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Check if a given array contains duplicate elements within k distance from each other

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Given an unsorted array that may contain duplicates. Also given a number k which is smaller than the size of the array. Write a function that returns true if the array contains duplicates within k distance.
Examples: 

Input: k = 3, arr[] = {1, 2, 3, 4, 1, 2, 3, 4}
Output: false
All duplicates are more than k distance away.

Input: k = 3, arr[] = {1, 2, 3, 1, 4, 5}
Output: true
1 is repeated at distance 3.

Input: k = 3, arr[] = {1, 2, 3, 4, 5}
Output: false

Input: k = 3, arr[] = {1, 2, 3, 4, 4}
Output: true
Recommended Practice

A Simple Solution is to run two loops. The outer loop picks every element ‘arr[i]’ as a starting element, and the inner loop compares all elements which are within k distance of ‘arr[i]’. The time complexity of this solution is O(k * n). 

Implementation:

C++

// C++ program to Check if a given array contains duplicate
// elements within k distance from each other
#include <bits/stdc++.h>
using namespace std;
 
bool checkDuplicatesWithinK(int arr[], int n, int k)
{
    // traversing the input array
    for (int i = 0; i < n; i++) {
        int j = i + 1;
        int range = k;
        // searching in next k-1 elements if its duplicate
        // is present or not
        while (range > 0 and j < n) {
            if (arr[i] == arr[j])
                return true;
            j++;
            range--;
        }
    }
    return false;
}
 
// Driver method to test above method
int main()
{
    int arr[] = { 10, 5, 3, 4, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (checkDuplicatesWithinK(arr, n, 3))
        cout << "Yes";
    else
        cout << "No";
}
 
// This article is contributed by Arpit Jain

                    

C

// C program to Check if a given array contains duplicate
// elements within k distance from each other
 
#include <stdbool.h>
#include <stdio.h>
 
bool checkDuplicatesWithinK(int arr[], int n, int k)
{
    // traversing the input array
    for (int i = 0; i < n; i++) {
        int j = i + 1;
        int range = k;
        // searching in next k-1 elements if its duplicate
        // is present or not
        while (range > 0 && j < n) {
            if (arr[i] == arr[j])
                return true;
            j++;
            range--;
        }
    }
    return false;
}
 
// Driver method to test above method
int main()
{
    int arr[] = { 10, 5, 3, 4, 3, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    if (checkDuplicatesWithinK(arr, n, 3))
        printf("Yes");
    else
        printf("No");
}
 
// This code is contributed by Aditya Kumar (adityakumar129)

                    

Java

public class GFG
{
   
  /* Java program to Check if a given array contains
    duplicate elements within k distance from each other */
  public static boolean
    checkDuplicatesWithinK(int[] arr, int n, int k)
  {
 
    // traversing the input array
    for (int i = 0; i < n; i++) {
      int j = i + 1;
      int range = k;
      // searching in next k-1 elements if its
      // duplicate is present or not
      while (range > 0 && j < n) {
        if (arr[i] == arr[j]) {
          return true;
        }
        j++;
        range--;
      }
    }
    return false;
  }
 
  // Driver method to test above method
  public static void main(String[] args)
  {
    int[] arr = { 10, 5, 3, 4, 3, 5, 6 };
    int n = arr.length;
    if (checkDuplicatesWithinK(arr, n, 3)) {
      System.out.print("Yes");
    }
    else {
      System.out.print("No");
    }
  }
}
 
// This article is contributed by Aarti_Rathi

                    

Python3

# Python3 program to Check if a given array contains duplicate
# elements within k distance from each other
def checkDuplicatesWithinK(arr,  n,  k):
 
    # traversing the input array
    for i in range(n):
        j = i + 1
        range_ = k
         
        # searching in next k-1 elements if its duplicate
        # is present or not
        while (range_ > 0 and j < n):
            if (arr[i] == arr[j]):
                return True
            j += 1
            range_ -= 1
 
    return False
 
 
# Driver method to test above method
 
arr = [10, 5, 3, 4, 3, 5, 6]
n = len(arr)
if (checkDuplicatesWithinK(arr, n, 3) == True):
    print("Yes")
else:
    print("No")
 
 
# This article is contributed by Abhijeet Kumar(abhijeet19403)

                    

C#

/* C# program to Check if a given
array contains duplicate elements
within k distance from each other */
using System;
using System.Collections.Generic;
 
class GFG {
    static bool checkDuplicatesWithinK(int[] arr, int n,
                                       int k)
    {
        // traversing the input array
        for (int i = 0; i < n; i++) {
            int j = i + 1;
            int range = k;
            // searching in next k-1 elements if its
            // duplicate is present or not
            while (range > 0 && j < n) {
                if (arr[i] == arr[j])
                    return true;
                j++;
                range--;
            }
        }
        return false;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 10, 5, 3, 4, 3, 5, 6 };
        int n = arr.Length;
        if (checkDuplicatesWithinK(arr, n, 3))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code has been contributed by Aarti_Rathi

                    

Javascript

class GFG
{
    // javascript program to Check if a given array contains
    //     duplicate elements within k distance from each other
    static checkDuplicatesWithinK(arr, n, k)
    {
        // traversing the input array
        for (var i=0; i < n; i++)
        {
            var j = i + 1;
            var range = k;
            // searching in next k-1 elements if its
            // duplicate is present or not
            while (range > 0 && j < n)
            {
                if (arr[i] == arr[j])
                {
                    return true;
                }
                j++;
                range--;
            }
        }
        return false;
    }
     
    // Driver method to test above method
    static main(args)
    {
        var arr = [10, 5, 3, 4, 3, 5, 6];
        var n = arr.length;
        if (GFG.checkDuplicatesWithinK(arr, n, 3))
        {
            console.log("Yes");
        }
        else
        {
            console.log("No");
        }
    }
}
GFG.main([]);
 
// This code is contributed by aadityaburujwale.

