diff --git a/README.md b/README.md
index ab4a28d..2646ab3 100644
--- a/README.md
+++ b/README.md
@@ -1,11 +1,11 @@
-# practice
+# leetcode_javascript_and_python
+
目的
-1. 主要用來練習JS，和Python
+1. 主要用來練習JavaScript和Python
2. 題目來源:
- LeetCode
- CodeWars
- HackerRank
-雖是JavaScript，但實際上使用Node.js，因此不需要打開瀏覽器便可執行JS的環境
-CMD打上node {檔案名稱.js}，EG.node index.js
-而Python，則是打上 python3 {檔案名稱.py} EG.python3 index.py
\ No newline at end of file
+使用Docker對應Image和腳本來執行
+JavaScript 使用node
diff --git a/javascript/LeetCode/Array/2442.js b/javascript/LeetCode/Array/2442.js
new file mode 100644
index 0000000..5b93fff
--- /dev/null
+++ b/javascript/LeetCode/Array/2442.js
@@ -0,0 +1,23 @@
+/**
+ * 2442. Count Number of Distinct Integers After Reverse Operations
+ *
+ * 計算陣列元素digits反轉後，加上原有陣列會有幾個數字是唯一值
+ *
+ * @param {number[]} nums
+ * @return {number}
+ */
+var countDistinctIntegers = function(nums) {
+ // O(N)
+ nums.push(...nums.map(num =>
+ parseInt(num.toString().split('').reverse().join(''))
+ ));
+ return new Set(nums).size;
+};
+let nums = [1,13,10,12,31];
+/*
+Output: 6
+Explanation: After including the reverse of each number, the resulting array is [1,13,10,12,31,1,31,1,21,13].
+The reversed integers that were added to the end of the array are underlined. Note that for the integer 10, after reversing it, it becomes 01 which is just 1.
+The number of distinct integers in this array is 6 (The numbers 1, 10, 12, 13, 21, and 31).
+*/
+console.log(countDistinctIntegers(nums));
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diff --git a/javascript/LeetCode/String/1653.js b/javascript/LeetCode/String/1653.js
new file mode 100644
index 0000000..ace1096
--- /dev/null
+++ b/javascript/LeetCode/String/1653.js
@@ -0,0 +1,47 @@
+/**
+ * 1653. Minimum Deletions to Make String Balanced
+ *
+ * 參數s中只有'a' & 'b'這兩個字母。
+ * 刪除任一字母使s balanced，若不存在一對index (i,j) 使得 i < j 且 s[i] = 'b' 且 s[j] = 'a'，則s 是balanced。
+ * 回傳至少須刪除幾次(操作幾次)才能使s balanced
+ *
+ *
+ * @param {string} s
+ * @return {number}
+ */
+var minimumDeletions = function(s) {
+ // balanced string中，b不能出現在a之後
+ // no such 'b' at s[i] where s[j] is 'a' and i < j
+
+ // TC:O(N)
+ // 計算a,b各自出現幾次
+ let countA = 0,countB = 0;
+ let minDel = s.length;
+ // 先計算a出現幾次
+ for(let i = 0;i < s.length;++i) {
+ if(s[i] === "a"){
+ countA++;
+ }
+ }
+ // 之後再次迴圈，若遇到a則--
+ for(let i = 0; i < s.length;++i) {
+ if(s[i] === "a"){
+ countA--;
+ }
+ // 不斷更新比較雙方次數
+ minDel = Math.min(minDel,countA + countB);
+
+ // 遇到b，++
+ if(s[i] === "b"){
+ countB++;
+ }
+ }
+ return minDel;
+};
+let s = "aababbab";
+/*Output: 2
+Explanation: You can either:
+Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
+Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
+*/
+console.log(minimumDeletions(s));
diff --git a/javascript/LeetCode/String/3760.js b/javascript/LeetCode/String/3760.js
new file mode 100644
index 0000000..e4813ba
--- /dev/null
+++ b/javascript/LeetCode/String/3760.js
@@ -0,0 +1,31 @@
+/**
+ * 3760. Maximum Substrings With Distinct Start
+ * Difficulty:Medium
+ *
+ * @param {string} s
+ * @return {number}
+ */
+var maxDistinct = function(s) {
+ // 計算字串中，若每個開頭是跟另一substring開頭不同的字母，可以有幾種組合
+ // 計算每個字母出現次數
+
+ // let map = new Map();
+ // for(let i = 0; i < s.length;++i) {
+ // map = map.has(s[i]) ? map.set(s[i], map.get(s[i]) + 1) : map.set(s[i], 1);
+ // }
+ // return map.size;
+
+ // solution 2.
+ /**
+ * TC: O(N) =>
+ * 將s弄成陣列，須loop所有元素，因此O(N)
+ * new Set(...) 將每個元素插入set，add是O(1)但要做n次，因此O(N)
+ * size 讀取長度，因此O(1)
+ *
+ * new Set([...s]) 需要loop並插入所有元素，所以整體是O(n)
+ * */
+ return new Set([...s]).size;
+};
+let s = "abab";
+// 2
+console.log(maxDistinct(s))
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diff --git a/javascript/LeetCode/String/3884.js b/javascript/LeetCode/String/3884.js
new file mode 100644
index 0000000..21be1c9
--- /dev/null
+++ b/javascript/LeetCode/String/3884.js
@@ -0,0 +1,26 @@
+/**
+ * 3884. First Matching Character From Both Ends
+ *
+ * Return the smallest index i such that s[i] == s[s.length - i - 1].
