update: 添加问题“2302.统计得分小于K的子数组数目”的代码和题解(#905)

LetMeFly666 · LetMeFly666 · commit 45bc686c5383 · 2025-04-30T17:21:27.000+08:00
Signed-off-by: LetMeFly666 &lt;814114971@qq.com&gt;
diff --git a/Codes/2302-count-subarrays-with-score-less-than-k.cpp b/Codes/2302-count-subarrays-with-score-less-than-k.cpp
@@ -0,0 +1,56 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-29 23:46:23
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-29 23:58:18
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+typedef long long ll;
+
+class Solution {
+public:
+    long long countSubarrays(vector<int>& nums, long long k) {
+        vector<ll> suffix(nums.size() + 1);
+        for (int i = 0; i < nums.size(); i++) {
+            suffix[i + 1] = suffix[i] + nums[i];
+        }
+        ll ans = 0;
+        for (int l = 0, r = 0; r < nums.size(); r++) {
+            // while ((suffix[r + 1] - suffix[l]) * (r - l + 1) >= k && l < r) {  // 不可，否则可能因为l=r退出循环，导致[i, i]大于k但仍被统计
+            while ((suffix[r + 1] - suffix[l]) * (r - l + 1) >= k) {
+                l++;
+            }
+            ans += r - l + 1;
+            // cout << "[" << l << ", " << r << "]" << endl;
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+[2,1,4,3,5]
+10
+
+6
+*/
+/*
+[9,5,3,8,4,7,2,7,4,5,4,9,1,4,8,10,8,10,4,7]
+4
+
+3
+*/
+int main() {
+    string s;
+    int k;
+    while (cin >> s >> k) {
+        vector<int> v = stringToVector(s);
+        Solution sol;
+        cout << sol.countSubarrays(v, k) << endl;
+    }
+    return 0;
+}
+#endif
diff --git a/Codes/2302-count-subarrays-with-score-less-than-k.go b/Codes/2302-count-subarrays-with-score-less-than-k.go
@@ -0,0 +1,21 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-29 23:46:45
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-30 10:55:53
+ */
+package main
+
+func countSubarrays(nums []int, k int64) (ans int64) {
+    cnt := int64(0)
+    l := 0
+    for r, t := range nums {
+        cnt += int64(t)
+        for cnt * int64(r - l + 1) >= k {
+            cnt -= int64(nums[l])
+            l++
+        }
+        ans += int64(r - l + 1)
+    }
+    return
+}
diff --git a/Codes/2302-count-subarrays-with-score-less-than-k.java b/Codes/2302-count-subarrays-with-score-less-than-k.java
@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-29 23:46:43
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-30 10:41:19
+ */
+class Solution {
+    public long countSubarrays(int[] nums, long k) {
+        long ans = 0, cnt = 0;
+        for (int l = 0, r = 0; r < nums.length; r++) {
+            cnt += nums[r];
+            while (cnt * (r - l + 1) >= k) {
+                cnt -= nums[l++];
+            }
+            ans += (r - l + 1);
+        }
+        return ans;
+    }
+}
diff --git a/Codes/2302-count-subarrays-with-score-less-than-k.py b/Codes/2302-count-subarrays-with-score-less-than-k.py
@@ -0,0 +1,18 @@
+'''
+Author: LetMeFly
+Date: 2025-04-29 23:46:38
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-04-30 10:39:13
+'''
+from typing import List
+
+class Solution:
+    def countSubarrays(self, nums: List[int], k: int) -> int:
+        ans = cnt = l = 0
+        for r in range(len(nums)):
+            cnt += nums[r]
+            while cnt * (r - l + 1) >= k:
+                cnt -= nums[l]
+                l += 1
+            ans += (r - l + 1)
+        return ans
diff --git a/Codes/2302-count-subarrays-with-score-less-than-k_notSuffix.cpp b/Codes/2302-count-subarrays-with-score-less-than-k_notSuffix.cpp
@@ -0,0 +1,26 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-30 10:29:37
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-30 10:34:24
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+typedef long long ll;
+
+class Solution {
+public:
+    long long countSubarrays(vector<int>& nums, long long k) {
+        ll ans = 0, cnt = 0;
+        for (int l = 0, r = 0; r < nums.size(); r++) {
+            cnt += nums[r];
+            while (cnt * (r - l + 1) >= k) {
