feat: add solutions to lc problem: No.0689 (doocs#1982)

No.0689.Maximum Sum of 3 Non-Overlapping Subarrays

Author: yanglbme

solution/0600-0699/0689.Maximum Sum of 3 Non-Overlapping Subarrays/README.md

@@ -42,8 +42,22 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：滑动窗口**
+
 滑动窗口，枚举第三个子数组的位置，同时维护前两个无重叠子数组的最大和及其位置。
 
+时间复杂度 $O(n)$，其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
+
+**方法二：预处理前后缀 + 枚举中间子数组**
+
+我们可以预处理得到数组 $nums$ 的前缀和数组 $s$，其中 $s[i] = \sum_{j=0}^{i-1} nums[j]$，那么对于任意的 $i$，$j$，$s[j] - s[i]$ 就是子数组 $[i, j)$ 的和。
+
+接下来，我们使用动态规划的方法，维护两个长度为 $n$ 的数组 $pre$ 和 $suf$，其中 $pre[i]$ 表示 $[0, i]$ 范围内长度为 $k$ 的子数组的最大和及其起始位置，$suf[i]$ 表示 $[i, n)$ 范围内长度为 $k$ 的子数组的最大和及其起始位置。
+
+然后，我们枚举中间子数组的起始位置 $i$，那么三个子数组的和就是 $pre[i-1][0] + suf[i+k][0] + (s[i+k] - s[i])$，其中 $pre[i-1][0]$ 表示 $[0, i-1]$ 范围内长度为 $k$ 的子数组的最大和，$suf[i+k][0]$ 表示 $[i+k, n)$ 范围内长度为 $k$ 的子数组的最大和，$(s[i+k] - s[i])$ 表示 $[i, i+k)$ 范围内长度为 $k$ 的子数组的和。我们找出和的最大值对应的三个子数组的起始位置即可。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -77,6 +91,36 @@ class Solution:
         return ans
 ```
 
+```python
+class Solution:
+    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
+        n = len(nums)
+        s = list(accumulate(nums, initial=0))
+        pre = [[] for _ in range(n)]
+        suf = [[] for _ in range(n)]
+        t = idx = 0
+        for i in range(n - k + 1):
+            if (cur := s[i + k] - s[i]) > t:
+                pre[i + k - 1] = [cur, i]
+                t, idx = pre[i + k - 1]
+            else:
+                pre[i + k - 1] = [t, idx]
+        t = idx = 0
+        for i in range(n - k, -1, -1):
+            if (cur := s[i + k] - s[i]) >= t:
+                suf[i] = [cur, i]
+                t, idx = suf[i]
+            else:
+                suf[i] = [t, idx]
+        t = 0
+        ans = []
+        for i in range(k, n - 2 * k + 1):
+            if (cur := s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0]) > t:
+                ans = [pre[i - 1][1], i, suf[i + k][1]]
+                t = cur
+        return ans
+```
+
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
@@ -116,6 +160,49 @@ class Solution {
 }
 ```
 
+```java
+class Solution {
+    public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
+        int n = nums.length;
+        int[] s = new int[n + 1];
+        for (int i = 0; i < n; ++i) {
+            s[i + 1] = s[i] + nums[i];
+        }
+        int[][] pre = new int[n][0];
+        int[][] suf = new int[n][0];
+        for (int i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
+            int cur = s[i + k] - s[i];
+            if (cur > t) {
+                pre[i + k - 1] = new int[] {cur, i};
+                t = cur;
+                idx = i;
+            } else {
+                pre[i + k - 1] = new int[] {t, idx};
+            }
+        }
+        for (int i = n - k, t = 0, idx = 0; i >= 0; --i) {
+            int cur = s[i + k] - s[i];
+            if (cur >= t) {
+                suf[i] = new int[] {cur, i};
+                t = cur;
+                idx = i;
+            } else {
+                suf[i] = new int[] {t, idx};
+            }
+        }
+        int[] ans = new int[0];
+        for (int i = k, t = 0; i < n - 2 * k + 1; ++i) {
+            int cur = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
+            if (cur > t) {
+                ans = new int[] {pre[i - 1][1], i, suf[i + k][1]};
+                t = cur;
+            }
+        }
+        return ans;
+    }
+}
+```
+
 ### **C++**
 
 ```cpp
@@ -154,6 +241,55 @@ public:
 };
 ```
 
