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 3Input: 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 Kint 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 kvoid 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 codeint 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 Kimport 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 Kimport math as mt# Counts distinct elements in window# of size Kdef 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 kdef countDistinct(arr, N, K): # Traverse through every window for i in range(N - K + 1): print(countWindowDistinct(arr[i:K + i], K))# Driver's Codeif __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 Kusing 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 codeint 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 approachimport 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 Kfrom collections import defaultdictdef 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 codeif __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 approachusing 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.
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