# # Leetcode 第294场周赛题解

## # Problem A - 字母在字符串中的百分比 (opens new window)

### # 方法一：模拟

• 时间复杂度 $\mathcal{O}(|S|)$
• 空间复杂度 $\mathcal{O}(1)$

class Solution:
def percentageLetter(self, s: str, letter: str) -> int:
return s.count(letter) * 100 // len(s)

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class Solution {
public:
int percentageLetter(string s, char letter) {
int n = s.size();
int cnt = 0;
for (char c : s)
if (c == letter)
cnt++;
return cnt * 100 / n;
}
};

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## # Problem B - 装满石头的背包的最大数量 (opens new window)

### # 方法一：排序 + 贪心

• 时间复杂度 $\mathcal{O}(N\log N)$
• 空间复杂度 $\mathcal{O}(1)$

class Solution:
def maximumBags(self, capacity: List[int], rocks: List[int], additionalRocks: int) -> int:
rem = sorted([c - r for c, r in zip(capacity, rocks)])
for i, ri in enumerate(rem):
else:
return i
return len(rocks)

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class Solution {
public:
int maximumBags(vector<int>& capacity, vector<int>& rocks, int additionalRocks) {
int n = capacity.size();
vector<int> rem(n);
for (int i = 0; i < n; ++i)
rem[i] = capacity[i] - rocks[i];
sort(rem.begin(), rem.end());
for (int i = 0; i < n; ++i) {
else
return i;
}
return n;
}
};

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## # Problem C - 表示一个折线图的最少线段数 (opens new window)

### # 方法一：排序

• 时间复杂度 $\mathcal{O}(N\log N)$
• 空间复杂度 $\mathcal{O}(1)$

class Solution {
public:
int minimumLines(vector<vector<int>>& stockPrices) {
int n = stockPrices.size();
if (n <= 2) return n - 1;
sort(stockPrices.begin(), stockPrices.end());
int ans = 1;
for (int i = 1; i + 1 < n; ++i) {
int lx = stockPrices[i][0] - stockPrices[i - 1][0];
int ly = stockPrices[i][1] - stockPrices[i - 1][1];
int rx = stockPrices[i + 1][0] - stockPrices[i][0];
int ry = stockPrices[i + 1][1] - stockPrices[i][1];
if (1LL * lx * ry != 1LL * ly * rx) ans++;
}
return ans;
}
};

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## # Problem D - 巫师的总力量和 (opens new window)

### # 方法一：单调栈 + 前缀和

• 时间复杂度为 $\mathcal{O}(N)$
• 空间复杂度 $\mathcal{O}(N)$

using ll = long long;
const ll MOD = 1e9 + 7;

class Solution {
public:
int totalStrength(vector<int>& strength) {
int n = strength.size();
vector<int> l(n, -1), r(n, n);
stack<int> s;
for (int i = 0; i < n; ++i) {
while (!s.empty() && strength[i] < strength[s.top()]) {
r[s.top()] = i;
s.pop();
}
s.push(i);
}

s = stack<int>();
for (int i = n - 1; i >= 0; --i) {
while (!s.empty() && strength[i] <= strength[s.top()]) {
l[s.top()] = i;
s.pop();
}
s.push(i);
}

vector<ll> pre(n + 1), apre(n + 1), dpre(n + 2);
for (int i = 0; i < n; i++) {
pre[i + 1] = (pre[i] + strength[i]) % MOD;
apre[i + 1] = (apre[i] + 1LL * (i + 1) * strength[i]) % MOD;
}
for (int i = n; i >= 1; --i) {
dpre[i] = (dpre[i + 1] + 1LL * (n - i + 1) * strength[i - 1]) % MOD;
}

ll ans = 0;
for (int i = 0; i < n; ++i) {
ll left = i - l[i], right = r[i] - i;
ll lsum =
apre[i] - apre[l[i] + 1] - (pre[i] - pre[l[i] + 1]) * (l[i] + 1);
lsum = lsum * right % MOD;
if (lsum < 0) lsum += MOD;
ll rsum =
dpre[i + 2] - dpre[r[i] + 1] - (pre[r[i]] - pre[i + 1]) * (n - r[i]);
rsum = rsum * left % MOD;
if (rsum < 0) rsum += MOD;
ans += (lsum + rsum + left * right % MOD * strength[i] % MOD) % MOD * strength[i] % MOD;
ans %= MOD;
}
return ans;
}
};

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