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 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; } |
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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 |
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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 |
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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*/ |
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Output:
20
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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