Count distinct elements in every window of size k

Given an array of size N and an integer K, return the count of distinct numbers in all windows of size K. 

Examples:

Input: arr[] = {1, 2, 1, 3, 4, 2, 3}, K = 4
Output: 3 4 4 3
Explanation: First window is {1, 2, 1, 3}, count of distinct numbers is 3
                      Second window is {2, 1, 3, 4} count of distinct numbers is 4
                      Third window is {1, 3, 4, 2} count of distinct numbers is 4
                      Fourth window is {3, 4, 2, 3} count of distinct numbers is 3

Input: arr[] = {1, 2, 4, 4}, K = 2
Output: 2 2 1
Explanation: First window is {1, 2}, count of distinct numbers is 2
                      First window is {2, 4}, count of distinct numbers is 2
                      First window is {4, 4}, count of distinct numbers is 1

This problem has appeared in Microsoft Interview Question.

Naive Approach for finding the count of distinct numbers in all windows of size K:

Traverse the given array considering every window of size K in it and keeping a count on the distinct elements of the window

Follow the given steps to solve the problem:

• For every index, i from 0 to N – K, traverse the array from i to i + k using another loop. This is the window
• Traverse the window, from i to that index and check if the element is present or not
• If the element is not present in the prefix of the array, i.e no duplicate element is present from i to index-1, then increase the count, else ignore it
• Print the count

Below is the implementation of the above approach:

C++

// C++ program to count distinct
// elements in every window of size K

#include <bits/stdc++.h>
using namespace std;

// Counts distinct elements in window of size K
int countWindowDistinct(int win[], int K)
{
    int dist_count = 0;

    // Traverse the window
    for (int i = 0; i < K; i++) {
        // Check if element arr[i] exists in arr[0..i-1]
        int j;
        for (j = 0; j < i; j++)
            if (win[i] == win[j])
                break;
        if (j == i)
            dist_count++;
    }
    return dist_count;
}

// Counts distinct elements in all windows of size k
void countDistinct(int arr[], int N, int K)
{
    // Traverse through every window
    for (int i = 0; i <= N - K; i++)
        cout << countWindowDistinct(arr + i, K) << endl;
}

// Driver's code
int main()
{
    int arr[] = {1, 2, 1, 3, 4, 2, 3}, K = 4;
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    countDistinct(arr, N, K);
    return 0;
}

Java

// Java program to count distinct elements in every
// window of size K

import java.util.Arrays;

class Test {

    // Counts distinct elements in window of size K
    static int countWindowDistinct(int win[], int K)
    {
        int dist_count = 0;

        // Traverse the window
        for (int i = 0; i < K; i++) {
            // Check if element arr[i] exists in arr[0..i-1]
            int j;
            for (j = 0; j < i; j++)
                if (win[i] == win[j])
                    break;
            if (j == i)
                dist_count++;
        }
        return dist_count;
    }

    // Counts distinct elements in all windows of size K
    static void countDistinct(int arr[], int N, int K)
    {
        // Traverse through every window
        for (int i = 0; i <= N - K; i++)
            System.out.println(countWindowDistinct(
                Arrays.copyOfRange(arr, i, arr.length), K));
    }

    // Driver's code
    public static void main(String args[])
    {
        int arr[] = {1, 2, 1, 3, 4, 2, 3}, K = 4;

          // Function call
        countDistinct(arr, arr.length, K);
    }
}

Python3

# Python3 program to count distinct
# elements in every window of size K

import math as mt

# Counts distinct elements in window
# of size K


def countWindowDistinct(win, K):

    dist_count = 0

    # Traverse the window
    for i in range(K):

        # Check if element arr[i] exists
        # in arr[0..i-1]
        j = 0
        while j < i:
            if (win[i] == win[j]):
                break
            else:
                j += 1
        if (j == i):
            dist_count += 1

