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feat: add solutions to lc problem: No.1425
No.1425.Constrained Subsequence Sum
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‎README.md

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@@ -59,6 +59,7 @@
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- [滑动窗口最大值](/solution/0200-0299/0239.Sliding%20Window%20Maximum/README.md) - `单调队列`
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- [满足不等式的最大值](/solution/1400-1499/1499.Max%20Value%20of%20Equation/README.md) - `单调队列`
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- [和至少为 K 的最短子数组](/solution/0800-0899/0862.Shortest%20Subarray%20with%20Sum%20at%20Least%20K/README.md) - `单调队列`
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- [带限制的子序列和](/solution/1400-1499/1425.Constrained%20Subsequence%20Sum/README.md) - `动态规划`、`单调队列优化`
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### 3. 搜索
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‎README_EN.md

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@@ -57,6 +57,7 @@ Complete solutions to [LeetCode](https://leetcode.com/problemset/all/), [LCOF](h
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- [Sliding Window Maximum](/solution/0200-0299/0239.Sliding%20Window%20Maximum/README_EN.md) - `Monotonic Queue`
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- [Max Value of Equation](/solution/1400-1499/1499.Max%20Value%20of%20Equation/README_EN.md) - `Monotonic Queue`
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- [Shortest Subarray with Sum at Least K](/solution/0800-0899/0862.Shortest%20Subarray%20with%20Sum%20at%20Least%20K/README_EN.md) - `Monotonic Queue`
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- [带限制的子序列和](/solution/1400-1499/1425.Constrained%20Subsequence%20Sum/README.md) - `Dynamic Programming`, `Monotonic Queue`
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### 3. Search
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‎solution/1400-1499/1425.Constrained Subsequence Sum/README.md

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<!-- 这里可写通用的实现逻辑 -->
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**方法一：动态规划 + 单调队列**
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
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n = len(nums)
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dp = [0] * n
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ans = float('-inf')
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q = deque()
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for i, v in enumerate(nums):
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if q and i - q[0] > k:
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q.popleft()
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dp[i] = max(0, 0 if not q else dp[q[0]]) + v
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while q and dp[q[-1]] <= dp[i]:
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q.pop()
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q.append(i)
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ans = max(ans, dp[i])
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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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```java
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class Solution {
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public int constrainedSubsetSum(int[] nums, int k) {
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int n = nums.length;
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int[] dp = new int[n];
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int ans = Integer.MIN_VALUE;
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Deque<Integer> q = new ArrayDeque<>();
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for (int i = 0; i < n; ++i) {
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if (!q.isEmpty() && i - q.peek() > k) {
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q.poll();
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}
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dp[i] = Math.max(0, q.isEmpty() ? 0 : dp[q.peek()]) + nums[i];
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while (!q.isEmpty() && dp[q.peekLast()] <= dp[i]) {
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q.pollLast();
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}
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q.offer(i);
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ans = Math.max(ans, dp[i]);
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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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class Solution {
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public:
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int constrainedSubsetSum(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> dp(n);
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int ans = INT_MIN;
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deque<int> q;
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for (int i = 0; i < n; ++i)
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{
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if (!q.empty() && i - q.front() > k) q.pop_front();
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dp[i] = max(0, q.empty() ? 0 : dp[q.front()]) + nums[i];
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ans = max(ans, dp[i]);
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while (!q.empty() && dp[q.back()] <= dp[i]) q.pop_back();
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q.push_back(i);
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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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func constrainedSubsetSum(nums []int, k int) int {
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n := len(nums)
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dp := make([]int, n)
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ans := math.MinInt32
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q := []int{}
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for i, v := range nums {
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if len(q) > 0 && i-q[0] > k {
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q = q[1:]
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}
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dp[i] = v
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if len(q) > 0 && dp[q[0]] > 0 {
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dp[i] += dp[q[0]]
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}
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for len(q) > 0 && dp[q[len(q)-1]] < dp[i] {
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q = q[:len(q)-1]
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}
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q = append(q, i)
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ans = max(ans, dp[i])
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1425.Constrained Subsequence Sum/README_EN.md

