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

A simple solution is to consider all sub-arrays of B[]. For every sub-array, check if elements same sub-array in A[] are distinct or not. If distinct, then compare sum with result and update result. 

Time complexity: O(n²), as we will be using nested loops for checking every sub-array.
Auxiliary Space: O(1), as we are not using any extra space.

An efficient solution is to use hashing. 

1. Create an empty hash table.
2. 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.
3. Add B[i] to current sum and update result if current sum becomes more.
4. 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*/

Javascript

<script>

// Javascript 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.
function returnMaxSum(A, B, n)
{
    // Map is used to store elements
    // and their counts.
    var mp = new Set();

    var result = 0; // Initialize result

    // calculating the maximum possible
    // sum for each subarray containing
    // unique elements.
    var curr_sum = 0, curr_begin = 0;
    for (var i = 0; i < n; ++i) {

        // Remove all duplicate
        // instances of A[i] in
        // current window.
        while (mp.has(A[i])) {
            mp.delete(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
var A = [0, 1, 2, 3, 0, 1, 4];
var B = [9, 8, 1, 2, 3, 4, 5];
var n = A.length;
document.write( returnMaxSum(A, B, n));

// This code is contributed by importantly.
</script>

Output

20

Time complexity: O(n), as we are using a loop to traverse N times and unordered map operations will take constant time. Where N is the number of elements in the array.
Auxiliary Space: O(N), as we are using extra space for the map. Where N is the number of elements in the array.

Note that every element of array is inserted and removed at most once from array.

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