feat: add js/ts solutions to lc problem: No.1508

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README.md

@@ -31,7 +31,7 @@ tags:
 
 <pre>
 <strong>输入：</strong>nums = [1,2,3,4], n = 4, left = 1, right = 5
-<strong>输出：</strong>13 
+<strong>输出：</strong>13
 <strong>解释：</strong>所有的子数组和为 1, 3, 6, 10, 2, 5, 9, 3, 7, 4 。将它们升序排序后，我们得到新的数组 [1, 2, 3, 3, 4, 5, 6, 7, 9, 10] 。下标从 le = 1 到 ri = 5 的和为 1 + 2 + 3 + 3 + 4 = 13 。
 </pre>
 
@@ -161,6 +161,54 @@ func rangeSum(nums []int, n int, left int, right int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function rangeSum(nums: number[], n: number, left: number, right: number): number {
+    let arr = Array((n * (n + 1)) / 2).fill(0);
+    const mod = 10 ** 9 + 7;
+
+    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
+        for (let j = i; j < n; j++, k++) {
+            s += nums[j];
+            arr[k] = s;
+        }
+    }
+
+    let ans = 0;
+    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
+    for (const x of arr) {
+        ans += x;
+    }
+
+    return ans % mod;
+}
+```
+
+#### JavaScript
+
+```js
+function rangeSum(nums, n, left, right) {
+    let arr = Array((n * (n + 1)) / 2).fill(0);
+    const mod = 10 ** 9 + 7;
+
+    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
+        for (let j = i; j < n; j++, k++) {
+            s += nums[j];
+            arr[k] = s;
+        }
+    }
+
+    let ans = 0;
+    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
+    for (const x of arr) {
+        ans += x;
+    }
+
+    return ans % mod;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README_EN.md

@@ -30,8 +30,8 @@ tags:
 
 <pre>
 <strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 1, right = 5
-<strong>Output:</strong> 13 
-<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13. 
+<strong>Output:</strong> 13
+<strong>Explanation:</strong> All subarray sums are 1, 3, 6, 10, 2, 5, 9, 3, 7, 4. After sorting them in non-decreasing order we have the new array [1, 2, 3, 3, 4, 5, 6, 7, 9, 10]. The sum of the numbers from index le = 1 to ri = 5 is 1 + 2 + 3 + 3 + 4 = 13.
 </pre>
 
 <p><strong class="example">Example 2:</strong></p>
@@ -65,7 +65,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Sorting
+
+According to the problem statement, generate the `arr` array, sort it, and then sum all the elements in the range $[left-1,.. right-1]$ to get the result.
+
+Time complexity is $O(n^2 \times \log n)$, and space complexity is $O(n^2)$. Here, $n$ is the length of the array given in the problem.
 
 <!-- tabs:start -->
 
@@ -155,6 +159,54 @@ func rangeSum(nums []int, n int, left int, right int) (ans int) {
 }
 ```
 
+#### TypeScript
+
+```ts
+function rangeSum(nums: number[], n: number, left: number, right: number): number {
+    let arr = Array((n * (n + 1)) / 2).fill(0);
+    const mod = 10 ** 9 + 7;
+
+    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
+        for (let j = i; j < n; j++, k++) {
+            s += nums[j];
+            arr[k] = s;
+        }
+    }
+
+    let ans = 0;
+    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
+    for (const x of arr) {
+        ans += x;
+    }
+
+    return ans % mod;
+}
+```
+
+#### JavaScript
+
+```js
+function rangeSum(nums, n, left, right) {
+    let arr = Array((n * (n + 1)) / 2).fill(0);
+    const mod = 10 ** 9 + 7;
+
+    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
+        for (let j = i; j < n; j++, k++) {
+            s += nums[j];
+            arr[k] = s;
+        }
+    }
+
+    let ans = 0;
+    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
+    for (const x of arr) {
+        ans += x;
+    }
+
+    return ans % mod;
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/Solution.js

@@ -0,0 +1,19 @@
+function rangeSum(nums, n, left, right) {
+    let arr = Array((n * (n + 1)) / 2).fill(0);
+    const mod = 10 ** 9 + 7;
+
+    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
+        for (let j = i; j < n; j++, k++) {
+            s += nums[j];
+            arr[k] = s;
+        }
+    }
+
+    let ans = 0;
+    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
+    for (const x of arr) {
+        ans += x;
+    }
+
+    return ans % mod;
+}

solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/Solution.ts

@@ -0,0 +1,19 @@
+function rangeSum(nums: number[], n: number, left: number, right: number): number {
+    let arr = Array((n * (n + 1)) / 2).fill(0);
+    const mod = 10 ** 9 + 7;
+
+    for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
+        for (let j = i; j < n; j++, k++) {
+            s += nums[j];
+            arr[k] = s;
+        }
+    }
+
+    let ans = 0;
+    arr = arr.sort((a, b) => a - b).slice(left - 1, right);
+    for (const x of arr) {
+        ans += x;
+    }
+
+    return ans % mod;
+}