Maximum Unique Element in every subarray of size K

• Difficulty Level : Hard
• Last Updated : 21 May, 2021

Given an array and an integer K. We need to find the maximum of every segment of length K which has no duplicates in that segment.

Examples:

Input : a[] = {1, 2, 2, 3, 3},
K = 3.
Output : 1 3 2
For segment (1, 2, 2), Maximum = 1.
For segment (2, 2, 3), Maximum = 3.
For segment (2, 3, 3), Maximum = 2.

Input : a[] = {3, 3, 3, 4, 4, 2},
K = 4.
Output : 4 Nothing 3

A simple solution is to run two loops. For every subarray, find all distinct elements and print maximum unique elements.

An efficient solution is to use the sliding window technique. We have two structures in every window.
1) A hash table to store counts of all elements in the current window.
2) A self-balancing BST (implemented using set in C++ STL and TreeSet in Java). The idea is to quickly find the maximum element and update the maximum elements.
We process the first K-1 elements and store their counts in the hash table. We also store unique elements inset. Now we, one by one, process the last element of every window. If the current element is unique, we add it to the set. We also increase its count. After processing the last element, we print the maximum from the set. Before starting the next iteration, we remove the first element of the previous window.

C++

 // C++ code to calculate maximum unique// element of every segment of array#include using namespace std; void find_max(int A[], int N, int K){    // Storing counts of first K-1 elements    // Also storing distinct elements.    map Count;    for (int i = 0; i < K - 1; i++)        Count[A[i]]++;    set Myset;    for (auto x : Count)        if (x.second == 1)            Myset.insert(x.first);     // Before every iteration of this loop,    // we maintain that K-1 elements of current    // window are processed.    for (int i = K - 1; i < N; i++) {         // Process K-th element of current window        Count[A[i]]++;        if (Count[A[i]] == 1)            Myset.insert(A[i]);        else            Myset.erase(A[i]);         // If there are no distinct        // elements in current window        if (Myset.size() == 0)            printf("Nothing\n");         // Set is ordered and last element        // of set gives us maximum element.        else            printf("%d\n", *Myset.rbegin());         // Remove first element of current        // window before next iteration.        int x = A[i - K + 1];        Count[x]--;        if (Count[x] == 1)            Myset.insert(x);        if (Count[x] == 0)            Myset.erase(x);    }} // Driver codeint main(){    int a[] = { 1, 2, 2, 3, 3 };    int n = sizeof(a) / sizeof(a);    int k = 3;    find_max(a, n, k);    return 0;}

Java

 // Java code to calculate maximum unique// element of every segment of arrayimport java.io.*;import java.util.*;class GFG {     static void find_max(int[] A, int N, int K)    {        // Storing counts of first K-1 elements        // Also storing distinct elements.        HashMap Count = new HashMap<>();        for (int i = 0; i < K - 1; i++)            if (Count.containsKey(A[i]))                Count.put(A[i], 1 + Count.get(A[i]));            else                Count.put(A[i], 1);         TreeSet Myset = new TreeSet();        for (Map.Entry x : Count.entrySet()) {            if (Integer.parseInt(String.valueOf(x.getValue())) == 1)                Myset.add(Integer.parseInt(String.valueOf(x.getKey())));        }         // Before every iteration of this loop,        // we maintain that K-1 elements of current        // window are processed.        for (int i = K - 1; i < N; i++) {             // Process K-th element of current window            if (Count.containsKey(A[i]))                Count.put(A[i], 1 + Count.get(A[i]));            else                Count.put(A[i], 1);             if (Integer.parseInt(String.valueOf(Count.get(A[i]))) == 1)                Myset.add(A[i]);            else                Myset.remove(A[i]);             // If there are no distinct            // elements in current window            if (Myset.size() == 0)                System.out.println("Nothing");             // Set is ordered and last element            // of set gives us maximum element.            else                System.out.println(Myset.last());             // Remove first element of current            // window before next iteration.            int x = A[i - K + 1];            Count.put(x, Count.get(x) - 1);             if (Integer.parseInt(String.valueOf(Count.get(x))) == 1)                Myset.add(x);            if (Integer.parseInt(String.valueOf(Count.get(x))) == 0)                Myset.remove(x);        }    }     // Driver code    public static void main(String args[])    {        int[] a = { 1, 2, 2, 3, 3 };        int n = a.length;        int k = 3;        find_max(a, n, k);    }} // This code is contributed by rachana soma

Python3

 # Python3 code to calculate maximum unique# element of every segment of arraydef find_max(A, N, K):         # Storing counts of first K-1 elements    # Also storing distinct elements.    Count = dict()    for i in range(K - 1):        Count[A[i]] = Count.get(A[i], 0) + 1     Myset = dict()    for x in Count:        if (Count[x] == 1):            Myset[x] = 1     # Before every iteration of this loop,    # we maintain that K-1 elements of current    # window are processed.    for i in range(K - 1, N):         # Process K-th element of current window        Count[A[i]] = Count.get(A[i], 0) + 1         if (Count[A[i]] == 1):            Myset[A[i]] = 1        else:            del Myset[A[i]]         # If there are no distinct        # elements in current window        if (len(Myset) == 0):            print("Nothing")         # Set is ordered and last element        # of set gives us maximum element.        else:            maxm = -10**9            for i in Myset:                maxm = max(i, maxm)            print(maxm)         # Remove first element of current        # window before next iteration.        x = A[i - K + 1]        if x in Count.keys():            Count[x] -= 1            if (Count[x] == 1):                Myset[x] = 1            if (Count[x] == 0):                del Myset[x] # Driver codea = [1, 2, 2, 3, 3 ]n = len(a)k = 3find_max(a, n, k) # This code is contributed# by mohit kumar

C#

 using System;using System.Collections.Generic; public class GFG{  static void find_max(int[] A, int N, int K)  {     // Storing counts of first K-1 elements    // Also storing distinct elements.    Dictionary count = new Dictionary();     for (int i = 0; i < K - 1; i++)    {      if(count.ContainsKey(A[i]))      {        count[A[i]]++;      }      else      {        count.Add(A[i], 1);      }    }    HashSet Myset = new HashSet();    foreach(KeyValuePair x in count)    {      if(x.Value == 1)      {        Myset.Add(x.Key);      }    }     // Before every iteration of this loop,    // we maintain that K-1 elements of current    // window are processed.    for (int i = K - 1; i < N; i++)    {       // Process K-th element of current window      if (count.ContainsKey(A[i]))      {        count[A[i]]++;      }      else      {        count.Add(A[i], 1);      }      if(count[A[i]] == 1)      {        Myset.Add(A[i]);      }      else      {        Myset.Remove(A[i]);      }       // If there are no distinct      // elements in current window      if (Myset.Count == 0)        Console.Write("Nothing\n");       // Set is ordered and last element      // of set gives us maximum element.      else      {        List myset = new List(Myset);        Console.WriteLine(myset[myset.Count - 1]);      }       // Remove first element of current      // window before next iteration.      int x = A[i - K + 1];      count[x]--;      if(count[x] == 1)      {        Myset.Add(x);      }      if(count[x] == 0)      {        Myset.Remove(x);      }    }   }   // Driver code  static public void Main ()  {    int[] a = { 1, 2, 2, 3, 3 };    int n=a.Length;    int k = 3;    find_max(a, n, k);  }} // This code is contributed by rag2127

Javascript



Output:

1
3
2

Time Complexity: O(N Log K)

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