PROBLEM LINK:
Author: Deepjal Chhetri
Tester: Istvan Nagy
Editorialist: Oleksandr Kulkov
DIFFICULTY:
MEDIUM
PREREQUISITES:
Segment trees
PROBLEM:
You are to answer the following queries on an array (initially, all elements are zero):
- 1 x y - Change x'th element to y
- 2 l r - Count the number of subarrays [l_1, r_1] of subarray [l, r] (that is, l\le l_1\le r_1\le r), whose maximum element lies in range [L, R]. Numbers L and R are the same in all queries.
QUICK EXPLANATION:
Write the number of subarrays whose maximum element lies in the range [L, R] as:
|\{\text{subarrays with max}\in[L,R]\}|=|\{\text{subarrays with max} < R+1\}|-|\{\text{subarrays with max} < L\}|
Observe that:
|\{\text{subarrays with max} < X\}|=|\{\text{subarrays with all elements} < X\}|
For X=L and X=R+1, maintain an array whose element at position i is 0 if a[i]\ge X and 1 if a[i] < X. The query 2 now reduces to: â€ścount number of subarrays on segment [l,r] consisting entirely of onesâ€ť. This new problem can be solved using segment tree.
EXPLANATION:
Refer to the â€śQuick explanationâ€ť section for the first part of the solution. The problem is now reduced to answering the following queries for fixed X:
- 1 x y - Change x'th element to y (y\in\{0, 1\})
- 2 l r - Count the number of subarrays [l_1, r_1] of subarray [l, r] (that is, l\le l_1\le r_1\le r) consisting of 1's. Number X is the same in all queries.
Use segment tree to solve this problem. In each node, store the following information:
struct Node {
long long num_ones_subarrays;
int len, len_ones_left, len_ones_right;
};
Here,
- len is the length of the range [l;r] represented by the node (len=r-l+1)
- num_ones_subarrays is the number of subarrays consisting of ones in the range represented by the node
- len_ones_left is the length of the largest prefix consisting of ones in the range represented by the node
- len_ones_right is the length of the largest suffix consisting of ones in the range represented by the node
In order to solve the problem we should show that the stored information is self-sufficient, that is, we can merge two nodes and recalculate all 4 variables. So, suppose that nodes a and b are being merged into node res. Then:
- Length of res range is the sum of lengths of ranges a and b.
- The number of ones subarrays in the res range can be calculated as the sum of number of subarrays lying completely in a range (a.num_ones_subarrays), in b range (b.num_ones_subarrays) and the number of subarrays that cross the boundary between a and b. As the subarrays should be non-empty and should consist of ones, it is a.len_ones_right\times b.len_ones_left.
- The length of the largest prefix of ones in the res range is len_ones_left if there is at least one zero in the range a, that is, if a.len_ones_left \neq a.len. Otherwise, it is a.len_ones_left+b.len_ones_left.
- The length of the largest suffix is computed similarly.
This corresponds to the following Node
merge code:
Node merge(Node a, Node b) {
Node res;
res.len = a.len + b.len;
res.num_ones_subarrays =
a.num_ones_subarrays +
b.num_ones_subarrays +
a.len_ones_right * (li)b.len_ones_left;
res.len_ones_left = a.len_ones_left;
if (a.len_ones_left == a.len)
res.len_ones_left += b.len_ones_left;
res.len_ones_right = b.len_ones_right;
if (b.len_ones_right == b.len)
res.len_ones_right += a.len_ones_right;
return res;
}
The time complexity of the solution is O((N+Q)\log N) from the underlying segment tree (it is possible to make it O(N+Q\log N) by building the segment tree in linear time). The memory complexity of the solution is O(N).
The implementation with 2 separate segment trees for different X=L and X=R+1 is enough to get AC, yet it is possible to make it faster via combining the trees into one.
As the input is quite large, care should be taken to avoid unnecessary input lag. For example, using C++ iostreams consider adding:
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
AUTHORâ€™S AND TESTERâ€™S SOLUTIONS:
Authorâ€™s solution can be found here.
Testerâ€™s solution can be found here.
Editorialistâ€™s solution can be found here.