Given an array of N integers and a number M. The task is to find out the maximum number of unique integers among all possible contiguous subarrays of size M.
Examples:
Input : arr[] = {5, 3, 5, 2, 3, 2}, M = 3
Output : 3
Explanation:
In the sample test case, there are 4 subarrays of size 3.
s1 = (5, 3, 5)- Has 2 unique numbers.
s2 = (3, 5, 2)- Has 3 unique numbers.
s3 = (5, 2, 3)- Has 3 unique numbers.
s4 = (2, 3, 2)- Has 2 unique numbers.
In these subarrays, there are 2, 3, 3, 2 unique numbers, respectively.
The maximum amount of unique numbers among all possible contiguous subarrays is 3.Input : arr[] = {5, 5, 5, 5, 5, 5}, M = 3
Output : 1
Naive Approach:
- Generate all subarrays of size M.
- Count unique number for each subarray.
- Check whether it is greater than the previous maximum unique number or not, if yes, replace it with the previous maximum unique number.
- Continue until we generate all possible subarrays.
Below is the implementation of the above approach:
C++
// A C++ programme to find maximum distinct elements// in a subarray of size k#include<bits/stdc++.h>using namespace std;//Function to find maximum unique element in //a subarray of size kint maxUniqueNum(int a[],int N,int M){ int maxUnique=0; //search every subarray of size k //and find how many unique element present for(int i=0;i<=N-M;i++) { //create an empty set to store the unique elements set<int> s; for(int j=0;j<M;j++) { //insert all elements //duplicate elements are not stored in set s.insert(a[i+j]); } //update the maxUnique if(s.size()>maxUnique) { maxUnique=s.size(); } } return maxUnique;} int main(){ int arr[] = {5, 3, 5, 2, 3, 2}; int M=3,N=sizeof(arr)/sizeof(arr[0]); cout<<maxUniqueNum(arr,N,M)<<endl; } |
Java
// Java Program to find maximum number of// Unique integers in Sub-Array// of given sizeimport java.util.*;class GFG { // Function to find maximum number of // Unique integers in Sub-Array // of given size public static int maxUniqueNum(int arr[], int N, int M) { int maxUnique = 0; // Generate all subarrays of size M for (int i = 0; i <= N - M; i++) { int currentUnique = 0; HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for (int k = i; k < i + M; k++) { // if the key is new to the map, // push the key in map and increment // count for unique number if (!map.containsKey(arr[k])) { map.put(arr[i], 1); currentUnique++; } } if (currentUnique > maxUnique) maxUnique = currentUnique; } return maxUnique; } // Driver Code public static void main(String[] args) { int[] arr = { 5, 3, 5, 2, 3, 2 }; int N = 6; int M = 3; System.out.println(maxUniqueNum(arr, N, M)); }} |
Python3
# A python3 programme to find maximum # distinct elements in a subarray of size k# Function to find maximum unique # element in a subarray of size kdef maxUniqueNum(a, N, M): maxUnique = 0 # search every subarray of size k and # find how many unique element present for i in range(N - M + 1): # create an empty set to store # the unique elements s = set() for j in range(M): # insert all elements # duplicate elements are not # stored in set s.add(a[i + j]) # update the maxUnique if(len(s) > maxUnique): maxUnique = len(s) return maxUnique# Driver Code if __name__ == '__main__': arr = [5, 3, 5, 2, 3, 2] M = 3 N = len(arr) print(maxUniqueNum(arr, N, M))# This code is contributed by# Sanjit_Prasad |
C#
// C# Program to find maximum number of// Unique integers in Sub-Array// of given sizeusing System;using System.Collections.Generic;class GFG { // Function to find maximum number of // Unique integers in Sub-Array // of given size public static int maxUniqueNum(int []arr, int N, int M) { int maxUnique = 0; // Generate all subarrays of size M for (int i = 0; i <= N - M; i++) { int currentUnique = 0; Dictionary<int,int> map = new Dictionary<int,int>(); for (int k = i; k < i + M; k++) { // if the key is new to the map, // push the key in map and increment // count for unique number if (!map.ContainsKey(arr[k])) { map.Remove(arr[i]); map.Add(arr[i], 1); currentUnique++; continue; } } if (currentUnique > maxUnique) maxUnique = currentUnique; } return maxUnique; } // Driver Code public static void Main(String[] args) { int[] arr = { 5, 3, 5, 2, 3, 2 }; int N = 6; int M = 3; Console.WriteLine(maxUniqueNum(arr, N, M)); }}// This code has been contributed by 29AjayKumar |
