### ___Abc123___'s blog

By ___Abc123___, history, 12 days ago,

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

• 0

 » 12 days ago, # |   0 Auto comment: topic has been updated by ___Abc123___ (previous revision, new revision, compare).
 » 12 days ago, # | ← Rev. 2 →   +3 is this not just a standard seg/fenwick tree problem try learning one of these datastructures, then it will be a template problem
 » 12 days ago, # |   0 People there have given great explanations for this problem already.
 » 12 days ago, # | ← Rev. 2 →   0 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])......
•  » » 12 days ago, # ^ |   0 interestingly i have actually seen this problem beforetry thinking about it as how many times each element in the range contributes to the final sum
•  » » » 12 days ago, # ^ |   0 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
•  » » » » 12 days ago, # ^ |   0 Another way to look at it would be to only check the contribution of each individual element in $[L,R]$. Here the $ith$ elements contribution would be $arr[i]*(r-i+1)*(i-l+1)$ as these many subarrays in $[L,R]$ contain it. Now you can see that you only need the sum of $arr[i]*i, arr[i]*i*i, arr[i]$ to find the contribution of all elements which you can find using a segment tree.
•  » » » » 11 days ago, # ^ |   0 heres the soltuionyou 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 endsubtract the sum between r+1 to n from l and r and you get the answer to your query this works cuz math
 » 12 days ago, # |   0 Auto comment: topic has been updated by ___Abc123___ (previous revision, new revision, compare).
 » 12 days ago, # |   0
 » 12 days ago, # |   0 Try learning about Fenwick Tree(Binary Indexed Tree) or Segment trees! :)
 » 12 days ago, # |   0 this can be solved using Segment trees in a fast time Click on this and go learn it