Maximum possible sum of a window in an array such that elements of same window in other array are unique

Given two arrays A and B of equal number of elements. Task is to find the maximum sum possible of a window in array B such that elements of same window in A[] are unique.

Examples:

Input : A = [0, 1, 2, 3, 0, 1, 4] 
        B = [9, 8, 1, 2, 3, 4, 5]
Output : sum = 20
The maximum sum possible in B[] such that 
all corresponding elements in A[] are unique 
is (9+8+1+2) = 20.

Input : A = [0, 1, 2, 0, 2]
        B = [5, 6, 7, 8, 2]
Output :sum = 21

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

A simple solution is to consider all subarrays of B[]. For every subarray, check if elements same subarray in A[] are distinct or not. If distinct, then compare sum with result and update result.
Time complexity of this solution is O(n²)

An efficient solution is to use hashing.

Create an empty hash table.
Traverse array elements. Do following for every element A[i].
- While A[i] is present in hash table, keep removing elements from beginning of current window and keep subtracting window beginning element of B[] from current sum.
Add B[i] to current sum and update result if current sum becomes more.
Return result.

Below is the implementation of above steps.

C++

// C++ program to find the maximum 
// possible sum of a window in one 
// array such that elements in same 
// window of other array are unique. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Function to return maximum sum of window 
// in B[] according to given constraints. 
int returnMaxSum(int A[], int B[], int n) 
{ 
    // Map is used to store elements 
    // and their counts. 
    unordered_set<int> mp; 
  
    int result = 0; // Initialize result 
  
    // calculating the maximum possible 
    // sum for each subarray containing 
    // unique elements. 
    int curr_sum = 0, curr_begin = 0; 
    for (int i = 0; i < n; ++i) { 
  
        // Remove all duplicate 
        // instances of A[i] in 
        // current window. 
        while (mp.find(A[i]) != mp.end()) { 
            mp.erase(A[curr_begin]); 
            curr_sum -= B[curr_begin]; 
            curr_begin++; 
        } 
  
        // Add current instance of A[i] 
        // to map and to current sum. 
        mp.insert(A[i]); 
        curr_sum += B[i]; 
  
        // Update result if current 
        // sum is more. 
        result = max(result, curr_sum); 
    } 
  
    return result; 
} 
  
// Driver code 
int main() 
{ 
    int A[] = { 0, 1, 2, 3, 0, 1, 4 }; 
    int B[] = { 9, 8, 1, 2, 3, 4, 5 }; 
    int n = sizeof(A)/sizeof(A[0]); 
    cout << returnMaxSum(A, B, n); 
    return 0; 
} 

Java

// Java program to find the maximum 
// possible sum of a window in one 
// array such that elements in same 
// window of other array are unique. 
import java.util.HashSet; 
import java.util.Set; 
  
public class MaxPossibleSuminWindow 
{ 
    // Function to return maximum sum of window 
    // in A[] according to given constraints. 
    static int returnMaxSum(int A[], int B[], int n) 
    { 
  
        // Map is used to store elements 
        // and their counts. 
        Set<Integer> mp = new HashSet<Integer>(); 
  
        int result = 0; // Initialize result 
  
        // calculating the maximum possible 
        // sum for each subarray containing 
        // unique elements. 
        int curr_sum = 0, curr_begin = 0; 
        for (int i = 0; i < n; ++i) 
        { 
            // Remove all duplicate 
            // instances of A[i] in 
            // current window. 
            while (mp.contains(A[i])) 
            { 
                mp.remove(A[curr_begin]); 
                curr_sum -= B[curr_begin]; 
                curr_begin++; 
            } 
  
            // Add current instance of A[i] 
            // to map and to current sum. 
            mp.add(A[i]); 
            curr_sum += B[i]; 
  
            // Update result if current 
            // sum is more. 
            result = Integer.max(result, curr_sum); 
  
        } 
        return result; 
    } 
  
    //Driver Code to test above method 
    public static void main(String[] args) 
    { 
        int A[] = { 0, 1, 2, 3, 0, 1, 4 }; 
        int B[] = { 9, 8, 1, 2, 3, 4, 5 }; 
        int n = A.length; 
        System.out.println(returnMaxSum(A, B, n)); 
    } 
} 
// This code is contributed by Sumit Ghosh 

Python3

# Python3 program to find the maximum  
# possible sum of a window in one  
# array such that elements in same  
# window of other array are unique.  
  
# Function to return maximum sum of window  
# in B[] according to given constraints.  
def returnMaxSum(A, B, n):  
   
    # Map is used to store elements  
    # and their counts.  
    mp = set() 
    result = 0 # Initialize result  
  
    # calculating the maximum possible  
    # sum for each subarray containing  
    # unique elements.  
    curr_sum = curr_begin = 0 
    for i in range(0, n):   
  
        # Remove all duplicate instances  
        # of A[i] in current window.  
        while A[i] in mp:   
            mp.remove(A[curr_begin])  
            curr_sum -= B[curr_begin]  
            curr_begin += 1
           
        # Add current instance of A[i]  
        # to map and to current sum.  
        mp.add(A[i])  
        curr_sum += B[i]  
  
        # Update result if current  
        # sum is more.  
        result = max(result, curr_sum)  
       
    return result  
  
# Driver code  
if __name__ == "__main__": 
   
    A = [0, 1, 2, 3, 0, 1, 4]   
    B = [9, 8, 1, 2, 3, 4, 5]  
    n = len(A)  
    print(returnMaxSum(A, B, n)) 
  
# This code is contributed by Rituraj Jain 

C#

// C# program to find the maximum 
// possible sum of a window in one 
// array such that elements in same 
// window of other array are unique.  
using System; 
using System.Collections.Generic; 
  
public class MaxPossibleSuminWindow 
{ 
      
    // Function to return maximum sum of window 
    // in A[] according to given constraints. 
    static int returnMaxSum(int []A, int []B, int n) 
    { 
  
        // Map is used to store elements 
        // and their counts. 
        HashSet<int> mp = new HashSet<int>(); 
  
        int result = 0; // Initialize result 
  
        // calculating the maximum possible 
        // sum for each subarray containing 
        // unique elements. 
        int curr_sum = 0, curr_begin = 0; 
        for (int i = 0; i < n; ++i) 
        { 
            // Remove all duplicate 
            // instances of A[i] in 
            // current window. 
            while (mp.Contains(A[i])) 
            { 
                mp.Remove(A[curr_begin]); 
                curr_sum -= B[curr_begin]; 
                curr_begin++; 
            } 
  
            // Add current instance of A[i] 
            // to map and to current sum. 
            mp.Add(A[i]); 
            curr_sum += B[i]; 
  
            // Update result if current 
            // sum is more. 
            result = Math.Max(result, curr_sum); 
  
        } 
        return result; 
    } 
  
    // Driver Code  
    public static void Main(String[] args) 
    { 
        int []A = { 0, 1, 2, 3, 0, 1, 4 }; 
        int []B = { 9, 8, 1, 2, 3, 4, 5 }; 
        int n = A.Length; 
        Console.WriteLine(returnMaxSum(A, B, n)); 
    } 
}  
  
/* This code has been contributed  
by PrinciRaj1992*/

Output:

Time complexity of this solution is O(n). Note that every element of array is inserted and removed at most once from array.

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Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

C++

Java

Python3

C#

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