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feat: add solutions to lc problem: No.1918
No.1918.Kth Smallest Subarray Sum
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‎solution/1900-1999/1918.Kth Smallest Subarray Sum/README.md

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@@ -61,22 +61,159 @@
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：二分查找 + 双指针**
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我们注意到，题目中数组元素均为正整数，子数组的和 $s$ 越大，那么数组中子数组和小于等于 $s$ 的个数就越多。这存在一个单调性，因此我们可以考虑使用使用二分查找的方法来求解。
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我们二分枚举子数组的和，初始化左右边界分别为数组 $nums$ 中的最小值以及所有元素之和。每次我们计算数组中子数组和小于等于当前枚举值的个数，如果个数大于等于 $k$，则说明当前枚举值 $s$ 可能是第 $k$ 小的子数组和，我们缩小右边界，否则我们增大左边界。枚举结束后，左边界即为第 $k$ 小的子数组和。
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问题转换为计算一个数组中，有多少个子数组的和小于等于 $s$，我们可以通过函数 $f(s)$ 来计算。
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函数 $f(s)$ 的计算方法如下：
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- 初始化双指针 $j$ 和 $i$，分别指向当前窗口的左右边界，初始时 $j = i = 0$。初始化窗口内元素的和 $t = 0$。
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- 用变量 $cnt$ 记录子数组和小于等于 $s$ 的个数，初始时 $cnt = 0$。
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- 遍历数组 $nums$，每次遍历到一个元素 $nums[i]$，我们将其加入窗口，即 $t = t + nums[i]$。如果此时 $t \gt s$，我们需要不断地将窗口的左边界右移，直到 $t \le s$ 为止，即不断地执行 $t -= nums[j]$，并且 $j = j + 1$。接下来我们更新 $cnt$，即 $cnt = cnt + i - j + 1$。继续遍历下一个元素，直到遍历完整个数组。
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最后将 $cnt$ 作为函数 $f(s)$ 的返回值。
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时间复杂度 $O(n \times \log S)$，空间复杂度 $O(1)$。其中 $n$ 为数组 $nums$ 的长度，而 $S$ 为数组 $nums$ 中所有元素之和。
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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 kthSmallestSubarraySum(self, nums: List[int], k: int) -> int:
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def f(s):
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t = j = 0
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cnt = 0
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for i, x in enumerate(nums):
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t += x
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while t > s:
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t -= nums[j]
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j += 1
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cnt += i - j + 1
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return cnt >= k
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l, r = min(nums), sum(nums)
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return l + bisect_left(range(l, r + 1), True, key=f)
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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 kthSmallestSubarraySum(int[] nums, int k) {
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int l = 1 << 30, r = 0;
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for (int x : nums) {
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l = Math.min(l, x);
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r += x;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (f(nums, mid) >= k) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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private int f(int[] nums, int s) {
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int t = 0, j = 0;
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int cnt = 0;
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for (int i = 0; i < nums.length; ++i) {
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t += nums[i];
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while (t > s) {
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t -= nums[j++];
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}
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cnt += i - j + 1;
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}
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return cnt;
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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 kthSmallestSubarraySum(vector<int>& nums, int k) {
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int l = 1 << 30, r = 0;
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for (int& x : nums) {
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l = min(l, x);
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r += x;
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}
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auto f = [&](int s) {
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int cnt = 0, t = 0;
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for (int i = 0, j = 0; i < nums.size(); ++i) {
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t += nums[i];
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while (t > s) {
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t -= nums[j++];
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}
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cnt += i - j + 1;
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}
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return cnt;
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};
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while (l < r) {
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int mid = (l + r) >> 1;
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if (f(mid) >= k) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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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 kthSmallestSubarraySum(nums []int, k int) int {
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l, r := 1<<30, 0
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for _, x := range nums {
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l = min(l, x)
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r += x
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}
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f := func(s int) (cnt int) {
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t := 0
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for i, j := 0, 0; i < len(nums); i++ {
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t += nums[i]
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for t > s {
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t -= nums[j]
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j++
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}
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cnt += i - j + 1
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}
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return
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}
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for l < r {
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mid := (l + r) >> 1
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if f(mid) >= k {
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r = mid
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} else {
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l = mid + 1
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}
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}
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return l
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}
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func min(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/1900-1999/1918.Kth Smallest Subarray Sum/README_EN.md