Javascript
<script>// JavaScript Program to find maximum number of// Unique integers in Sub-Array// of given size// Function to find maximum number of // Unique integers in Sub-Array // of given sizefunction maxUniqueNum(arr,N,M){ let maxUnique = 0; // Generate all subarrays of size M for (let i = 0; i <= N - M; i++) { let currentUnique = 0; let map = new Map(); for (let k = i; k < i + M; k++) { // if the key is new to the map, // push the key in map and increment // count for unique number if (!map.has(arr[k])) { map.set(arr[i], 1); currentUnique++; } } if (currentUnique > maxUnique) maxUnique = currentUnique; } return maxUnique;}// Driver Codelet arr=[5, 3, 5, 2, 3, 2 ];let N = 6;let M = 3;document.write(maxUniqueNum(arr, N, M)); // This code is contributed by unknown2108</script> |
Output:
3
Time Complexity : O(M * N)
Auxiliary Space : O(M)
Efficient Solution An efficient solution is to use window sliding technique. We maintain a single hash table for storing unique elements of every window.
1) Store counts of first M elements in a hash map.
2) Traverse from (M+1)-th element and for every element, add it to hash map and remove first element of previous window.
Below is the implementation of the above approach:
C++
// An efficient Approach to count distinct elements in// every window of size k#include<bits/stdc++.h>using namespace std;//Function to find maximum unique element in//a subarray of size kint max_U_element(int a[],int N,int M){ //map to store the unique elements and their size map<int,int> hash; //Number of unique elements in an window int dist_count=0; int res=0; //Maximum unique element in a window //store all elements till size k i.e. //storing first window for(int i=0;i<M;i++) { //found an unique element if(hash.find(a[i])==hash.end()) { hash.insert(make_pair(a[i],1)); dist_count++; } //an Duplicate element inserting else { //Update the size of that element hash[a[i]]++; } } res=dist_count; //Traverse till the end of array for(int i=M;i<N;i++) { //Remove first element from map if(hash[a[i-M]]==1) { //when element present only one time // in window so delete this hash.erase(a[i-M]); dist_count--; } else { //when multiple time element has occurred // in window so decrease size by one hash[a[i-M]]--; } //Add new element to map //If element is unique to map //increment count if(hash.find(a[i])==hash.end()) { hash.insert(make_pair(a[i],1)); dist_count++; } //Duplicate element found //update the size of that element else { hash[a[i]]++; } //Update the res res=max(res,dist_count); } return res;}//Driver codeint main(){ int arr[] = {1, 2, 1, 3, 4, 2, 3}; int M=4,N=sizeof(arr)/sizeof(arr[0]); cout<<max_U_element(arr,N,M)<<endl;} |
Java
// An efficient Java program to count distinct elements in// every window of size kimport java.util.HashMap;class maxUniqueNumWindow { static int maxUniqueNum(int arr[], int M) { // Creates an empty hashMap hM HashMap<Integer, Integer> hM = new HashMap<Integer, Integer>(); // 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 < M; i++) { if (hM.get(arr[i]) == null) { hM.put(arr[i], 1); dist_count++; } else { int count = hM.get(arr[i]); hM.put(arr[i], count + 1); } } int res = dist_count; // Traverse through the remaining array for (int i = M; i < arr.length; i++) { // Remove first element of previous window // If there was only one occurrence, then // reduce distinct count. if (hM.get(arr[i - M]) == 1) { hM.remove(arr[i - M]); dist_count--; } else // reduce count of the removed element { int count = hM.get(arr[i - M]); hM.put(arr[i - M], count - 1); } // Add new element of current window // If this element appears first time, // increment distinct element count if (hM.get(arr[i]) == null) { hM.put(arr[i], 1); dist_count++; } else // Increment distinct element count { int count = hM.get(arr[i]); hM.put(arr[i], count + 1); } res = Math.max(res, dist_count); } return res; } // Driver method public static void main(String arg[]) { int arr[] = { 1, 2, 1, 3, 4, 2, 3 }; int M = 4; System.out.println(maxUniqueNum(arr, M)); }} |