+                cnt -= nums[l++];
+            }
+            ans += (r - l + 1);
+        }
+        return ans;
+    }
+};
diff --git a/README.md b/README.md
@@ -705,6 +705,7 @@
 |2296.设计一个文本编辑器|困难|<a href="https://leetcode.cn/problems/design-a-text-editor/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/02/27/LeetCode%202296.%E8%AE%BE%E8%AE%A1%E4%B8%80%E4%B8%AA%E6%96%87%E6%9C%AC%E7%BC%96%E8%BE%91%E5%99%A8/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/145898693" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/design-a-text-editor/solutions/3427866/letmefly-2296she-ji-yi-ge-wen-ben-bian-j-bgln/" target="_blank">LeetCode题解</a>|
 |2299.强密码检验器II|简单|<a href="https://leetcode.cn/problems/strong-password-checker-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/01/19/LeetCode%202299.%E5%BC%BA%E5%AF%86%E7%A0%81%E6%A3%80%E9%AA%8C%E5%99%A8II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128738747" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/strong-password-checker-ii/solutions/2070064/letmefly-2299qiang-mi-ma-jian-yan-qi-ii-60u29/" target="_blank">LeetCode题解</a>|
 |2300.咒语和药水的成功对数|中等|<a href="https://leetcode.cn/problems/successful-pairs-of-spells-and-potions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/11/10/LeetCode%202300.%E5%92%92%E8%AF%AD%E5%92%8C%E8%8D%AF%E6%B0%B4%E7%9A%84%E6%88%90%E5%8A%9F%E5%AF%B9%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134332611" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/successful-pairs-of-spells-and-potions/solutions/2520817/letmefly-2300zhou-yu-he-yao-shui-de-chen-xi4c/" target="_blank">LeetCode题解</a>|
+|2302.统计得分小于K的子数组数目|困难|<a href="https://leetcode.cn/problems/count-subarrays-with-score-less-than-k/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/04/30/LeetCode%202302.%E7%BB%9F%E8%AE%A1%E5%BE%97%E5%88%86%E5%B0%8F%E4%BA%8EK%E7%9A%84%E5%AD%90%E6%95%B0%E7%BB%84%E6%95%B0%E7%9B%AE/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/147637062" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/count-subarrays-with-score-less-than-k/solutions/3666361/letmefly-2302tong-ji-de-fen-xiao-yu-k-de-3vkc/" target="_blank">LeetCode题解</a>|
 |2303.计算应缴税款总额|简单|<a href="https://leetcode.cn/problems/calculate-amount-paid-in-taxes/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/01/23/LeetCode%202303.%E8%AE%A1%E7%AE%97%E5%BA%94%E7%BC%B4%E7%A8%8E%E6%AC%BE%E6%80%BB%E9%A2%9D/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128751683" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/calculate-amount-paid-in-taxes/solutions/2073612/letmefly-2303ji-suan-ying-jiao-shui-kuan-lzsi/" target="_blank">LeetCode题解</a>|
 |2304.网格中的最小路径代价|中等|<a href="https://leetcode.cn/problems/minimum-path-cost-in-a-grid/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/11/22/LeetCode%202304.%E7%BD%91%E6%A0%BC%E4%B8%AD%E7%9A%84%E6%9C%80%E5%B0%8F%E8%B7%AF%E5%BE%84%E4%BB%A3%E4%BB%B7/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/134563145" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/minimum-path-cost-in-a-grid/solutions/2537837/letmefly-2304wang-ge-zhong-de-zui-xiao-l-bjad/" target="_blank">LeetCode题解</a>|
 |2309.兼具大小写的最好英文字母|简单|<a href="https://leetcode.cn/problems/greatest-english-letter-in-upper-and-lower-case/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/01/27/LeetCode%202309.%E5%85%BC%E5%85%B7%E5%A4%A7%E5%B0%8F%E5%86%99%E7%9A%84%E6%9C%80%E5%A5%BD%E8%8B%B1%E6%96%87%E5%AD%97%E6%AF%8D/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/128769360" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/greatest-english-letter-in-upper-and-lower-case/solutions/2077745/letmefly-2309jian-ju-da-xiao-xie-de-zui-3hgag/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 2302.统计得分小于K的子数组数目.md b/Solutions/LeetCode 2302.统计得分小于K的子数组数目.md
@@ -0,0 +1,181 @@
+---
+title: 2302.统计得分小于 K 的子数组数目：滑动窗口（不需要前缀和）
+date: 2025-04-30 17:01:00
+tags: [题解, LeetCode, 困难, 数组, 二分查找, 前缀和, 滑动窗口]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】2302.统计得分小于 K 的子数组数目：滑动窗口（不需要前缀和）
