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feat: add typescript solution to lc problem: No.2060 (#607)
No.2060.Check if an Original String Exists Given Two Encoded Strings
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‎solution/2000-2099/2060.Check if an Original String Exists Given Two Encoded Strings/README.md

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Original file line numberDiff line numberDiff line change
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105105

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<!-- 这里可写通用的实现逻辑 -->
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动态规划
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108110
<!-- tabs:start -->
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### **Python3**
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```
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### **TypeScript**
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```ts
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function possiblyEquals(s1: string, s2: string): boolean {
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const n = s1.length, m = s2.length;
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let dp: Array<Array<Set<number>>> = Array.from({ length: n + 1 }, v => Array.from({ length: m + 1}, w => new Set()));
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dp[0][0].add(0);
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for (let i = 0; i <= n; i++) {
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for (let j = 0; j <= m; j++) {
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for (let delta of dp[i][j]) {
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// s1为数字
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let num = 0;
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if (delta <= 0) {
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for (let p = i; i < Math.min(i + 3, n); p++) {
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if (isDigit(s1[p])) {
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num = num * 10 + Number(s1[p]);
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dp[p + 1][j].add(delta + num);
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} else {
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break;
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}
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}
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}
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// s2为数字
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num = 0;
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if (delta >= 0) {
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for (let q = j; q < Math.min(j + 3, m); q++) {
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if (isDigit(s2[q])) {
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num = num * 10 + Number(s2[q]);
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dp[i][q + 1].add(delta - num);
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} else {
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break;
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}
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}
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}
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// 数字匹配s1为字母
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if (i < n && delta < 0 && !isDigit(s1[i])) {
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dp[i + 1][j].add(delta + 1);
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}
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// 数字匹配s2为字母
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if (j < m && delta > 0 && !isDigit(s2[j])) {
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dp[i][j + 1].add(delta - 1);
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}
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// 两个字母匹配
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if (i < n && j < m && delta == 0 && s1[i] == s2[j]) {
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dp[i + 1][j + 1].add(0);
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}
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}
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}
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}
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return dp[n][m].has(0);
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};
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function isDigit(char: string): boolean {
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return (/^\d{1}$/g).test(char);
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}
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```
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### **...**
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```

‎solution/2000-2099/2060.Check if an Original String Exists Given Two Encoded Strings/README_EN.md

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## Solutions
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Dynamic Programming
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<!-- tabs:start -->
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### **Python3**
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```
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### **TypeScript**
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```ts
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function possiblyEquals(s1: string, s2: string): boolean {
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const n = s1.length, m = s2.length;
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let dp: Array<Array<Set<number>>> = Array.from({ length: n + 1 }, v => Array.from({ length: m + 1}, w => new Set()));
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dp[0][0].add(0);
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for (let i = 0; i <= n; i++) {
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for (let j = 0; j <= m; j++) {
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for (let delta of dp[i][j]) {
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// s1为数字
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let num = 0;
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if (delta <= 0) {
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for (let p = i; i < Math.min(i + 3, n); p++) {
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if (isDigit(s1[p])) {
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num = num * 10 + Number(s1[p]);
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dp[p + 1][j].add(delta + num);
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} else {
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break;
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}
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}
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}
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// s2为数字
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num = 0;
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if (delta >= 0) {
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for (let q = j; q < Math.min(j + 3, m); q++) {
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if (isDigit(s2[q])) {
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num = num * 10 + Number(s2[q]);
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dp[i][q + 1].add(delta - num);
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} else {
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break;
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}
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}
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}
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// 数字匹配s1为字母
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if (i < n && delta < 0 && !isDigit(s1[i])) {
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dp[i + 1][j].add(delta + 1);
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}
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// 数字匹配s2为字母
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if (j < m && delta > 0 && !isDigit(s2[j])) {
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dp[i][j + 1].add(delta - 1);
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}
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// 两个字母匹配
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if (i < n && j < m && delta == 0 && s1[i] == s2[j]) {
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dp[i + 1][j + 1].add(0);
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}
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}
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}
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}
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return dp[n][m].has(0);
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};
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function isDigit(char: string): boolean {
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return (/^\d{1}$/g).test(char);
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}
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```
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121187
### **...**
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123189
```

‎solution/2000-2099/2060.Check if an Original String Exists Given Two Encoded Strings/Solution.ts

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function possiblyEquals(s1: string, s2: string): boolean {
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const n = s1.length, m = s2.length;
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let dp: Array<Array<Set<number>>> = Array.from({ length: n + 1 }, v => Array.from({ length: m + 1}, w => new Set()));
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dp[0][0].add(0);
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for (let i = 0; i <= n; i++) {
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for (let j = 0; j <= m; j++) {
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for (let delta of dp[i][j]) {
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// s1为数字
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let num = 0;
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if (delta <= 0) {
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for (let p = i; i < Math.min(i + 3, n); p++) {
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if (isDigit(s1[p])) {
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num = num * 10 + Number(s1[p]);
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dp[p + 1][j].add(delta + num);
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} else {
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break;
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}
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}
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}
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// s2为数字
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num = 0;
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if (delta >= 0) {
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for (let q = j; q < Math.min(j + 3, m); q++) {
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if (isDigit(s2[q])) {
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num = num * 10 + Number(s2[q]);
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dp[i][q + 1].add(delta - num);
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} else {
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break;
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}
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}
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}
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// 数字匹配s1为字母
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if (i < n && delta < 0 && !isDigit(s1[i])) {
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dp[i + 1][j].add(delta + 1);
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}
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// 数字匹配s2为字母
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if (j < m && delta > 0 && !isDigit(s2[j])) {
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dp[i][j + 1].add(delta - 1);
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}
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// 两个字母匹配
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if (i < n && j < m && delta == 0 && s1[i] == s2[j]) {
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dp[i + 1][j + 1].add(0);
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}
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}
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}
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}
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return dp[n][m].has(0);
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};
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function isDigit(char: string): boolean {
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return (/^\d{1}$/g).test(char);
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}

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