feat: add solutions to lc problem: No.2838 (#1550)

No.2838.Maximum Coins Heroes Can Collect

Author: yanglbme

solution/2800-2899/2838.Maximum Coins Heroes Can Collect/README.md

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+# [2838. Maximum Coins Heroes Can Collect](https://leetcode.cn/problems/maximum-coins-heroes-can-collect)
+
+[English Version](/solution/2800-2899/2838.Maximum%20Coins%20Heroes%20Can%20Collect/README_EN.md)
+
+## 题目描述
+
+<!-- 这里写题目描述 -->
+
+<p>There is a battle and <code>n</code> heroes are trying to defeat <code>m</code> monsters. You are given two <strong>1-indexed</strong> arrays of <strong>positive</strong> integers <code><font face="monospace">heroes</font></code> and <code><font face="monospace">monsters</font></code> of length <code>n</code> and <code>m</code>, respectively. <code><font face="monospace">heroes</font>[i]</code> is the power of <code>i<sup>th</sup></code> hero, and <code><font face="monospace">monsters</font>[i]</code> is the power of <code>i<sup>th</sup></code> monster.</p>
+
+<p>The <code>i<sup>th</sup></code> hero can defeat the <code>j<sup>th</sup></code> monster if <code>monsters[j] &lt;= heroes[i]</code>.</p>
+
+<p>You are also given a <strong>1-indexed</strong> array <code>coins</code> of length <code>m</code> consisting of <strong>positive</strong> integers. <code>coins[i]</code> is the number of coins that each hero earns after defeating the <code>i<sup>th</sup></code> monster.</p>
+
+<p>Return<em> an array </em><code>ans</code><em> of length </em><code>n</code><em> where </em><code>ans[i]</code><em> is the <strong>maximum</strong> number of coins that the </em><code>i<sup>th</sup></code><em> hero can collect from this battle</em>.</p>
+
+<p><strong>Notes</strong></p>
+
+<ul>
+	<li>The health of a hero doesn&#39;t get reduced after defeating a monster.</li>
+	<li>Multiple heroes can defeat a monster, but each monster can be defeated by a given hero only once.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> heroes = [1,4,2], monsters = [1,1,5,2,3], coins = [2,3,4,5,6]
+<strong>Output:</strong> [5,16,10]
+<strong>Explanation: </strong>For each hero, we list the index of all the monsters he can defeat:
+1<sup>st</sup> hero: [1,2] since the power of this hero is 1 and monsters[1], monsters[2] &lt;= 1. So this hero collects coins[1] + coins[2] = 5 coins.
+2<sup>nd</sup> hero: [1,2,4,5] since the power of this hero is 4 and monsters[1], monsters[2], monsters[4], monsters[5] &lt;= 4. So this hero collects coins[1] + coins[2] + coins[4] + coins[5] = 16 coins.
+3<sup>rd</sup> hero: [1,2,4] since the power of this hero is 2 and monsters[1], monsters[2], monsters[4] &lt;= 2. So this hero collects coins[1] + coins[2] + coins[4] = 10 coins.
+So the answer would be [5,16,10].</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> heroes = [5], monsters = [2,3,1,2], coins = [10,6,5,2]
+<strong>Output:</strong> [23]
+<strong>Explanation:</strong> This hero can defeat all the monsters since monsters[i] &lt;= 5. So he collects all of the coins: coins[1] + coins[2] + coins[3] + coins[4] = 23, and the answer would be [23].
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> heroes = [4,4], monsters = [5,7,8], coins = [1,1,1]
+<strong>Output:</strong> [0,0]
+<strong>Explanation:</strong> In this example, no hero can defeat a monster. So the answer would be [0,0],
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= n == heroes.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= m == monsters.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>coins.length == m</code></li>
+	<li><code>1 &lt;= heroes[i], monsters[i], coins[i] &lt;= 10<sup>9</sup></code></li>
+</ul>
+
+## 解法
+
+<!-- 这里可写通用的实现逻辑 -->
+
+**方法一：排序 + 前缀和 + 二分查找**
+
+我们可以将怪物和金币按照怪物的战斗力从小到大排序，然后使用前缀和计算每个英雄打败前 $i$ 个怪物可以获得的金币总数。
+
+接下来，对于每个英雄，我们可以使用二分查找找到他可以打败的最大的怪物，然后使用前缀和计算他可以获得的金币总数即可。
+
