -1;
}
let arr = [ 1, 2, 4, 3, 4, 5, 6, 3 ];
let n = arr.length;
console.log(findduplicate(arr, n));
// This code is contributed by akashish__
Complexity Analysis:
- Time Complexity: O(n*n)
- Auxiliary Space: O(1)
Efficient Approach:
We will use the concept that all elements here are between 1 and n-1.
So we will perform these steps to find the Duplicate element
- Consider a pointer 'p' which is currently at index 0.
- Run a while loop until the pointer p reaches the value n.
- if the value of a[p] is -1 then increment the pointer by 1 and skip the iteration
- Else,go to the position of the element to which the current pointer is pointing i.e. at index a[p].
- Now if the value at index a[p] i.e. a[a[p]] is -1 then break the loop as the element a[p] is the duplicate one.
- Otherwise store the value of a[a[p]] in a[p] i.e. a[p]=a[a[p]] and put -1 in a[a[p]] i.e. a[a[p]]=-1.
Code:
C++
#include <iostream>
using namespace std;
void find_duplicate(int a[], int n)
{
int p = 0;
while (p != n) {
if (a[p] == -1) {
p++;
}
else {
if (a[a[p] - 1] == -1) {
cout << a[p] << endl;
break;
}
else {
a[p] = a[a[p] - 1];
a[a[p] - 1] = -1;
}
}
}
}
int main()
{
int a[] = { 1, 2, 4, 3, 4, 5, 6, 3 };
int n = sizeof(a) / sizeof(a[0]);
find_duplicate(a, n);
return 0;
}
// This Code is Contributed by Abhishek Purohit
// Java Program for the above approach
import java.io.*;
class GFG {
static void find_duplicate(int a[], int n)
{
int p = 0;
while (p != n) {
if (a[p] == -1) {
p++;
}
else {
if (a[a[p] - 1] == -1) {
System.out.println(a[p]);
break;
}
else {
a[p] = a[a[p] - 1];
a[a[p] - 1] = -1;
}
}
}
}
public static void main(String[] args)
{
int[] a = { 1, 2, 4, 3, 4, 5, 6, 3 };
int n = a.length;
find_duplicate(a, n);
}
}
// This Code is Contributed by lokesh (lokeshmvs21).
class GFG :
@staticmethod
def find_duplicate( a, n) :
p = 0
while (p != n) :
if (a[p] == -1) :
p += 1
else :
if (a[a[p] - 1] == -1) :
print(a[p])
break
else :
a[p] = a[a[p] - 1]
a[a[p] - 1] = -1
@staticmethod
def main( args) :
a = [1, 2, 4, 3, 4, 5, 6, 3]
n = len(a)
GFG.find_duplicate(a, n)
if __name__=="__main__":
GFG.main([])
# This code is contributed by aadityaburujwale.
// Include namespace system
using System;
public class GFG
{
public static void find_duplicate(int[] a, int n)
{
var p = 0;
while (p != n)
{
if (a[p] == -1)
{
p++;
}
else
{
if (a[a[p] - 1] == -1)
{
Console.WriteLine(a[p]);
break;
}
else
{
a[p] = a[a[p] - 1];
a[a[p] - 1] = -1;
}
}
}
}
public static void Main(String[] args)
{
int[] a = {1, 2, 4, 3, 4, 5, 6, 3};
var n = a.Length;
GFG.find_duplicate(a, n);
}
}
// This code is contributed by aadityaburujwale.
function find_duplicate( a, n)
{
let p = 0;
while (p != n) {
if (a[p] == -1) {
p++;
}
else {
if (a[a[p] - 1] == -1) {
console.log(a[p]);
break;
}
else {
a[p] = a[a[p] - 1];
a[a[p] - 1] = -1;
}
}
}
}
let a = [1, 2, 4, 3, 4, 5, 6, 3 ];
let n = a.length;
find_duplicate(a, n);
// This Code is Contributed by akashish__
Complexity Analysis:
- Time Complexity: O(n)
- Auxiliary Space: O(1)
Another Approach:- Use Hashing to find duplicate element.
Implementation:-
C++
#include <bits/stdc++.h>
using namespace std;
//function to find duplicate
int find_duplicate(int a[], int n)
{
//map to store frequency
unordered_map<int,int> mm;
//iterating over array
for(int i=0;i<n;i++){
//storing frequency
mm[a[i]]++;
//checking for duplicacy
if(mm[a[i]]>1)return a[i];
}
}
int main()
{
int a[] = { 1, 2, 4, 3, 4, 5, 6, 3 };
int n = sizeof(a) / sizeof(a[0]);
cout<<find_duplicate(a, n);
return 0;
}
// This Code is Contributed by shubhamrajput6156
// Java code for the above approach:
import java.io.*;
import java.util.*;
class GFG {
// function to find duplicate
static int find_duplicate(int[] a, int n)
{
// map to store frequency
Map<Integer, Integer> mm = new HashMap<>();
// iterating over array
for (int i = 0; i < n; i++) {
// storing frequency
if (mm.containsKey(a[i])) {
int val = mm.get(a[i]);
mm.put(a[i], val + 1);
}
else {
mm.put(a[i], 1);
}
// checking for duplicacy
if (mm.get(a[i]) > 1)
return a[i];
}
return -1;
}
public static void main(String[] args)
{
int[] a = { 1, 2, 4, 3, 4, 5, 6, 3 };
int n = a.length;
System.out.print(find_duplicate(a, n));
}
}
// This code is contributed by sankar.
# function to find duplicate
def find_duplicate(a, n):
# Dictionary to store frequency
mm = {}
# Iterating over array
for i in range(n):
# Storing frequency
if a[i] in mm:
mm[a[i]] += 1
else:
mm[a[i]] = 1
# Checking for duplicacy
if mm[a[i]] > 1:
return a[i]
# driver code
a = [1, 2, 4, 3, 4, 5, 6, 3]
n = len(a)
print(find_duplicate(a, n))
# This code is contributed by redmoonz.
using System;
using System.Collections.Generic;
class GFG
{
// function to find duplicate
static int find_duplicate(int[] a, int n)
{
// map to store frequency
Dictionary<int,int> mm = new Dictionary<int, int>();
// iterating over array
for(int i = 0; i < n; i++)
{
// storing frequency
if (mm.ContainsKey(a[i]))
{
var val = mm[a[i]];
mm.Remove(a[i]);
mm.Add(a[i], val + 1);
}
else
{
mm.Add(a[i], 1);
}
// checking for duplicacy
if(mm[a[i]] > 1)
return a[i];
}
return -1;
}
static void Main(string[] args)
{
int[] a = { 1, 2, 4, 3, 4, 5, 6, 3 };
int n = a.Length;
Console.Write(find_duplicate(a, n));
}
}
function findDuplicate(a) {
// Create an empty object to store frequency
let frequency = {};
// Iterate over the array
for (let i = 0; i < a.length; i++)
{
// If the element is present in the frequency object, increment the frequency
if (a[i] in frequency)
{
frequency[a[i]]++;
}
else
{
// If not, set the frequency to 1
frequency[a[i]] = 1;
}
// Check if the frequency of the current element is greater than 1
if (frequency[a[i]] > 1)
{
// If yes, return the element
return a[i];
}
}
}
// Driver code
let arr = [1, 2, 4, 3, 4, 5, 6, 3];
console.log(findDuplicate(arr));
// This code is contributed by lokeshpotta20.
Output:- 4
Time Complexity:- O(N)
Space Complexity:- O(N)
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