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Commit eed4b10

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feat: solve No.2374,2390
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# 2374. Node With Highest Edge Score
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- Difficulty: Medium.
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- Related Topics: Hash Table, Graph.
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- Similar Questions: Two Sum, Sort Characters By Frequency, Sort Array by Increasing Frequency.
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## Problem
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You are given a directed graph with `n` nodes labeled from `0` to `n - 1`, where each node has **exactly one** outgoing edge.
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The graph is represented by a given **0-indexed** integer array `edges` of length `n`, where `edges[i]` indicates that there is a **directed** edge from node `i` to node `edges[i]`.
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The **edge score** of a node `i` is defined as the sum of the **labels** of all the nodes that have an edge pointing to `i`.
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Return **the node with the highest **edge score****. If multiple nodes have the same **edge score**, return the node with the **smallest** index.
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Example 1:
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![](https://assets.leetcode.com/uploads/2022/06/20/image-20220620195403-1.png)
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```
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Input: edges = [1,0,0,0,0,7,7,5]
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Output: 7
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Explanation:
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- The nodes 1, 2, 3 and 4 have an edge pointing to node 0. The edge score of node 0 is 1 +たす 2 +たす 3 +たす 4 =わ 10.
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- The node 0 has an edge pointing to node 1. The edge score of node 1 is 0.
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- The node 7 has an edge pointing to node 5. The edge score of node 5 is 7.
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- The nodes 5 and 6 have an edge pointing to node 7. The edge score of node 7 is 5 + 6 = 11.
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Node 7 has the highest edge score so return 7.
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```
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Example 2:
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![](https://assets.leetcode.com/uploads/2022/06/20/image-20220620200212-3.png)
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```
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Input: edges = [2,0,0,2]
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Output: 0
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Explanation:
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- The nodes 1 and 2 have an edge pointing to node 0. The edge score of node 0 is 1 + 2 = 3.
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- The nodes 0 and 3 have an edge pointing to node 2. The edge score of node 2 is 0 + 3 = 3.
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Nodes 0 and 2 both have an edge score of 3. Since node 0 has a smaller index, we return 0.
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```
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**Constraints:**
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- `n == edges.length`
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- `2 <= n <= 105`
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- `0 <= edges[i] < n`
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- `edges[i] != i`
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## Solution
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```javascript
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/**
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* @param {number[]} edges
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* @return {number}
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*/
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var edgeScore = function(edges) {
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var sumMap = Array(edges.length).fill(0);
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var maxSumNode = 0;
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for (var i = 0; i < edges.length; i++) {
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sumMap[edges[i]] += i;
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if (sumMap[edges[i]] > sumMap[maxSumNode]
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|| (sumMap[edges[i]] === sumMap[maxSumNode] && maxSumNode > edges[i])) {
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maxSumNode = edges[i];
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}
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}
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return maxSumNode;
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};
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n).
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* Space complexity : O(n).
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# 2390. Removing Stars From a String
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- Difficulty: Medium.
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- Related Topics: String, Stack, Simulation.
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- Similar Questions: Backspace String Compare, Remove All Adjacent Duplicates In String.
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## Problem
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You are given a string `s`, which contains stars `*`.
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In one operation, you can:
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- Choose a star in `s`.
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- Remove the closest **non-star** character to its **left**, as well as remove the star itself.
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Return **the string after **all** stars have been removed**.
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**Note:**
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- The input will be generated such that the operation is always possible.
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- It can be shown that the resulting string will always be unique.
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Example 1:
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```
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Input: s = "leet**cod*e"
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Output: "lecoe"
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Explanation: Performing the removals from left to right:
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- The closest character to the 1st star is 't' in "leet**cod*e". s becomes "lee*cod*e".
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- The closest character to the 2nd star is 'e' in "lee*cod*e". s becomes "lecod*e".
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- The closest character to the 3rd star is 'd' in "lecod*e". s becomes "lecoe".
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There are no more stars, so we return "lecoe".
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```
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Example 2:
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```
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Input: s = "erase*****"
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Output: ""
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Explanation: The entire string is removed, so we return an empty string.
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```
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**Constraints:**
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- `1 <= s.length <= 105`
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- `s` consists of lowercase English letters and stars `*`.
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- The operation above can be performed on `s`.
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## Solution
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```javascript
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/**
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* @param {string} s
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* @return {string}
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*/
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var removeStars = function(s) {
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var stack = [];
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for (var i = 0; i < s.length; i++) {
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if (s[i] === '*') {
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stack.pop();
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} else {
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stack.push(s[i]);
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}
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}
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return stack.join('');
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};
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```
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**Explain:**
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nope.
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**Complexity:**
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* Time complexity : O(n).
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* Space complexity : O(n).

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