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

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

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 Θ(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.

This article is contributed by Anuj. Please write comments if you find anything incorrect, or if you want to share more information about the topic discussed above.