feat: add ts solutions to lc problems: No.2543,2547

yanglbme · yanglbme · commit e70c6557d30e · 2023-01-26T20:28:00.000+08:00
* No.2543.Check if Point Is Reachable
* No.2547.Minimum Cost to Split an Array
diff --git a/solution/2500-2599/2543.Check if Point Is Reachable/README.md b/solution/2500-2599/2543.Check if Point Is Reachable/README.md
@@ -117,6 +117,19 @@ func gcd(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function isReachable(targetX: number, targetY: number): boolean {
+    const x = gcd(targetX, targetY);
+    return (x & (x - 1)) === 0;
+}
+
+function gcd(a: number, b: number): number {
+    return b == 0 ? a : gcd(b, a % b);
+}
+```
+
 ### **...**
 
 ```
diff --git a/solution/2500-2599/2543.Check if Point Is Reachable/README_EN.md b/solution/2500-2599/2543.Check if Point Is Reachable/README_EN.md
@@ -97,6 +97,19 @@ func gcd(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function isReachable(targetX: number, targetY: number): boolean {
+    const x = gcd(targetX, targetY);
+    return (x & (x - 1)) === 0;
+}
+
+function gcd(a: number, b: number): number {
+    return b == 0 ? a : gcd(b, a % b);
+}
+```
+
 ### **...**
 
 ```
diff --git a/solution/2500-2599/2543.Check if Point Is Reachable/Solution.ts b/solution/2500-2599/2543.Check if Point Is Reachable/Solution.ts
@@ -0,0 +1,8 @@
+function isReachable(targetX: number, targetY: number): boolean {
+    const x = gcd(targetX, targetY);
+    return (x & (x - 1)) === 0;
+}
+
+function gcd(a: number, b: number): number {
+    return b == 0 ? a : gcd(b, a % b);
+}
diff --git a/solution/2500-2599/2547.Minimum Cost to Split an Array/README.md b/solution/2500-2599/2547.Minimum Cost to Split an Array/README.md
@@ -236,6 +236,38 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function minCost(nums: number[], k: number): number {
+    const n = nums.length;
+    const f = new Array(n).fill(0);
+    const dfs = (i: number) => {
+        if (i >= n) {
+            return 0;
+        }
+        if (f[i]) {
+            return f[i];
+        }
+        const cnt = new Array(n).fill(0);
+        let one = 0;
+        let ans = 1 << 30;
+        for (let j = i; j < n; ++j) {
+            const x = ++cnt[nums[j]];
+            if (x == 1) {
+                ++one;
+            } else if (x == 2) {
+                --one;
+            }
+            ans = Math.min(ans, k + j - i + 1 - one + dfs(j + 1));
+        }
+        f[i] = ans;
+        return f[i];
+    };
+    return dfs(0);
+}
+```
+
 ### **...**
 
 ```
diff --git a/solution/2500-2599/2547.Minimum Cost to Split an Array/README_EN.md b/solution/2500-2599/2547.Minimum Cost to Split an Array/README_EN.md
@@ -220,6 +220,38 @@ func min(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function minCost(nums: number[], k: number): number {
+    const n = nums.length;
+    const f = new Array(n).fill(0);
+    const dfs = (i: number) => {
+        if (i >= n) {
+            return 0;
+        }
+        if (f[i]) {
+            return f[i];
+        }
+        const cnt = new Array(n).fill(0);
+        let one = 0;
+        let ans = 1 << 30;
+        for (let j = i; j < n; ++j) {
+            const x = ++cnt[nums[j]];
+            if (x == 1) {
+                ++one;
+            } else if (x == 2) {
+                --one;
+            }
+            ans = Math.min(ans, k + j - i + 1 - one + dfs(j + 1));
+        }
+        f[i] = ans;
+        return f[i];
+    };
+    return dfs(0);
+}
+```
+
 ### **...**
 
 ```
diff --git a/solution/2500-2599/2547.Minimum Cost to Split an Array/Solution.ts b/solution/2500-2599/2547.Minimum Cost to Split an Array/Solution.ts
@@ -0,0 +1,27 @@
+function minCost(nums: number[], k: number): number {
+    const n = nums.length;
+    const f = new Array(n).fill(0);
+    const dfs = (i: number) => {
+        if (i >= n) {
+            return 0;
+        }
+        if (f[i]) {
+            return f[i];
+        }
+        const cnt = new Array(n).fill(0);
+        let one = 0;
+        let ans = 1 << 30;
+        for (let j = i; j < n; ++j) {
+            const x = ++cnt[nums[j]];
+            if (x == 1) {
+                ++one;
+            } else if (x == 2) {
+                --one;
+            }
+            ans = Math.min(ans, k + j - i + 1 - one + dfs(j + 1));
+        }
+        f[i] = ans;
+        return f[i];
+    };
+    return dfs(0);
+}
