add solution for #678 Valid Parenthesis String.

problems/valid-parenthesis-string/README.md

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+# Valid Parenthesis String
+
+LeetCode #: [678](https://leetcode.com/problems/valid-parenthesis-string/)
+
+Difficulty: Medium.
+
+Topics: String.
+
+## Problem
+
+Given a string containing only three types of characters: '(', ')' and '*', write a function to check whether this string is valid. We define the validity of a string by these rules:
+
+- Any left parenthesis `'('` must have a corresponding right parenthesis `')'`.
+- Any right parenthesis `')'` must have a corresponding left parenthesis `'('`.
+- Left parenthesis `'('` must go before the corresponding right parenthesis `')'`.
+- `'*'` could be treated as a single right parenthesis `')'` or a single left parenthesis `'('` or an empty string.
+- An empty string is also valid.
+
+Example 1:
+
+```text
+Input: "()"
+Output: True
+```
+
+Example 2:
+
+```text
+Input: "(*)"
+Output: True
+```
+
+Example 3:
+
+```text
+Input: "(*))"
+Output: True
+```
+
+Note:
+
+- The string size will be in the range [1, 100].
+
+## Solution Explanation
+
+We need to check for pairs of matching left and right parenthesis. For this kind of question, we usually just need to keep track of number of left parentheses.
+
+However, with the possible occurrence of the character `'*'`, there could be different number of left parentheses. We track it by using `minLeft` (minimum number of left paretheses) and `maxLeft` (possible maximum number of left paretheses). For example:
+
+```text
+Input: ((
+minLeft: 2
+maxLeft: 2
+
+Input: ((**
+minLeft: 2
+maxLeft: 4
+```
+
+When we encounter a `')'`, we just need to minus 1 accordingly from `minLeft` and `maxLeft`.
+
+The result is false when:
+
+- `maxLeft` is less than 0 (i.e. the number of `')'` is more than `'('` and `'*'`); or,
+- `minLeft` is not equal to 0 in the end (i.e. there is no exact matching pairs of left and right parentheses).
+
+## Complexity Analysis
+
+Space complexity: O(1) constant space with `minLeft` and `maxLeft` only.
+
+Time complexity: O(n) by looping through each character once.

problems/valid-parenthesis-string/checkValidString.js

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+/**
+ * @param {string} s
+ * @return {boolean}
+ */
+const checkValidString = function (s) {
+  let minLeft = 0
+  let maxLeft = 0
+
+  for (let i = 0; i < s.length; i++) {
+    const char = s[i]
+
+    if (char === '(') {
+      minLeft++
+      maxLeft++
+    }
+
+    if (char === '*') {
+      minLeft = Math.max(0, minLeft - 1)
+      maxLeft++
+    }
+
+    if (char === ')') {
+      minLeft = Math.max(0, minLeft - 1)
+      maxLeft--
+
+      if (maxLeft < 0) {
+        return false
+      }
+    }
+  }
+
+  return minLeft === 0
+}
+
+module.exports = checkValidString

problems/valid-parenthesis-string/checkValidString.test.js

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+const checkValidString = require('./checkValidString')
+
+test('Example 1', () => {
+  const input = '()'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(true)
+})
+
+test('Example 2', () => {
+  const input = '(*)'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(true)
+})
+
+test('Example 3', () => {
+  const input = '(*))'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(true)
+})
+
+test('empty string is valid', () => {
+  const input = ''
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(true)
+})
+
+test('one (', () => {
+  const input = '('
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(false)
+})
+
+test('one )', () => {
+  const input = ')'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(false)
+})
+
+test('(()', () => {
+  const input = '(()'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(false)
+})
+
+test('())', () => {
+  const input = '())'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(false)
+})
+
+test('*)', () => {
+  const input = '*)'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(true)
+})
+
+test('(*', () => {
+  const input = '(*'
+
+  const result = checkValidString(input)
+
+  expect(result).toBe(true)
+})