    return dist_count


# Counts distinct elements in all
# windows of size k
def countDistinct(arr, N, K):

    # Traverse through every window
    for i in range(N - K + 1):
        print(countWindowDistinct(arr[i:K + i], K))


# Driver's Code
if __name__=='__main__':
  arr = [1, 2, 1, 3, 4, 2, 3]
  K = 4
  N = len(arr)

  # Function call
  countDistinct(arr, N, K)

# This code is contributed by
# Mohit kumar 29

C#

// Simple C# program to count distinct
// elements in every window of size K
using System;
using System.Collections.Generic;

class GFG {
    // Counts distinct elements in
    // window of size K
    static int countWindowDistinct(int[] win, int K)
    {
        int dist_count = 0;

        // Traverse the window
        for (int i = 0; i < K; i++) {
            // Check if element arr[i]
            // exists in arr[0..i-1]
            int j;
            for (j = 0; j < i; j++)
                if (win[i] == win[j])
                    break;
            if (j == i)
                dist_count++;
        }
        return dist_count;
    }

    // Counts distinct elements in
    // all windows of size k
    static void countDistinct(int[] arr, int N, int K)
    {
        // Traverse through every window
        for (int i = 0; i <= N - K; i++) {
            int[] newArr = new int[K];
            Array.Copy(arr, i, newArr, 0, K);
            Console.WriteLine(
                countWindowDistinct(newArr, K));
        }
    }

    // Driver's Code
    public static void Main(String[] args)
    {
        int[] arr = {1, 2, 1, 3, 4, 2, 3};
        int K = 4;

          // Function call
        countDistinct(arr, arr.Length, K);
    }
}

// This code is contributed by Princi Singh

Javascript

// Javascript program to count distinct elements in every
// window of size k

// Counts distinct elements in window of size k
    function countWindowDistinct(win, k)
    {
        let dist_count = 0;

        // Traverse the window
        for (let i = 0; i < k; i++) {
            // Check if element arr[i] exists in arr[0..i-1]
            let j;
            for (j = 0; j < i; j++)
                if (win[i] == win[j])
                    break;
            if (j == i)
                dist_count++;
        }
        return dist_count;
    }

    // Counts distinct elements in all windows of size k
    function countDistinct(arr, N, K)
    {
        // Traverse through every window
        for (let i = 0; i <= N - K; i++)
            document.write(countWindowDistinct(arr.slice(i, arr.length), K) + "<br/>");
    }


// Driver program

      let arr = [1, 2, 1, 3, 4, 2, 3], K = 4;

        countDistinct(arr, arr.length, K);

 // This code is contributed by target_2.

Output

3
4
4
3

Time complexity: O(N * K²)
Auxiliary Space: O(1) 

Count distinct numbers in all windows of size K using hashing:

So, there is an efficient solution using hashing, though hashing requires extra O(n) space but the time complexity will improve. The trick is to use the count of the previous window while sliding the window. To do this a hash map can be used that stores elements of the current window. The hash-map is also operated on by simultaneous addition and removal of an element while keeping track of distinct elements. The problem deals with finding the count of distinct elements in a window of length k, at any step while shifting the window and discarding all the computation done in the previous step, even though k – 1 elements are same from the previous adjacent window. For example, assume that elements from index i to i + k – 1 are stored in a Hash Map as an element-frequency pair. So, while updating the Hash Map in range i + 1 to i + k, reduce the frequency of the i-th element by 1 and increase the frequency of (i + k)-th element by 1. 
Insertion and deletion from the HashMap takes constant time.

Follow the given steps to solve the problem:

• Create an empty hash map. Let the hash map be hm.
• Initialize the count of distinct elements as dist_count to 0.
• Traverse through the first window and insert elements of the first window to hm. The elements are used as key and their counts as the value in hm. Also, keep updating dist_count
• Print distinct count for the first window.
• Traverse through the remaining array (or other windows).
• Remove the first element of the previous window. 
  • If the removed element appeared only once, remove it from hm and decrease the distinct count, i.e. do “dist_count–“
  • else (appeared multiple times in hm), then decrement its count in hm
• Add the current element (last element of the new window) 
  • If the added element is not present in hm, add it to hm and increase the distinct count, i.e. do “dist_count++”
  • Else (the added element appeared multiple times), increment its count in hm

Below is the illustration of the above approach:

Consider the array arr[] = {1, 2, 1, 3, 4, 2, 3} and K = 4

Initial State

Checking in the first window

In all the remaining steps, remove the first element of the previous window and add new element of current window.