+ * 找出最小index，須符合s[i] === s[s.length - i - 1]這條件，若沒有則-1
+ *
+ * @param {string} s
+ * @return {number}
+ */
+var firstMatchingIndex = function(s) {
+ // TC: O(N)
+ // SC: O(1)
+ let i = 0,j = s.length - 1;
+ while(i <= j){
+ if(s[i] === s[j]){
+ // 左邊index一定是最小的
+ return i;
+ }
+ i++; // 左邊index ++
+ j--; // 右邊index --
+ }
+ return -1;
+};
+let s = "abcacbd";
+// 1
+console.log(firstMatchingIndex(s));
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diff --git a/javascript/index.js b/javascript/index.js
index f11eb90..db0c900 100644
--- a/javascript/index.js
+++ b/javascript/index.js
@@ -1,9 +1,7 @@
// debugger
-import { format } from 'node:path';
import {ExecutionTimer} from './time.js';
import assert from 'node:assert/strict';
import { count } from 'node:console';
-import { lchown } from 'node:fs';
/*
22. Generate Parentheses
@@ -108,7 +106,7 @@ var findLongestWord = function (s, dictionary) {
* Given two non-negative integers num1 and num2 represented as strings, return the product of num1 and num2, also represented as a string.
* Note: You must not use any built-in BigInteger library or convert the inputs to integer directly.
*
- * Input num1 and num2 are 非負數以字串方式呈現
+ * Input num1 and num2 非負數以字串方式呈現
* Output num1 * num2(以字串方式呈現)
* 不能使用內建含式或直接把Input轉成數字
* -------------------------------------------
@@ -131,21 +129,23 @@ var findLongestWord = function (s, dictionary) {
* @return {string}
*/
var multiply = function (num1, num2) {
+ /**
+ * 不能使用內建涵式或轉換型態
+ */
- let pattern = /^[0-9]+$/;
-
- if (!num1.match(pattern) || !num2.match(pattern) || Number(num1) === 0 || Number(num2) === 0) {
- return;
+ let answer = Array(num1.length + num2.length).fill(0);
+ console.log(answer)
+ for(let i = num1.length - 1;i >= 0;i--){
+
}
-
-
-
};
// const num1 = "2", num2 = "3";
// "6"
// const num1 = "123", num2 = "456";
// "56088"
-// console.log(multiply(num1, num2));
+const num1 = "123456789",num2 = "987654321";
+// "121932631112635269"
+// console.log(multiply(num1, num2));ㄋㄋ
@@ -1258,24 +1258,61 @@ var minRemoval = function(nums, k) {
// console.log(minRemoval(nums,k));
-/**
- * 1653. Minimum Deletions to Make String Balanced
+/***
+ * 890. Find and Replace Pattern
*
- * 參數s中只有'a' & 'b'這兩個字母。
- * 刪除任一字母使s balanced，若不存在一對index (i,j) 使得 i < j 且 s[i] = 'b' 且 s[j] = 'a'，則s 是balanced。
- * 回傳最小須刪除幾次才能使s balanced
- *
- * @param {string} s
- * @return {number}
+ * @param {string[]} words
+ * @param {string} pattern
+ * @return {string[]}
*/
-var minimumDeletions = function(s) {
-
-};
-// let s = "aababbab";
-/*Output: 2
-Explanation: You can either:
-Delete the characters at 0-indexed positions 2 and 6 ("aababbab" -> "aaabbb"), or
-Delete the characters at 0-indexed positions 3 and 6 ("aababbab" -> "aabbbb").
-*/
-// console.log(minimumDeletions(s));
+var findAndReplacePattern = function(words, pattern) {
+ // solution 1.
+ // TC: O(n * m)
+ // let result = [];
+ // for(let i = 0;i < words.length;i++) {
+ // if(checkEqual(words[i],pattern)){
+ // result.push(words[i]);
+ // }
+ // }
+ // return result;
+
+ // /**
+ // * @param {string} a
+ // * @param {string} b
+ // * @return {boolean}
+ // */
+ // function checkEqual(a,b) {
+ // for(let i = 0;i < a.length;i++) {
+ // if(a.indexOf(a[i]) !== b.indexOf(b[i])){
+ // return false;
+ // }
+ // }
+ // return true;
+ // }
+
+ // solution 2.
+ // hash map
+ let result = [];
+ for(const a of words) {
+ if(checkEqual(a,pattern)){
+ result.push(a);
+ }
+ // console.log(a)
+ }
+
+ function checkEqual(a,b){
+ let map = new Map();
+ for(let i = 0;i < a.length;++i) {
+ if(!map.has(a[i])){
+ map.set(i,a[i]);
+ }
+ if(map.get(a[i]) ){
+
+ }
+ }
+ }
+};
+let word = ["abc","deq","mee","aqq","dkd","ccc"], pattern = "abb";
+// ["mee","aqq"]
+// console.log(findAndReplacePattern(word,pattern));
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