feat: add solutions to lc problem: No.1031

No.1031.Maximum Sum of Two Non-Overlapping Subarrays

Author: yanglbme

solution/1000-1099/1031.Maximum Sum of Two Non-Overlapping Subarrays/README.md

@@ -59,11 +59,11 @@
 
 接下来，我们分两种情况枚举：
 
-假设 $firstLen$ 个元素的子数组在 $secondLen$ 个元素的子数组的左边，那么我们可以枚举 $secondLen$ 个元素的子数组的左端点 $i$，用变量 $t$ 维护左边 $firstLen$ 个元素的子数组的最大和，那么答案就是 $t + s[i + secondLen] - s[i]$。枚举完所有的 $i$，就可以得到候选答案。
+假设 $firstLen$ 个元素的子数组在 $secondLen$ 个元素的子数组的左边，那么我们可以枚举 $secondLen$ 个元素的子数组的左端点 $i$，用变量 $t$ 维护左边 $firstLen$ 个元素的子数组的最大和，那么当前最大和就是 $t + s[i + secondLen] - s[i]$。其中 $s[i + secondLen] - s[i]$ 表示 $secondLen$ 个元素的子数组的和。枚举完所有的 $i$，就得到了第一种情况下的最大和。
 
-假设 $secondLen$ 个元素的子数组在 $firstLen$ 个元素的子数组的左边，那么我们可以枚举 $firstLen$ 个元素的子数组的左端点 $i$，用变量 $t$ 维护左边 $secondLen$ 个元素的子数组的最大和，那么答案就是 $t + s[i + firstLen] - s[i]$。枚举完所有的 $i$，就可以得到候选答案。
+假设 $secondLen$ 个元素的子数组在 $firstLen$ 个元素的子数组的左边，那么我们可以枚举 $firstLen$ 个元素的子数组的左端点 $i$，用变量 $t$ 维护左边 $secondLen$ 个元素的子数组的最大和，那么当前最大和就是 $t + s[i + firstLen] - s[i]$。其中 $s[i + firstLen] - s[i]$ 表示 $firstLen$ 个元素的子数组的和。枚举完所有的 $i$，就得到了第二种情况下的最大和。
 
-最后，我们取两种情况下的候选答案的最大值即可。
+取两种情况下的最大值作为答案即可。
 
 时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
 
@@ -172,6 +172,32 @@ func max(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function maxSumTwoNoOverlap(
+    nums: number[],
+    firstLen: number,
+    secondLen: number,
+): number {
+    const n = nums.length;
+    const s: number[] = new Array(n + 1).fill(0);
+    for (let i = 0; i < n; ++i) {
+        s[i + 1] = s[i] + nums[i];
+    }
+    let ans = 0;
+    for (let i = firstLen, t = 0; i + secondLen - 1 < n; ++i) {
+        t = Math.max(t, s[i] - s[i - firstLen]);
+        ans = Math.max(ans, t + s[i + secondLen] - s[i]);
+    }
+    for (let i = secondLen, t = 0; i + firstLen - 1 < n; ++i) {
+        t = Math.max(t, s[i] - s[i - secondLen]);
+        ans = Math.max(ans, t + s[i + firstLen] - s[i]);
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```

solution/1000-1099/1031.Maximum Sum of Two Non-Overlapping Subarrays/README_EN.md

@@ -47,6 +47,20 @@
 
 ## Solutions
 
+**Approach 1: Prefix Sum + Enumeration**
+
+First, we preprocess the prefix sum array $s$ of the array `nums`, where $s[i]$ indicates the sum of the first $i$ elements in `nums`.
+
+Then, we enumerate in two cases:
+
+Suppose $firstLen$ elements of the subarray are on the left side of the $secondLen$ elements of the subarray, then we can enumerate the left endpoint $i$ of the $secondLen$ elements of the subarray, use the variable $t$ to maintain the maximum sum of the left $firstLen$ elements of the subarray, then the current maximum sum is $t + s[i + secondLen] - s[i]$. Where $s[i + secondLen] - s[i]$ represents the sum of the $secondLen$ elements of the subarray. After enumerating all $i$, we get the maximum sum of the first case.
+
+Suppose the $secondLen$ elements of the subarray are on the left side of the $firstLen$ elements of the subarray, then we can enumerate the left endpoint $i$ of the $firstLen$ elements of the subarray, use the variable $t$ to maintain the maximum sum of the left $secondLen$ elements of the subarray, then the current maximum sum is $t + s[i + firstLen] - s[i]$. Where $s[i + firstLen] - s[i]$ represents the sum of the $firstLen$ elements of the subarray. After enumerating all $i$, we get the maximum sum of the second case.
+
+Take the maximum value of the two cases as the answer.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the length of the array `nums`.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -148,6 +162,32 @@ func max(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function maxSumTwoNoOverlap(
+    nums: number[],
+    firstLen: number,
+    secondLen: number,
+): number {
+    const n = nums.length;
+    const s: number[] = new Array(n + 1).fill(0);
+    for (let i = 0; i < n; ++i) {
+        s[i + 1] = s[i] + nums[i];
+    }
+    let ans = 0;
+    for (let i = firstLen, t = 0; i + secondLen - 1 < n; ++i) {
+        t = Math.max(t, s[i] - s[i - firstLen]);
+        ans = Math.max(ans, t + s[i + secondLen] - s[i]);
+    }
+    for (let i = secondLen, t = 0; i + firstLen - 1 < n; ++i) {
+        t = Math.max(t, s[i] - s[i - secondLen]);
+        ans = Math.max(ans, t + s[i + firstLen] - s[i]);
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```

solution/1000-1099/1031.Maximum Sum of Two Non-Overlapping Subarrays/Solution.ts

@@ -0,0 +1,21 @@
+function maxSumTwoNoOverlap(
+    nums: number[],
+    firstLen: number,
+    secondLen: number,
+): number {
+    const n = nums.length;
+    const s: number[] = new Array(n + 1).fill(0);
+    for (let i = 0; i < n; ++i) {
+        s[i + 1] = s[i] + nums[i];
+    }
+    let ans = 0;
+    for (let i = firstLen, t = 0; i + secondLen - 1 < n; ++i) {
+        t = Math.max(t, s[i] - s[i - firstLen]);
+        ans = Math.max(ans, t + s[i + secondLen] - s[i]);
+    }
+    for (let i = secondLen, t = 0; i + firstLen - 1 < n; ++i) {
+        t = Math.max(t, s[i] - s[i - secondLen]);
+        ans = Math.max(ans, t + s[i + firstLen] - s[i]);
+    }
+    return ans;
+}