feat: add solutions to lc problem: No.2071

solution/2000-2099/2071.Maximum Number of Tasks You Can Assign/README.md

@@ -296,6 +296,69 @@ func maxTaskAssign(tasks []int, workers []int, pills int, strength int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxTaskAssign(
+    tasks: number[],
+    workers: number[],
+    pills: number,
+    strength: number,
+): number {
+    tasks.sort((a, b) => a - b);
+    workers.sort((a, b) => a - b);
+
+    const n = tasks.length;
+    const m = workers.length;
+
+    const check = (x: number): boolean => {
+        const dq = new Array<number>(x);
+        let head = 0;
+        let tail = 0;
+        const empty = () => head === tail;
+        const pushBack = (val: number) => {
+            dq[tail++] = val;
+        };
+        const popFront = () => {
+            head++;
+        };
+        const popBack = () => {
+            tail--;
+        };
+        const front = () => dq[head];
+
+        let i = 0;
+        let p = pills;
+
+        for (let j = m - x; j < m; j++) {
+            while (i < x && tasks[i] <= workers[j] + strength) {
+                pushBack(tasks[i]);
+                i++;
+            }
+
+            if (empty()) return false;
+
+            if (front() <= workers[j]) {
+                popFront();
+            } else {
+                if (p === 0) return false;
+                p--;
+                popBack();
+            }
+        }
+        return true;
+    };
+
+    let [left, right] = [0, Math.min(n, m)];
+    while (left < right) {
+        const mid = (left + right + 1) >> 1;
+        if (check(mid)) left = mid;
+        else right = mid - 1;
+    }
+    return left;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2000-2099/2071.Maximum Number of Tasks You Can Assign/README_EN.md

@@ -84,7 +84,21 @@ The last pill is not given because it will not make any worker strong enough for
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Greedy + Binary Search
+
+Sort the tasks in ascending order of completion time and the workers in ascending order of ability.
+
+Suppose the number of tasks we want to assign is $x$. We can greedily assign the first $x$ tasks to the $x$ workers with the highest strength. If it is possible to complete $x$ tasks, then it is also possible to complete $x-1$, $x-2$, $x-3$, ..., $1$, $0$ tasks. Therefore, we can use binary search to find the maximum $x$ such that it is possible to complete $x$ tasks.
+
+We define a function $check(x)$ to determine whether it is possible to complete $x$ tasks.
+
+The implementation of $check(x)$ is as follows:
+
+Iterate through the $x$ workers with the highest strength in ascending order. Let the current worker being processed be $j$. The current available tasks must satisfy $tasks[i] \leq workers[j] + strength$.
+
+If the smallest required strength task $task[i]$ among the current available tasks is less than or equal to $workers[j]$, then worker $j$ can complete task $task[i]$ without using a pill. Otherwise, the current worker must use a pill. If there are pills remaining, use one pill and complete the task with the highest required strength among the current available tasks. Otherwise, return `false`.
+
+The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$, where $n$ is the number of tasks.
 
 <!-- tabs:start -->
 
@@ -272,6 +286,69 @@ func maxTaskAssign(tasks []int, workers []int, pills int, strength int) int {
 }
 ```
 
+#### TypeScript
+
+```ts
+function maxTaskAssign(
+    tasks: number[],
+    workers: number[],
+    pills: number,
+    strength: number,
+): number {
+    tasks.sort((a, b) => a - b);
+    workers.sort((a, b) => a - b);
+
+    const n = tasks.length;
+    const m = workers.length;
+
+    const check = (x: number): boolean => {
+        const dq = new Array<number>(x);
+        let head = 0;
+        let tail = 0;
+        const empty = () => head === tail;
+        const pushBack = (val: number) => {
+            dq[tail++] = val;
+        };
+        const popFront = () => {
+            head++;
+        };
+        const popBack = () => {
+            tail--;
+        };
+        const front = () => dq[head];
+
+        let i = 0;
+        let p = pills;
+
+        for (let j = m - x; j < m; j++) {
+            while (i < x && tasks[i] <= workers[j] + strength) {
+                pushBack(tasks[i]);
+                i++;
+            }
+
+            if (empty()) return false;
+
+            if (front() <= workers[j]) {
+                popFront();
+            } else {
+                if (p === 0) return false;
+                p--;
+                popBack();
+            }
+        }
+        return true;
+    };
+
+    let [left, right] = [0, Math.min(n, m)];
+    while (left < right) {
+        const mid = (left + right + 1) >> 1;
+        if (check(mid)) left = mid;
+        else right = mid - 1;
+    }
+    return left;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/2000-2099/2071.Maximum Number of Tasks You Can Assign/Solution.ts

@@ -0,0 +1,58 @@
+function maxTaskAssign(
+    tasks: number[],
+    workers: number[],
+    pills: number,
+    strength: number,
+): number {
+    tasks.sort((a, b) => a - b);
+    workers.sort((a, b) => a - b);
+
+    const n = tasks.length;
+    const m = workers.length;
+
+    const check = (x: number): boolean => {
+        const dq = new Array<number>(x);
+        let head = 0;
+        let tail = 0;
+        const empty = () => head === tail;
+        const pushBack = (val: number) => {
+            dq[tail++] = val;
+        };
+        const popFront = () => {
+            head++;
+        };
+        const popBack = () => {
+            tail--;
+        };
+        const front = () => dq[head];
+
+        let i = 0;
+        let p = pills;
+
+        for (let j = m - x; j < m; j++) {
+            while (i < x && tasks[i] <= workers[j] + strength) {
+                pushBack(tasks[i]);
+                i++;
+            }
+
+            if (empty()) return false;
+
+            if (front() <= workers[j]) {
+                popFront();
+            } else {
+                if (p === 0) return false;
+                p--;
+                popBack();
+            }
+        }
+        return true;
+    };
+
+    let [left, right] = [0, Math.min(n, m)];
+    while (left < right) {
+        const mid = (left + right + 1) >> 1;
+        if (check(mid)) left = mid;
+        else right = mid - 1;
+    }
+    return left;
+}