+```cpp
+class Solution {
+public:
+    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
+        int n = nums.size();
+        vector<int> s(n + 1, 0);
+        for (int i = 0; i < n; ++i) {
+            s[i + 1] = s[i] + nums[i];
+        }
+
+        vector<vector<int>> pre(n, vector<int>(2, 0));
+        vector<vector<int>> suf(n, vector<int>(2, 0));
+
+        for (int i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
+            int cur = s[i + k] - s[i];
+            if (cur > t) {
+                pre[i + k - 1] = {cur, i};
+                t = cur;
+                idx = i;
+            } else {
+                pre[i + k - 1] = {t, idx};
+            }
+        }
+
+        for (int i = n - k, t = 0, idx = 0; i >= 0; --i) {
+            int cur = s[i + k] - s[i];
+            if (cur >= t) {
+                suf[i] = {cur, i};
+                t = cur;
+                idx = i;
+            } else {
+                suf[i] = {t, idx};
+            }
+        }
+
+        vector<int> ans;
+        for (int i = k, t = 0; i < n - 2 * k + 1; ++i) {
+            int cur = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
+            if (cur > t) {
+                ans = {pre[i - 1][1], i, suf[i + k][1]};
+                t = cur;
+            }
+        }
+
+        return ans;
+    }
+};
+```
+
 ### **Go**
 
 ```go
@@ -189,6 +325,102 @@ func maxSumOfThreeSubarrays(nums []int, k int) []int {
 }
 ```
 
+```go
+func maxSumOfThreeSubarrays(nums []int, k int) (ans []int) {
+	n := len(nums)
+	s := make([]int, n+1)
+	for i := 0; i < n; i++ {
+		s[i+1] = s[i] + nums[i]
+	}
+
+	pre := make([][]int, n)
+	suf := make([][]int, n)
+
+	for i, t, idx := 0, 0, 0; i < n-k+1; i++ {
+		cur := s[i+k] - s[i]
+		if cur > t {
+			pre[i+k-1] = []int{cur, i}
+			t, idx = cur, i
+		} else {
+			pre[i+k-1] = []int{t, idx}
+		}
+	}
+
+	for i, t, idx := n-k, 0, 0; i >= 0; i-- {
+		cur := s[i+k] - s[i]
+		if cur >= t {
+			suf[i] = []int{cur, i}
+			t, idx = cur, i
+		} else {
+			suf[i] = []int{t, idx}
+		}
+	}
+
+	for i, t := k, 0; i < n-2*k+1; i++ {
+		cur := s[i+k] - s[i] + pre[i-1][0] + suf[i+k][0]
+		if cur > t {
+			ans = []int{pre[i-1][1], i, suf[i+k][1]}
+			t = cur
+		}
+	}
+
+	return
+}
+```
+
+### **TypeScript**
+
+```ts
+function maxSumOfThreeSubarrays(nums: number[], k: number): number[] {
+    const n: number = nums.length;
+    const s: number[] = Array(n + 1).fill(0);
+
+    for (let i = 0; i < n; ++i) {
+        s[i + 1] = s[i] + nums[i];
+    }
+
+    const pre: number[][] = Array(n)
+        .fill([])
+        .map(() => new Array(2).fill(0));
+    const suf: number[][] = Array(n)
+        .fill([])
+        .map(() => new Array(2).fill(0));
+
+    for (let i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
+        const cur: number = s[i + k] - s[i];
+        if (cur > t) {
+            pre[i + k - 1] = [cur, i];
+            t = cur;
+            idx = i;
+        } else {
+            pre[i + k - 1] = [t, idx];
+        }
+    }
+
+    for (let i = n - k, t = 0, idx = 0; i >= 0; --i) {
+        const cur: number = s[i + k] - s[i];
+        if (cur >= t) {
+            suf[i] = [cur, i];
+            t = cur;
+            idx = i;
+        } else {
+            suf[i] = [t, idx];
+        }
+    }
+
+    let ans: number[] = [];
+    for (let i = k, t = 0; i < n - 2 * k + 1; ++i) {
+        const cur: number = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
+        if (cur > t) {
+            ans = [pre[i - 1][1], i, suf[i + k][1]];
+            t = cur;
+        }
+    }
+
+    return ans;
+}
+```
+
 ### **...**
 
 ```