feat: add typescript solution to lc problem: No.2064 (doocs#612)

No.2064. Minimized Maximum of Products Distributed to Any Store

Author: zhaocchen

solution/2000-2099/2064.Minimized Maximum of Products Distributed to Any Store/README.md

@@ -66,6 +66,8 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+二分法
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -84,6 +86,29 @@
 
 ```
 
+### **TypeScript**
+
+```ts
+function minimizedMaximum(n: number, quantities: number[]): number {
+    const m = quantities.length;
+    let left = 1, right = Math.max(...quantities);
+    while (left <= right) {
+        let mid = (left + right) >> 1;
+        let sum = 0;
+        for (let num of quantities) {
+            let cur = Math.floor((num - 1) / mid) + 1;
+            sum += cur;
+        }
+        if (sum > n) {
+            left = mid + 1;
+        } else {
+            right = mid - 1;
+        }
+    }
+    return left;
+};
+```
+
 ### **...**
 
 ```

solution/2000-2099/2064.Minimized Maximum of Products Distributed to Any Store/README_EN.md

@@ -60,6 +60,8 @@ The maximum number of products given to any store is max(100000) = 100000.
 
 ## Solutions
 
+Binary search.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -74,6 +76,29 @@ The maximum number of products given to any store is max(100000) = 100000.
 
 ```
 
+### **TypeScript**
+
+```ts
+function minimizedMaximum(n: number, quantities: number[]): number {
+    const m = quantities.length;
+    let left = 1, right = Math.max(...quantities);
+    while (left <= right) {
+        let mid = (left + right) >> 1;
+        let sum = 0;
+        for (let num of quantities) {
+            let cur = Math.floor((num - 1) / mid) + 1;
+            sum += cur;
+        }
+        if (sum > n) {
+            left = mid + 1;
+        } else {
+            right = mid - 1;
+        }
+    }
+    return left;
+};
+```
+
 ### **...**
 
 ```

solution/2000-2099/2064.Minimized Maximum of Products Distributed to Any Store/Solution.ts

@@ -0,0 +1,18 @@
+function minimizedMaximum(n: number, quantities: number[]): number {
+    const m = quantities.length;
+    let left = 1, right = Math.max(...quantities);
+    while (left <= right) {
+        let mid = (left + right) >> 1;
+        let sum = 0;
+        for (let num of quantities) {
+            let cur = Math.floor((num - 1) / mid) + 1;
+            sum += cur;
+        }
+        if (sum > n) {
+            left = mid + 1;
+        } else {
+            right = mid - 1;
+        }
+    }
+    return left;
+};