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@@ -60,13 +60,132 @@ Ordering the sums from smallest to largest gives 3, 3, 5, 5, 6, 8, <u>10</u>, 11
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### **Python3**
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```python
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class Solution:
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def kthSmallestSubarraySum(self, nums: List[int], k: int) -> int:
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def f(s):
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t = j = 0
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cnt = 0
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for i, x in enumerate(nums):
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t += x
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while t > s:
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t -= nums[j]
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j += 1
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cnt += i - j + 1
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return cnt >= k
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l, r = min(nums), sum(nums)
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return l + bisect_left(range(l, r + 1), True, key=f)
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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 kthSmallestSubarraySum(int[] nums, int k) {
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int l = 1 << 30, r = 0;
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for (int x : nums) {
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l = Math.min(l, x);
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r += x;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (f(nums, mid) >= k) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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private int f(int[] nums, int s) {
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int t = 0, j = 0;
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int cnt = 0;
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for (int i = 0; i < nums.length; ++i) {
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t += nums[i];
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while (t > s) {
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t -= nums[j++];
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}
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cnt += i - j + 1;
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}
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return cnt;
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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 kthSmallestSubarraySum(vector<int>& nums, int k) {
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int l = 1 << 30, r = 0;
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for (int& x : nums) {
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l = min(l, x);
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r += x;
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}
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auto f = [&](int s) {
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int cnt = 0, t = 0;
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for (int i = 0, j = 0; i < nums.size(); ++i) {
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t += nums[i];
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while (t > s) {
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t -= nums[j++];
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}
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cnt += i - j + 1;
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}
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return cnt;
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};
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while (l < r) {
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int mid = (l + r) >> 1;
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if (f(mid) >= k) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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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 kthSmallestSubarraySum(nums []int, k int) int {
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l, r := 1<<30, 0
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for _, x := range nums {
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l = min(l, x)
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r += x
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}
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f := func(s int) (cnt int) {
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t := 0
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for i, j := 0, 0; i < len(nums); i++ {
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t += nums[i]
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for t > s {
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t -= nums[j]
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j++
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}
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cnt += i - j + 1
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}
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return
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}
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for l < r {
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mid := (l + r) >> 1
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if f(mid) >= k {
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r = mid
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} else {
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l = mid + 1
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}
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}
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return l
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}
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func min(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/1900-1999/1918.Kth Smallest Subarray Sum/Solution.cpp

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,30 @@
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class Solution {
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public:
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int kthSmallestSubarraySum(vector<int>& nums, int k) {
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int l = 1 << 30, r = 0;
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for (int& x : nums) {
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l = min(l, x);
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r += x;
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}
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auto f = [&](int s) {
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int cnt = 0, t = 0;
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for (int i = 0, j = 0; i < nums.size(); ++i) {
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t += nums[i];
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while (t > s) {
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t -= nums[j++];
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}
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cnt += i - j + 1;
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}
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return cnt;
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};
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while (l < r) {
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int mid = (l + r) >> 1;
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if (f(mid) >= k) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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};

‎solution/1900-1999/1918.Kth Smallest Subarray Sum/Solution.go

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func kthSmallestSubarraySum(nums []int, k int) int {
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l, r := 1<<30, 0
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for _, x := range nums {
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l = min(l, x)
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r += x
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}
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f := func(s int) (cnt int) {
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t := 0
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for i, j := 0, 0; i < len(nums); i++ {
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t += nums[i]
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for t > s {
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t -= nums[j]
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j++
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}
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cnt += i - j + 1
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}
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return
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}
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for l < r {
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mid := (l + r) >> 1
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if f(mid) >= k {
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r = mid
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} else {
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l = mid + 1
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}
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}
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return l
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}
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func min(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/1900-1999/1918.Kth Smallest Subarray Sum/Solution.java

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class Solution {
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public int kthSmallestSubarraySum(int[] nums, int k) {
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int l = 1 << 30, r = 0;
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for (int x : nums) {
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l = Math.min(l, x);
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r += x;
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}
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while (l < r) {
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int mid = (l + r) >> 1;
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if (f(nums, mid) >= k) {
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r = mid;
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} else {
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l = mid + 1;
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}
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}
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return l;
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}
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private int f(int[] nums, int s) {
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int t = 0, j = 0;
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int cnt = 0;
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for (int i = 0; i < nums.length; ++i) {
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t += nums[i];
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while (t > s) {
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t -= nums[j++];
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}
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cnt += i - j + 1;
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}
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return cnt;
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}
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}

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