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### **Python3**
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```python
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class Solution:
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def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
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n = len(nums)
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dp = [0] * n
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ans = float('-inf')
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q = deque()
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for i, v in enumerate(nums):
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if q and i - q[0] > k:
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q.popleft()
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dp[i] = max(0, 0 if not q else dp[q[0]]) + v
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while q and dp[q[-1]] <= dp[i]:
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q.pop()
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q.append(i)
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ans = max(ans, dp[i])
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return ans
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```
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### **Java**
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```java
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class Solution {
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public int constrainedSubsetSum(int[] nums, int k) {
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int n = nums.length;
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int[] dp = new int[n];
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int ans = Integer.MIN_VALUE;
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Deque<Integer> q = new ArrayDeque<>();
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for (int i = 0; i < n; ++i) {
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if (!q.isEmpty() && i - q.peek() > k) {
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q.poll();
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}
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dp[i] = Math.max(0, q.isEmpty() ? 0 : dp[q.peek()]) + nums[i];
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while (!q.isEmpty() && dp[q.peekLast()] <= dp[i]) {
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q.pollLast();
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}
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q.offer(i);
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ans = Math.max(ans, dp[i]);
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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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class Solution {
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public:
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int constrainedSubsetSum(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> dp(n);
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int ans = INT_MIN;
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deque<int> q;
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for (int i = 0; i < n; ++i)
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{
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if (!q.empty() && i - q.front() > k) q.pop_front();
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dp[i] = max(0, q.empty() ? 0 : dp[q.front()]) + nums[i];
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ans = max(ans, dp[i]);
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while (!q.empty() && dp[q.back()] <= dp[i]) q.pop_back();
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q.push_back(i);
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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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func constrainedSubsetSum(nums []int, k int) int {
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n := len(nums)
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dp := make([]int, n)
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ans := math.MinInt32
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q := []int{}
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for i, v := range nums {
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if len(q) > 0 && i-q[0] > k {
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q = q[1:]
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}
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dp[i] = v
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if len(q) > 0 && dp[q[0]] > 0 {
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dp[i] += dp[q[0]]
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}
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for len(q) > 0 && dp[q[len(q)-1]] < dp[i] {
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q = q[:len(q)-1]
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}
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q = append(q, i)
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ans = max(ans, dp[i])
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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### **...**

‎solution/1400-1499/1425.Constrained Subsequence Sum/Solution.cpp

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class Solution {
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public:
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int constrainedSubsetSum(vector<int>& nums, int k) {
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int n = nums.size();
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vector<int> dp(n);
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int ans = INT_MIN;
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deque<int> q;
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for (int i = 0; i < n; ++i)
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{
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if (!q.empty() && i - q.front() > k) q.pop_front();
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dp[i] = max(0, q.empty() ? 0 : dp[q.front()]) + nums[i];
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ans = max(ans, dp[i]);
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while (!q.empty() && dp[q.back()] <= dp[i]) q.pop_back();
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q.push_back(i);
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}
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return ans;
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}
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};

‎solution/1400-1499/1425.Constrained Subsequence Sum/Solution.go

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func constrainedSubsetSum(nums []int, k int) int {
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n := len(nums)
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dp := make([]int, n)
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ans := math.MinInt32
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q := []int{}
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for i, v := range nums {
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if len(q) > 0 && i-q[0] > k {
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q = q[1:]
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}
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dp[i] = v
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if len(q) > 0 && dp[q[0]] > 0 {
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dp[i] += dp[q[0]]
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}
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for len(q) > 0 && dp[q[len(q)-1]] < dp[i] {
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q = q[:len(q)-1]
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}
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q = append(q, i)
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ans = max(ans, dp[i])
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}
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return ans
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}
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func max(a, b int) int {
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if a > b {
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return a
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}
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return b
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}

‎solution/1400-1499/1425.Constrained Subsequence Sum/Solution.java

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class Solution {
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public int constrainedSubsetSum(int[] nums, int k) {
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int n = nums.length;
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int[] dp = new int[n];
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int ans = Integer.MIN_VALUE;
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Deque<Integer> q = new ArrayDeque<>();
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for (int i = 0; i < n; ++i) {
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if (!q.isEmpty() && i - q.peek() > k) {
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q.poll();
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}
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dp[i] = Math.max(0, q.isEmpty() ? 0 : dp[q.peek()]) + nums[i];
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while (!q.isEmpty() && dp[q.peekLast()] <= dp[i]) {
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q.pollLast();
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}
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q.offer(i);
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ans = Math.max(ans, dp[i]);
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}
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return ans;
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}
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}

‎solution/1400-1499/1425.Constrained Subsequence Sum/Solution.py

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class Solution:
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def constrainedSubsetSum(self, nums: List[int], k: int) -> int:
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n = len(nums)
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dp = [0] * n
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ans = float('-inf')
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q = deque()
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for i, v in enumerate(nums):
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if q and i - q[0] > k:
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q.popleft()
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dp[i] = max(0, 0 if not q else dp[q[0]]) + v
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while q and dp[q[-1]] <= dp[i]:
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q.pop()
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q.append(i)
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ans = max(ans, dp[i])
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return ans

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