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Commit 935ac39

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feat: add solutions to lc problem: No.0689 (#1982)
No.0689.Maximum Sum of 3 Non-Overlapping Subarrays
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‎solution/0600-0699/0689.Maximum Sum of 3 Non-Overlapping Subarrays/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:滑动窗口**
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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。
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时间复杂度 $O(n),ドル其中 $n$ 是数组 $nums$ 的长度。空间复杂度 $O(1)$。
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**方法二:预处理前后缀 + 枚举中间子数组**
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我们可以预处理得到数组 $nums$ 的前缀和数组 $s,ドル其中 $s[i] = \sum_{j=0}^{i-1} nums[j],ドル那么对于任意的 $i,ドル$j,ドル$s[j] - s[i]$ 就是子数组 $[i, j)$ 的和。
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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,ドル其中 $pre[i]$ 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。
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然后,我们枚举中间子数组的起始位置 $i,ドル那么三个子数组的和就是 $pre[i-1][0] + suf[i+k][0] + (s[i+k] - s[i]),ドル其中 $pre[i-1][0]$ 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。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 是数组 $nums$ 的长度。
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<!-- tabs:start -->
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### **Python3**
@@ -77,6 +91,36 @@ class Solution:
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return ans
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```
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```python
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class Solution:
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def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
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n = len(nums)
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s = list(accumulate(nums, initial=0))
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pre = [[] for _ in range(n)]
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suf = [[] for _ in range(n)]
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t = idx = 0
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for i in range(n - k + 1):
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if (cur := s[i + k] - s[i]) > t:
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pre[i + k - 1] = [cur, i]
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t, idx = pre[i + k - 1]
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else:
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pre[i + k - 1] = [t, idx]
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t = idx = 0
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for i in range(n - k, -1, -1):
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if (cur := s[i + k] - s[i]) >= t:
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suf[i] = [cur, i]
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t, idx = suf[i]
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else:
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suf[i] = [t, idx]
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t = 0
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ans = []
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for i in range(k, n - 2 * k + 1):
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if (cur := s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0]) > t:
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ans = [pre[i - 1][1], i, suf[i + k][1]]
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t = cur
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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}
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```
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```java
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class Solution {
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public int[] maxSumOfThreeSubarrays(int[] nums, int k) {
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int n = nums.length;
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int[] s = new int[n + 1];
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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int[][] pre = new int[n][0];
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int[][] suf = new int[n][0];
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for (int i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
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int cur = s[i + k] - s[i];
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if (cur > t) {
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pre[i + k - 1] = new int[] {cur, i};
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t = cur;
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idx = i;
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} else {
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pre[i + k - 1] = new int[] {t, idx};
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}
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}
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for (int i = n - k, t = 0, idx = 0; i >= 0; --i) {
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int cur = s[i + k] - s[i];
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if (cur >= t) {
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suf[i] = new int[] {cur, i};
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t = cur;
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idx = i;
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} else {
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suf[i] = new int[] {t, idx};
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}
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}
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int[] ans = new int[0];
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for (int i = k, t = 0; i < n - 2 * k + 1; ++i) {
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int cur = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
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if (cur > t) {
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ans = new int[] {pre[i - 1][1], i, suf[i + k][1]};
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t = cur;
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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};
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```
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```cpp
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class Solution {
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public:
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vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> s(n + 1, 0);
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for (int i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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vector<vector<int>> pre(n, vector<int>(2, 0));
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vector<vector<int>> suf(n, vector<int>(2, 0));
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for (int i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
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int cur = s[i + k] - s[i];
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if (cur > t) {
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pre[i + k - 1] = {cur, i};
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t = cur;
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idx = i;
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} else {
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pre[i + k - 1] = {t, idx};
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}
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}
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for (int i = n - k, t = 0, idx = 0; i >= 0; --i) {
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int cur = s[i + k] - s[i];
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if (cur >= t) {
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suf[i] = {cur, i};
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t = cur;
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idx = i;
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} else {
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suf[i] = {t, idx};
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}
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}
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vector<int> ans;
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for (int i = k, t = 0; i < n - 2 * k + 1; ++i) {
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int cur = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
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if (cur > t) {
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ans = {pre[i - 1][1], i, suf[i + k][1]};
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t = cur;
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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}
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```
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```go
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func maxSumOfThreeSubarrays(nums []int, k int) (ans []int) {
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n := len(nums)
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s := make([]int, n+1)
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for i := 0; i < n; i++ {
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s[i+1] = s[i] + nums[i]
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}
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pre := make([][]int, n)
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suf := make([][]int, n)
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for i, t, idx := 0, 0, 0; i < n-k+1; i++ {
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cur := s[i+k] - s[i]
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if cur > t {
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pre[i+k-1] = []int{cur, i}
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t, idx = cur, i
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} else {
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pre[i+k-1] = []int{t, idx}
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}
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}
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for i, t, idx := n-k, 0, 0; i >= 0; i-- {
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cur := s[i+k] - s[i]
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if cur >= t {
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suf[i] = []int{cur, i}
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t, idx = cur, i
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} else {
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suf[i] = []int{t, idx}
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}
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}
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for i, t := k, 0; i < n-2*k+1; i++ {
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cur := s[i+k] - s[i] + pre[i-1][0] + suf[i+k][0]
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if cur > t {
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ans = []int{pre[i-1][1], i, suf[i+k][1]}
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t = cur
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function maxSumOfThreeSubarrays(nums: number[], k: number): number[] {
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const n: number = nums.length;
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const s: number[] = Array(n + 1).fill(0);
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for (let i = 0; i < n; ++i) {
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s[i + 1] = s[i] + nums[i];
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}
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const pre: number[][] = Array(n)
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.fill([])
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.map(() => new Array(2).fill(0));
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const suf: number[][] = Array(n)
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.fill([])
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.map(() => new Array(2).fill(0));
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for (let i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
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const cur: number = s[i + k] - s[i];
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if (cur > t) {
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pre[i + k - 1] = [cur, i];
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t = cur;
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idx = i;
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} else {
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pre[i + k - 1] = [t, idx];
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}
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}
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for (let i = n - k, t = 0, idx = 0; i >= 0; --i) {
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const cur: number = s[i + k] - s[i];
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if (cur >= t) {
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suf[i] = [cur, i];
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t = cur;
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idx = i;
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} else {
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suf[i] = [t, idx];
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}
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}
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let ans: number[] = [];
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for (let i = k, t = 0; i < n - 2 * k + 1; ++i) {
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const cur: number = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
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if (cur > t) {
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ans = [pre[i - 1][1], i, suf[i + k][1]];
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t = cur;
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}
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}
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return ans;
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}
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```
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### **...**
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```

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