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Commit f54cc81

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feat: add js/ts solutions to lc problem: No.1508 (doocs#3361)
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‎solution/1500-1599/1508.Range Sum of Sorted Subarray Sums/README.md

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<pre>
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<strong>输入:</strong>nums = [1,2,3,4], n = 4, left = 1, right = 5
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<strong>输出:</strong>13
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<strong>输出:</strong>13
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<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 +たす 2 +たす 3 +たす 3 +たす 4 =わ 13 。
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</pre>
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}
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```
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#### TypeScript
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```ts
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function rangeSum(nums: number[], n: number, left: number, right: number): number {
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let arr = Array((n * (n + 1)) / 2).fill(0);
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const mod = 10 ** 9 + 7;
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for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
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for (let j = i; j < n; j++, k++) {
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s += nums[j];
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arr[k] = s;
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}
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}
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let ans = 0;
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arr = arr.sort((a, b) => a - b).slice(left - 1, right);
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for (const x of arr) {
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ans += x;
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}
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return ans % mod;
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}
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```
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#### JavaScript
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```js
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function rangeSum(nums, n, left, right) {
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let arr = Array((n * (n + 1)) / 2).fill(0);
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const mod = 10 ** 9 + 7;
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for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
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for (let j = i; j < n; j++, k++) {
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s += nums[j];
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arr[k] = s;
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}
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}
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let ans = 0;
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arr = arr.sort((a, b) => a - b).slice(left - 1, right);
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for (const x of arr) {
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ans += x;
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}
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return ans % mod;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

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

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<pre>
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<strong>Input:</strong> nums = [1,2,3,4], n = 4, left = 1, right = 5
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<strong>Output:</strong> 13
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<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 +たす 2 +たす 3 +たす 3 +たす 4 =わ 13.
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<strong>Output:</strong> 13
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<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 +たす 2 +たす 3 +たす 3 +たす 4 =わ 13.
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<!-- solution:start -->
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### Solution 1
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### Solution 1: Sorting
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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.
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Time complexity is $O(n^2 \times \log n),ドル and space complexity is $O(n^2)$. Here, $n$ is the length of the array given in the problem.
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function rangeSum(nums: number[], n: number, left: number, right: number): number {
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let arr = Array((n * (n + 1)) / 2).fill(0);
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const mod = 10 ** 9 + 7;
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for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
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for (let j = i; j < n; j++, k++) {
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s += nums[j];
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arr[k] = s;
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}
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}
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let ans = 0;
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arr = arr.sort((a, b) => a - b).slice(left - 1, right);
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for (const x of arr) {
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ans += x;
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}
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return ans % mod;
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}
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```
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#### JavaScript
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```js
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function rangeSum(nums, n, left, right) {
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let arr = Array((n * (n + 1)) / 2).fill(0);
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const mod = 10 ** 9 + 7;
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for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
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for (let j = i; j < n; j++, k++) {
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s += nums[j];
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arr[k] = s;
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}
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}
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let ans = 0;
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arr = arr.sort((a, b) => a - b).slice(left - 1, right);
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for (const x of arr) {
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ans += x;
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}
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return ans % mod;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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function rangeSum(nums, n, left, right) {
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let arr = Array((n * (n + 1)) / 2).fill(0);
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const mod = 10 ** 9 + 7;
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for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
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for (let j = i; j < n; j++, k++) {
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s += nums[j];
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arr[k] = s;
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}
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}
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let ans = 0;
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arr = arr.sort((a, b) => a - b).slice(left - 1, right);
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for (const x of arr) {
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ans += x;
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}
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return ans % mod;
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}
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function rangeSum(nums: number[], n: number, left: number, right: number): number {
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let arr = Array((n * (n + 1)) / 2).fill(0);
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const mod = 10 ** 9 + 7;
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for (let i = 0, s = 0, k = 0; i < n; i++, s = 0) {
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for (let j = i; j < n; j++, k++) {
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s += nums[j];
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arr[k] = s;
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}
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}
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let ans = 0;
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arr = arr.sort((a, b) => a - b).slice(left - 1, right);
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for (const x of arr) {
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ans += x;
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}
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return ans % mod;
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}

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