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.
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.
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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