Updating query a[i] = max(a[i], x) and Minimum query for intervals

Question

I want to know the fast algorithm to solve this simple problem:

You are given a sequence a[0], a[1],..., a[N - 1].
You are given Q queries. All queries are query 1 or 2. Process all queries.
1. You are given integers l, r, x. Update a[i] = max(a[i], x) for l <= i <= r.
2. You are given integers l, r. Find a[l] + a[l + 1] + ... + a[r].

I only have a naive O(NQ) algorithm, so please find more faster algorithm because I want to solve for N <= 200000, Q <= 200000.

This sounds a lot like homework – Filip Allberg Mar 15 '17 at 09:53 — Filip Allberg, Mar 15 '17 at 09:53
@FilipAllberg I think it is contest problem. – Petar Petrovic Mar 15 '17 at 10:12 — Petar Petrovic, Mar 15 '17 at 10:12

score 0 · Answer 1 · answered Mar 15 '17 at 12:36

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This problem can be easily solved by interval tree(or segment tree) with lazy tag.If you are not familiar with interval tree,you may refer to this article.The time complexity is O(Q*logN).

answered Mar 15 '17 at 12:36

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can you elaborate more? – Petar Petrovic Mar 16 '17 at 05:35

Updating query a[i] = max(a[i], x) and Minimum query for intervals

1 Answers1