+
+力扣题目链接：[https://leetcode.cn/problems/count-subarrays-with-score-less-than-k/](https://leetcode.cn/problems/count-subarrays-with-score-less-than-k/)
+
+<p>一个数组的 <strong>分数</strong>&nbsp;定义为数组之和 <strong>乘以</strong>&nbsp;数组的长度。</p>
+
+<ul>
+	<li>比方说，<code>[1, 2, 3, 4, 5]</code>&nbsp;的分数为&nbsp;<code>(1 + 2 + 3 + 4 + 5) * 5 = 75</code>&nbsp;。</li>
+</ul>
+
+<p>给你一个正整数数组&nbsp;<code>nums</code>&nbsp;和一个整数&nbsp;<code>k</code>&nbsp;，请你返回&nbsp;<code>nums</code>&nbsp;中分数&nbsp;<strong>严格小于&nbsp;</strong><code>k</code>&nbsp;的&nbsp;<strong>非空整数子数组数目</strong>。</p>
+
+<p><strong>子数组</strong> 是数组中的一个连续元素序列。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [2,1,4,3,5], k = 10
+<b>输出：</b>6
+<strong>解释：</strong>
+有 6 个子数组的分数小于 10 ：
+- [2] 分数为 2 * 1 = 2 。
+- [1] 分数为 1 * 1 = 1 。
+- [4] 分数为 4 * 1 = 4 。
+- [3] 分数为 3 * 1 = 3 。 
+- [5] 分数为 5 * 1 = 5 。
+- [2,1] 分数为 (2 + 1) * 2 = 6 。
+注意，子数组 [1,4] 和 [4,3,5] 不符合要求，因为它们的分数分别为 10 和 36，但我们要求子数组的分数严格小于 10 。</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<b>输入：</b>nums = [1,1,1], k = 5
+<b>输出：</b>5
+<strong>解释：</strong>
+除了 [1,1,1] 以外每个子数组分数都小于 5 。
+[1,1,1] 分数为 (1 + 1 + 1) * 3 = 9 ，大于 5 。
+所以总共有 5 个子数组得分小于 5 。
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= k &lt;= 10<sup>15</sup></code></li>
+</ul>
+
+Hard里面的简单题。
+
+## 解题方法：滑动窗口
+
+本题计算的是字数组和乘以数组长度，由于数组元素为正，所以数组中元素越多计算得到的值越大，**具有单调性**，可以使用滑动窗口。
+
+使用$cnt$记录窗口中元素总和，右指针依次遍历数组中每个元素作为窗口终点（遍历时将指向元素加入窗口）。
+
+对于每个右指针的位置：
+
+> 如果$cnt*len(subArray)\geq k$，则不断右移左指针。
+>
+> 左指针移动结束后所在位置就是第一个窗口“分数”小于“k”的位置，从左指针到右指针任一元素**开始**到右指针所指元素**结束**的子数组“分数”都小于“k”，都是合法子数组。总个数为$r-l+1$，累加到答案中即可。
+
++ 时间复杂度$O(len(nums))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-30 10:29:37
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-30 10:34:24
+ */
+typedef long long ll;
+
+class Solution {
+public:
+    long long countSubarrays(vector<int>& nums, long long k) {
+        ll ans = 0, cnt = 0;
+        for (int l = 0, r = 0; r < nums.size(); r++) {
+            cnt += nums[r];
+            while (cnt * (r - l + 1) >= k) {
+                cnt -= nums[l++];
+            }
+            ans += (r - l + 1);
+        }
+        return ans;
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+Author: LetMeFly
+Date: 2025-04-29 23:46:38
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-04-30 10:39:13
+'''
+from typing import List
+
+class Solution:
+    def countSubarrays(self, nums: List[int], k: int) -> int:
+        ans = cnt = l = 0
+        for r in range(len(nums)):
+            cnt += nums[r]
+            while cnt * (r - l + 1) >= k:
+                cnt -= nums[l]
+                l += 1
+            ans += (r - l + 1)
+        return ans
+```
+
+#### Java
+
+```java
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-29 23:46:43
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-30 10:41:19
+ */
+class Solution {
+    public long countSubarrays(int[] nums, long k) {
+        long ans = 0, cnt = 0;
+        for (int l = 0, r = 0; r < nums.length; r++) {
+            cnt += nums[r];
+            while (cnt * (r - l + 1) >= k) {
+                cnt -= nums[l++];
+            }
+            ans += (r - l + 1);
+        }
+        return ans;
+    }
+}
+```
+
+#### Go
+
+```go
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-04-29 23:46:45
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-04-30 10:55:53
+ */
+package main
+
+func countSubarrays(nums []int, k int64) (ans int64) {
+    cnt := int64(0)
+    l := 0
+    for r, t := range nums {
+        cnt += int64(t)
+        for cnt * int64(r - l + 1) >= k {
+            cnt -= int64(nums[l])
+            l++
+        }
+        ans += int64(r - l + 1)
+    }
+    return
+}
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/147637062)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/04/30/LeetCode%202302.%E7%BB%9F%E8%AE%A1%E5%BE%97%E5%88%86%E5%B0%8F%E4%BA%8EK%E7%9A%84%E5%AD%90%E6%95%B0%E7%BB%84%E6%95%B0%E7%9B%AE/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
