feat: add typescript solution to lc problem: No.2186

No.2186.Minimum Number of Steps to Make Two Strings Anagram II

Author: zhaocchen

solution/2100-2199/2186.Minimum Number of Steps to Make Two Strings Anagram II/README.md

@@ -66,6 +66,21 @@
 ### **TypeScript**
 
 ```ts
+function minSteps(s: string, t: string): number {
+    let count1 = new Array(128).fill(0);
+    let count2 = new Array(128).fill(0);
+    for (let char of s) {
+        count1[char.charCodeAt(0)]++;
+    }
+    for (let char of t) {
+        count2[char.charCodeAt(0)]++;
+    }
+    let ans = 0;
+    for (let i = 0; i < 128; i++) {
+        ans += Math.abs(count1[i] - count2[i]);
+    }
+    return ans;
+};
 ```
 
 ### **...**

solution/2100-2199/2186.Minimum Number of Steps to Make Two Strings Anagram II/README_EN.md

@@ -59,6 +59,21 @@ It can be shown that there is no way to make them anagrams of each other with le
 ### **TypeScript**
 
 ```ts
+function minSteps(s: string, t: string): number {
+    let count1 = new Array(128).fill(0);
+    let count2 = new Array(128).fill(0);
+    for (let char of s) {
+        count1[char.charCodeAt(0)]++;
+    }
+    for (let char of t) {
+        count2[char.charCodeAt(0)]++;
+    }
+    let ans = 0;
+    for (let i = 0; i < 128; i++) {
+        ans += Math.abs(count1[i] - count2[i]);
+    }
+    return ans;
+};
 ```
 
 ### **...**

solution/2100-2199/2186.Minimum Number of Steps to Make Two Strings Anagram II/Solution.ts

@@ -0,0 +1,15 @@
+function minSteps(s: string, t: string): number {
+    let count1 = new Array(128).fill(0);
+    let count2 = new Array(128).fill(0);
+    for (let char of s) {
+        count1[char.charCodeAt(0)]++;
+    }
+    for (let char of t) {
+        count2[char.charCodeAt(0)]++;
+    }
+    let ans = 0;
+    for (let i = 0; i < 128; i++) {
+        ans += Math.abs(count1[i] - count2[i]);
+    }
+    return ans;
+};