                    

Output
Yes

Time Complexity: O(N*K).
Auxiliary Space: O(1).

Another Solution using hashing:-
We can solve this problem in ?(n) time using Hashing. The idea is to add elements to the hash. We also remove elements that are at more than k distance from the current element. Following is a detailed algorithm.

  1. Create an empty hashtable. 
  2. Traverse all elements from left to right. Let the current element be ‘arr[i]’ 
    • If the current element ‘arr[i]’ is present in a hashtable, then return true. 
    • Else add arr[i] to hash and remove arr[i-k] from hash if i is greater than or equal to k

Implementation:

C++

#include<bits/stdc++.h>
using namespace std;
 
/* C++ program to Check if a given array contains duplicate
   elements within k distance from each other */
bool checkDuplicatesWithinK(int arr[], int n, int k)
{
    // Creates an empty hashset
    unordered_set<int> myset;
 
    // Traverse the input array
    for (int i = 0; i < n; i++)
    {
        // If already present n hash, then we found
        // a duplicate within k distance
        if (myset.find(arr[i]) != myset.end())
            return true;
 
        // Add this item to hashset
        myset.insert(arr[i]);
 
        // Remove the k+1 distant item
        if (i >= k)
            myset.erase(arr[i-k]);
    }
    return false;
}
 
// Driver method to test above method
int main ()
{
    int arr[] = {10, 5, 3, 4, 3, 5, 6};
    int n = sizeof(arr) / sizeof(arr[0]);
    if (checkDuplicatesWithinK(arr, n, 3))
        cout << "Yes";
    else
        cout << "No";
}
 
//This article is contributed by Chhavi

                    

Java

/* Java program to Check if a given array contains duplicate
   elements within k distance from each other */
import java.util.*;
 
class Main
{
    static boolean checkDuplicatesWithinK(int arr[], int k)
    {
        // Creates an empty hashset
        HashSet<Integer> set = new HashSet<>();
 
        // Traverse the input array
        for (int i=0; i<arr.length; i++)
        {
            // If already present n hash, then we found
            // a duplicate within k distance
            if (set.contains(arr[i]))
               return true;
 
            // Add this item to hashset
            set.add(arr[i]);
 
            // Remove the k+1 distant item
            if (i >= k)
              set.remove(arr[i-k]);
        }
        return false;
    }
 
    // Driver method to test above method
    public static void main (String[] args)
    {
        int arr[] = {10, 5, 3, 4, 3, 5, 6};
        if (checkDuplicatesWithinK(arr, 3))
           System.out.println("Yes");
        else
           System.out.println("No");
    }
}

                    

Python 3

# Python 3 program to Check if a given array
# contains duplicate elements within k distance
# from each other
def checkDuplicatesWithinK(arr, n, k):
 
    # Creates an empty list
    myset = []
 
    # Traverse the input array
    for i in range(n):
     
        # If already present n hash, then we
        # found a duplicate within k distance
        if arr[i] in myset:
            return True
 
        # Add this item to hashset
        myset.append(arr[i])
 
        # Remove the k+1 distant item
        if (i >= k):
            myset.remove(arr[i - k])
    return False
 
# Driver Code
if __name__ == "__main__":
     
    arr = [10, 5, 3, 4, 3, 5, 6]
    n = len(arr)
    if (checkDuplicatesWithinK(arr, n, 3)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by ita_c

                    

C#

/* C# program to Check if a given
array contains duplicate elements
within k distance from each other */
using System;
using System.Collections.Generic;
 
class GFG
{
    static bool checkDuplicatesWithinK(int []arr, int k)
    {
        // Creates an empty hashset
        HashSet<int> set = new HashSet<int>();
 
        // Traverse the input array
        for (int i = 0; i < arr.Length; i++)
        {
            // If already present n hash, then we found
            // a duplicate within k distance
            if (set.Contains(arr[i]))
                return true;
 
            // Add this item to hashset
            set.Add(arr[i]);
 
            // Remove the k+1 distant item
            if (i >= k)
                set.Remove(arr[i - k]);
        }
        return false;
    }
 
    // Driver code
    public static void Main (String[] args)
    {
        int []arr = {10, 5, 3, 4, 3, 5, 6};
        if (checkDuplicatesWithinK(arr, 3))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code has been contributed
// by 29AjayKumar

                    

Javascript

<script>
 
/* Javascript program to Check if a given array contains duplicate
   elements within k distance from each other */
    
     
    function checkDuplicatesWithinK(arr, n, k)
    {
        // Creates an empty hashset
        let myset = [];
         
        // Traverse the input array
        for(let i=0;i<n;i++)
        {
            // If already present n hash, then we found
            // a duplicate within k distance
            if(arr.includes(arr[i]))
            {
                return true;
            }
            // Add this item to hashset
            myset.add(arr[i]);
             
            // Remove the k+1 distant item
            if (i >= k)
            {
                index = array.indexOf(arr[i - k]);
                array.splice(index, 1);
            }
        }
        return false;
    }
    // Driver method to test above method
    let arr = [10, 5, 3, 4, 3, 5, 6];
    let n= arr.length;
    if (checkDuplicatesWithinK(arr, n, 3))
    {
        document.write("Yes");
    }
    else
    {
        document.write("No");
    }
     
     
    // This code is contributed by rag2127
     
</script>

                    

Output
Yes

Time Complexity: O(N).
Auxiliary Space: O(N) for using an unordered set.



Last Updated : 16 Dec, 2022
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