Checking in the second window

Checking in the third window

Checking the last window

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <iostream>
#include <unordered_map>
using namespace std;

void countDistinct(int arr[], int K, int N)
{
    // Creates an empty hashmap hm
    unordered_map<int, int> hm;

    // initialize distinct element count for current window
    int dist_count = 0;

    // Traverse the first window and store count
    // of every element in hash map
    for (int i = 0; i < K; i++) {
        if (hm[arr[i]] == 0) {
            dist_count++;
        }

        hm[arr[i]] += 1;
    }

    // Print count of first window
    cout << dist_count << endl;

    // Traverse through the remaining array
    for (int i = K; i < N; i++) {
        // Remove first element of previous window
        // If there was only one occurrence, then reduce distinct count.
        if (hm[arr[i - K]] == 1) {
            dist_count--;
        }
        // reduce count of the removed element
        hm[arr[i - K]] -= 1;

        // Add new element of current window
        // If this element appears first time,
        // increment distinct element count

        if (hm[arr[i]] == 0) {
            dist_count++;
        }
        hm[arr[i]] += 1;

        // Print count of current window
        cout << dist_count << endl;
    }
}

// Driver's code
int main()
{
    int arr[] = {1, 2, 1, 3, 4, 2, 3};
    int N = sizeof(arr) / sizeof(arr[0]);
    int K = 4;

      // Function call
    countDistinct(arr, K, N);

    return 0;
}
// This solution is contributed by Aditya Goel

Java

// Java program for the above approach

import java.util.HashMap;

class CountDistinctWindow {
    static void countDistinct(int arr[], int K)
    {
        // Creates an empty hashMap hM
        HashMap<Integer, Integer> hM
            = new HashMap<Integer, Integer>();

        // Traverse the first window and store count
        // of every element in hash map
        for (int i = 0; i < K; i++)
            hM.put(arr[i], hM.getOrDefault(arr[i], 0) + 1);

        // Print count of first window
        System.out.println(hM.size());

        // Traverse through the remaining array
        for (int i = K; i < arr.length; i++) {

            // Remove first element of previous window
            // If there was only one occurrence
            if (hM.get(arr[i - K]) == 1) {
                hM.remove(arr[i - K]);
            }

            else // reduce count of the removed element
                hM.put(arr[i - K], hM.get(arr[i - K]) - 1);

            // Add new element of current window
            // If this element appears first time,
            // set its count as 1,
            hM.put(arr[i], hM.getOrDefault(arr[i], 0) + 1);

            // Print count of current window
            System.out.println(hM.size());
        }
    }

    // Driver's code
    public static void main(String arg[])
    {
        int arr[] = { 1, 2, 1, 3, 4, 2, 3 };
        int K = 4;

        // Function call
        countDistinct(arr, K);
    }
}

Python3

# An efficient Python program to
# count distinct elements in
# every window of size K
from collections import defaultdict


def countDistinct(arr, K, N):

    # Creates an empty hashmap hm
    mp = defaultdict(lambda: 0)

    # initialize distinct element
    # count for current window
    dist_count = 0

    # Traverse the first window and store count
    # of every element in hash map
    for i in range(K):
        if mp[arr[i]] == 0:
            dist_count += 1
        mp[arr[i]] += 1

    # Print count of first window
    print(dist_count)

    # Traverse through the remaining array
    for i in range(K, N):

        # Remove first element of previous window
        # If there was only one occurrence,
        # then reduce distinct count.
        if mp[arr[i - K]] == 1:
            dist_count -= 1
        mp[arr[i - K]] -= 1