Python3
# An efficient Approach to count distinct elements in# every window of size k# Function to find maximum unique element in# a subarray of size kdef max_U_element(a, N, M): # map to store the unique elements and their size hsh = dict() # Number of unique elements in an window dist_count = 0 res = 0 # Maximum unique element in a window # store all elements till size k i.e. # storing first window for i in range(M): # found an unique element if(arr[i] not in hsh.keys()): hsh[a[i]] = 1 dist_count += 1 # an Duplicate element inserting else: # Update the size of that element hsh[a[i]] += 1 res = dist_count # Traverse till the end of array for i in range(M, N): # Remove first element from map if(a[i - M] in hsh.keys() and hsh[a[i - M]] == 1): # when element present only one time # in window so delete this del hsh[a[i-M]] dist_count -= 1 else: # when multiple time element has occurred # in window so decrease size by one hsh[a[i - M]] -= 1 # Add new element to map # If element is unique to map # increment count if(a[i] not in hsh.keys()): hsh[a[i]] = 1 dist_count += 1 # Duplicate element found # update the size of that element else: hsh[a[i]] += 1 # Update the res res = max(res, dist_count) return res# Driver codearr = [1, 2, 1, 3, 4, 2, 3]M = 4N = len(arr)print(max_U_element(arr, N, M))# This code is contributed by mohit kumar |
C#
// An efficient C# program to // count distinct elements in // every window of size kusing System;using System.Collections.Generic;class GFG { static int maxUniqueNum(int []arr, int M) { // 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 < M; i++) { if (!hM.ContainsKey(arr[i])) { hM.Add(arr[i], 1); dist_count++; } else { int count = hM[arr[i]]; hM[arr[i]] = count + 1; } } int res = dist_count; // Traverse through the remaining array for (int i = M; i < arr.Length; i++) { // Remove first element of previous window // If there was only one occurrence, then // reduce distinct count. if (hM[arr[i - M]] == 1) { hM.Remove(arr[i - M]); dist_count--; } // reduce count of the removed element else { int count = hM[arr[i - M]]; hM[arr[i - M]] = 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++; } // Increment distinct element count else { int count = hM[arr[i]]; hM[arr[i]] = count + 1; } res = Math.Max(res, dist_count); } return res; } // Driver Code public static void Main(String []arg) { int []arr = { 1, 2, 1, 3, 4, 2, 3 }; int M = 4; Console.WriteLine(maxUniqueNum(arr, M)); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// An efficient JavaScript program to// count distinct elements in// every window of size kfunction maxUniqueNum(arr,m){ // Creates an empty hashMap hM let hM = new Map(); // initialize distinct element count for // current window let dist_count = 0; // Traverse the first window and store count // of every element in hash map for (let i = 0; i < M; i++) { if (hM.get(arr[i]) == null) { hM.set(arr[i], 1); dist_count++; } else { let count = hM.get(arr[i]); hM.set(arr[i], count + 1); } } let res = dist_count; // Traverse through the remaining array for (let i = M; i < arr.length; i++) { // Remove first element of previous window // If there was only one occurrence, then // reduce distinct count. if (hM.get(arr[i - M]) == 1) { hM.delete(arr[i - M]); dist_count--; } else // reduce count of the removed element { let count = hM.get(arr[i - M]); hM.set(arr[i - M], count - 1); } // Add new element of current window // If this element appears first time, // increment distinct element count if (hM.get(arr[i]) == null) { hM.set(arr[i], 1); dist_count++; } else // Increment distinct element count { let count = hM.get(arr[i]); hM.set(arr[i], count + 1); } res = Math.max(res, dist_count); } return res;}// Driver methodlet arr=[1, 2, 1, 3, 4, 2, 3];let M = 4;document.write(maxUniqueNum(arr, M));// This code is contributed by patel2127</script> |
Output:
4
Time Complexity : O(N)
Auxiliary Space : O(M)
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