+时间复杂度 $O((m + n) \times \log n)$，空间复杂度 $O(m)$。其中 $m$ 和 $n$ 分别是怪物和英雄的数量。
+
+<!-- tabs:start -->
+
+### **Python3**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+class Solution:
+    def maximumCoins(
+        self, heroes: List[int], monsters: List[int], coins: List[int]
+    ) -> List[int]:
+        m = len(monsters)
+        idx = sorted(range(m), key=lambda i: monsters[i])
+        s = list(accumulate((coins[i] for i in idx), initial=0))
+        ans = []
+        for h in heroes:
+            i = bisect_right(idx, h, key=lambda i: monsters[i])
+            ans.append(s[i])
+        return ans
+```
+
+### **Java**
+
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+class Solution {
+    public long[] maximumCoins(int[] heroes, int[] monsters, int[] coins) {
+        int m = monsters.length;
+        Integer[] idx = new Integer[m];
+        for (int i = 0; i < m; ++i) {
+            idx[i] = i;
+        }
+
+        Arrays.sort(idx, Comparator.comparingInt(j -> monsters[j]));
+        long[] s = new long[m + 1];
+        for (int i = 0; i < m; ++i) {
+            s[i + 1] = s[i] + coins[idx[i]];
+        }
+        int n = heroes.length;
+        long[] ans = new long[n];
+        for (int k = 0; k < n; ++k) {
+            int i = search(monsters, idx, heroes[k]);
+            ans[k] = s[i];
+        }
+        return ans;
+    }
+
+    private int search(int[] nums, Integer[] idx, int x) {
+        int l = 0, r = idx.length;
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (nums[idx[mid]] > x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}
+```
+
+### **C++**
+
+```cpp
+class Solution {
+public:
+    vector<long long> maximumCoins(vector<int>& heroes, vector<int>& monsters, vector<int>& coins) {
+        int m = monsters.size();
+        vector<int> idx(m);
+        iota(idx.begin(), idx.end(), 0);
+        sort(idx.begin(), idx.end(), [&](int i, int j) {
+            return monsters[i] < monsters[j];
+        });
+        long long s[m + 1];
+        s[0] = 0;
+        for (int i = 1; i <= m; ++i) {
+            s[i] = s[i - 1] + coins[idx[i - 1]];
+        }
+        vector<long long> ans;
+        auto search = [&](int x) {
+            int l = 0, r = m;
+            while (l < r) {
+                int mid = (l + r) >> 1;
+                if (monsters[idx[mid]] > x) {
+                    r = mid;
+                } else {
+                    l = mid + 1;
+                }
+            }
+            return l;
+        };
+        for (int h : heroes) {
+            ans.push_back(s[search(h)]);
+        }
+        return ans;
+    }
+};
+```
+
+### **Go**
+
+```go
+func maximumCoins(heroes []int, monsters []int, coins []int) (ans []int64) {
+	m := len(monsters)
+	idx := make([]int, m)
+	for i := range idx {
+		idx[i] = i
+	}
+	sort.Slice(idx, func(i, j int) bool { return monsters[idx[i]] < monsters[idx[j]] })
+	s := make([]int64, m+1)
+	for i, j := range idx {
+		s[i+1] = s[i] + int64(coins[j])
+	}
+	for _, h := range heroes {
+		i := sort.Search(m, func(i int) bool { return monsters[idx[i]] > h })
+		ans = append(ans, s[i])
+	}
+	return
+}
+```
+
+### **TypeScript**
+
+```ts
+function maximumCoins(
+    heroes: number[],
+    monsters: number[],
+    coins: number[],
+): number[] {
+    const m = monsters.length;
+    const idx: number[] = Array.from({ length: m }, (_, i) => i);
+    idx.sort((i, j) => monsters[i] - monsters[j]);
+    const s: number[] = Array(m + 1).fill(0);
+    for (let i = 0; i < m; ++i) {
+        s[i + 1] = s[i] + coins[idx[i]];
+    }
+    const search = (x: number): number => {
+        let l = 0;
+        let r = m;
+        while (l < r) {
+            const mid = (l + r) >> 1;
+            if (monsters[idx[mid]] > x) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    };
+    return heroes.map(h => s[search(h)]);
+}
+```
+
+### **...**
+
+```
+
+```
+
+<!-- tabs:end -->