Hi, Here is the Question You are given an array A of length N and q queries where each query is one of the following types:
1 i val: Update the value at ith index to val (arr[i] = val) 2 L R: find the sum of all the subarray of arr[L....R]
Determine the value of query 2 Ex: N = 5
q 2;
arr = [2, 1, 4, 3, 1]
query = [(1, 2, 2), (2, 1, 3)]
Output = 26
How to approach this question
I have to find sum of every subarray of arr[L...R] ans will be arr[L] + arr[L...(L + 1)] + arr[L...(L +2)] +.....arr[L....R] + arr[L + 1] + (arr(L + 1)..(L + 2)] +.....(arr[(L + 1)...R])......
constraint 1 <= N, q <= 100000
Auto comment: topic has been updated by asharique22 (previous revision, new revision, compare).
is this not just a standard seg/fenwick tree problem try learning one of these datastructures, then it will be a template problem
I have to find sum of every subarray of arr[L...R] ans will be arr[L] + arr[L...(L + 1)] + arr[L...(L +2)] +.....arr[L....R] + arr[L + 1] + (arr(L + 1)..(L + 2)] +.....(arr[(L + 1)...R])......
interestingly i have actually seen this problem before
try thinking about it as how many times each element in the range contributes to the final sum
Yes I have done that final expression will be (r — l + 1)arr[l] + 2*(r — l)arr[l + 1] + 3*(r — l — 1)*arr[l + 2]+.....+(r — l + 1)arr[r].
still cant figure out what to do from here
heres the soltuion
you precompute an array containing the values arr[I] = nums[I] * (n-i)
now build a segtree on arr, for each query you would just query the sum between l and r and the sum from r+1 to the end
subtract the sum between r+1 to n from l and r and you get the answer to your query this works cuz math
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this can be solved using Segment trees in a fast time Click on this and go learn it