    # Add new element of current window
    # If this element appears first time,
    # increment distinct element count
        if mp[arr[i]] == 0:
            dist_count += 1
        mp[arr[i]] += 1

        # Print count of current window
        print(dist_count)

# Driver's code
if __name__=='__main__':
  arr = [1, 2, 1, 3, 4, 2, 3]
  N = len(arr)
  K = 4

  # Function call
  countDistinct(arr, K, N)

# This code is contributed by Shrikant13

C#

// C# program for the above approach

using System;
using System.Collections.Generic;

public class CountDistinctWindow {
    static void countDistinct(int[] arr, int K)
    {
        // Creates an empty hashMap hM
        Dictionary<int, int> hM
            = new Dictionary<int, int>();

        // initialize distinct element count for
        // current window
        int dist_count = 0;

        // Traverse the first window and store count
        // of every element in hash map
        for (int i = 0; i < K; i++) {
            if (!hM.ContainsKey(arr[i])) {
                hM.Add(arr[i], 1);
                dist_count++;
            }
            else {
                int count = hM[arr[i]];
                hM.Remove(arr[i]);
                hM.Add(arr[i], count + 1);
            }
        }

        // Print count of first window
        Console.WriteLine(dist_count);

        // Traverse through the remaining array
        for (int i = K; i < arr.Length; i++) {

            // Remove first element of previous window
            // If there was only one occurrence, then
            // reduce distinct count.
            if (hM[arr[i - K]] == 1) {
                hM.Remove(arr[i - K]);
                dist_count--;
            }
            else // reduce count of the removed element
            {
                int count = hM[arr[i - K]];
                hM.Remove(arr[i - K]);
                hM.Add(arr[i - K], count - 1);
            }

            // Add new element of current window
            // If this element appears first time,
            // increment distinct element count
            if (!hM.ContainsKey(arr[i])) {
                hM.Add(arr[i], 1);
                dist_count++;
            }
            else // Increment distinct element count
            {
                int count = hM[arr[i]];
                hM.Remove(arr[i]);
                hM.Add(arr[i], count + 1);
            }

            // Print count of current window
            Console.WriteLine(dist_count);
        }
    }

    // Driver's code
    public static void Main(String[] arg)
    {
        int[] arr = {1, 2, 1, 3, 4, 2, 3};
        int K = 4;

          // Function call
        countDistinct(arr, K);
    }
}

// This code contributed by Rajput-Ji

Javascript

// Javascript program to count distinct elements in
// every window of size k

   function countDistinct(arr, k)
    {
        // Creates an empty hashMap hM
        let hM = new Map();

        // Traverse the first window and store count
        // of every element in hash map
        for (let i = 0; i < k; i++)
        {
            if(hM.has(arr[i]))
            hM.set(arr[i], hM.get(arr[i])+1)
            else
            hM.set(arr[i], 1);
        }

        // Print count of first window
        document.write(hM.size + "<br/>");

        // Traverse through the remaining array
        for (let i = k; i < arr.length; i++) {

            // Remove first element of previous window
            // If there was only one occurrence
            if (hM.get(arr[i - k]) == 1) {
                hM.delete(arr[i - k]);
            }

            else // reduce count of the removed element
                hM.set(arr[i - k],  hM.get(arr[i - k]) - 1);

            // Add new element of current window
            // If this element appears first time,
            // set its count as 1,
            if(hM.has(arr[i]))
            hM.set(arr[i], hM.get(arr[i])+1)
            else
            hM.set(arr[i], 1);

            // Print count of current window
            document.write(hM.size + "<br/>");
        }
    }


// Driver code
    let arr = [1, 2, 1, 3, 4, 2, 3];
    let size = arr.length;
    let k = 4;
    countDistinct(arr, k, size);

    // This code is contributed by splevel62.

Output

3
4
4
3

Time complexity: O(N), A single traversal of the array is required.
Auxiliary Space: O(N), Since the hashmap requires linear space.

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