Last active
May 23, 2022 23:17
-
-
Save imaginate/af0f2f544bfa2976f0fa4f0a67b11d2e to your computer and use it in GitHub Desktop.
Leetcode 2281. Sum of Total Strength of Wizards - 1st Place Solution
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
class Solution | |
{ | |
public: | |
int totalStrength(vector<int>& a) | |
{ | |
int n = a.size(); | |
const long long MOD = 1000000007; | |
vector<long long> b(n + 1); | |
for (int i = 0; i < n; ++i) { | |
b[i + 1] = (b[i] + a[i]) % MOD; | |
} | |
struct Seg { | |
long long val; | |
long long count; | |
long long sum; | |
}; | |
vector<Seg> s; | |
s.push_back({0, 0}); | |
long long result = 0; | |
long long s1 = 0; | |
long long s2 = 0; | |
for (int i = n - 1; i >= 0; --i) { | |
Seg cur; | |
cur.val = a[i]; | |
cur.count = 1; | |
cur.sum = b[i + 1]; | |
while (s.back().val >= cur.val) { | |
const auto& ss = s.back(); | |
s1 = (s1 - ss.val * ss.sum) % MOD; | |
s2 = (s2 - ss.val * ss.count) % MOD; | |
cur.sum += ss.sum; | |
cur.count += ss.count; | |
cur.sum %= MOD; | |
s.pop_back(); | |
} | |
s1 = (s1 + cur.val * cur.sum) % MOD; | |
s2 = (s2 + cur.val * cur.count) % MOD; | |
s.push_back(cur); | |
result = (result + s1) % MOD; | |
result = (result - s2 * b[i]) % MOD; | |
} | |
result %= MOD; | |
result += MOD; | |
result %= MOD; | |
return (int)result; | |
} | |
}; |
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment