feat: add solutions to lc problem: No.3748 (#4842)

Author: yanglbme

solution/3700-3799/3748.Count Stable Subarrays/README.md

@@ -109,32 +109,220 @@ edit_url: https://github.com/doocs/leetcode/edit/main/solution/3700-3799/3748.Co
 
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-### 方法一
+### 方法一： 分段计数
+
+根据题目描述，稳定子数组的定义是不存在逆序对的子数组，即子数组中的元素是非降序排列的。因此，我们可以将数组划分为若干个非降序的段，用一个数组 $\text{seg}$ 来记录每个段的起始位置。同时，我们还需要一个前缀和数组 $\text{text{s}}$ 来记录每个段内稳定子数组的数量。
+
+然后，对于每个查询 $[l, r]$，可能存在 $3$ 种情况：
+
+1. 查询区间 $[l, r]$ 完全包含在某个段内，此时稳定子数组的数量可以直接通过公式计算得到，数量为 $\frac{(k + 1) \cdot k}{2}$，其中 $k = r - l + 1$。
+2. 查询区间 $[l, r]$ 跨越了多个段，此时我们需要分别计算左侧不完整段、右侧不完整段和中间完整段的稳定子数组数量，然后将它们相加得到最终结果。
+
+时间复杂度为 $O((n + q) \log n)$，其中 $n$ 是数组的长度，$q$ 是查询的数量。空间复杂度为 $O(n)$。
 
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 #### Python3
 
 ```python
-
+class Solution:
+    def countStableSubarrays(
+        self, nums: List[int], queries: List[List[int]]
+    ) -> List[int]:
+        s = [0]
+        l, n = 0, len(nums)
+        seg = []
+        for r, x in enumerate(nums):
+            if r == n - 1 or x > nums[r + 1]:
+                seg.append(l)
+                k = r - l + 1
+                s.append(s[-1] + (1 + k) * k // 2)
+                l = r + 1
+        ans = []
+        for l, r in queries:
+            i = bisect_right(seg, l)
+            j = bisect_right(seg, r) - 1
+            if i > j:
+                k = r - l + 1
+                ans.append((1 + k) * k // 2)
+            else:
+                a = seg[i] - l
+                b = r - seg[j] + 1
+                ans.append((1 + a) * a // 2 + s[j] - s[i] + (1 + b) * b // 2)
+        return ans
 ```
 
 #### Java
 
 ```java
-
+class Solution {
+    public long[] countStableSubarrays(int[] nums, int[][] queries) {
+        List<Integer> seg = new ArrayList<>();
+        List<Long> s = new ArrayList<>();
+        s.add(0L);
+
+        int l = 0;
+        int n = nums.length;
+        for (int r = 0; r < n; r++) {
+            if (r == n - 1 || nums[r] > nums[r + 1]) {
+                seg.add(l);
+                int k = r - l + 1;
+                s.add(s.getLast() + (long) k * (k + 1) / 2);
+                l = r + 1;
+            }
+        }
+
+        long[] ans = new long[queries.length];
+        for (int q = 0; q < queries.length; q++) {
+            int left = queries[q][0];
+            int right = queries[q][1];
+
+            int i = upperBound(seg, left);
+            int j = upperBound(seg, right) - 1;
+
+            if (i > j) {
+                int k = right - left + 1;
+                ans[q] = (long) k * (k + 1) / 2;
+            } else {
+                int a = seg.get(i) - left;
+                int b = right - seg.get(j) + 1;
+                ans[q] = (long) a * (a + 1) / 2 + s.get(j) - s.get(i) + (long) b * (b + 1) / 2;
+            }
+        }
+        return ans;
+    }
+
+    private int upperBound(List<Integer> list, int target) {
+        int l = 0, r = list.size();
+        while (l < r) {
+            int mid = (l + r) >> 1;
+            if (list.get(mid) > target) {
+                r = mid;
+            } else {
+                l = mid + 1;
+            }
+        }
+        return l;
+    }
+}
 ```
 
 #### C++
 
 ```cpp
-
+class Solution {
+public:
+    vector<long long> countStableSubarrays(vector<int>& nums, vector<vector<int>>& queries) {
+        int n = nums.size();
+        vector<int> seg;
+        vector<long long> s = {0};
+
+        int l = 0;
+        for (int r = 0; r < n; ++r) {
+            if (r == n - 1 || nums[r] > nums[r + 1]) {
+                seg.push_back(l);
+                long long k = r - l + 1;
+                s.push_back(s.back() + k * (k + 1) / 2);
+                l = r + 1;
+            }
+        }
+
+        vector<long long> ans;
+        for (auto& q : queries) {
+            int left = q[0], right = q[1];
+
+            int i = upper_bound(seg.begin(), seg.end(), left) - seg.begin();
+            int j = upper_bound(seg.begin(), seg.end(), right) - seg.begin() - 1;
+
+            if (i > j) {
+                long long k = right - left + 1;
+                ans.push_back(k * (k + 1) / 2);
+            } else {
+                long long a = seg[i] - left;
+                long long b = right - seg[j] + 1;
+                ans.push_back(a * (a + 1) / 2 + s[j] - s[i] + b * (b + 1) / 2);
+            }
+        }
+
+        return ans;
+    }
+};
 ```
 
 #### Go
 
 ```go
+func countStableSubarrays(nums []int, queries [][]int) []int64 {
+	n := len(nums)
+	seg := []int{}
+	s := []int64{0}
+
+	l := 0
+	for r := 0; r < n; r++ {
+		if r == n-1 || nums[r] > nums[r+1] {
+			seg = append(seg, l)
+			k := int64(r - l + 1)
+			s = append(s, s[len(s)-1]+k*(k+1)/2)
+			l = r + 1
+		}
+	}
+
+	ans := make([]int64, len(queries))
+	for idx, q := range queries {
+		left, right := q[0], q[1]
+
+		i := sort.SearchInts(seg, left+1)
+		j := sort.SearchInts(seg, right+1) - 1
+
+		if i > j {
+			k := int64(right - left + 1)
+			ans[idx] = k * (k + 1) / 2
+		} else {
+			a := int64(seg[i] - left)
+			b := int64(right - seg[j] + 1)
+			ans[idx] = a*(a+1)/2 + s[j] - s[i] + b*(b+1)/2
+		}
+	}
+
+	return ans
+}
+```
 
+#### TypeScript
+
+```ts
+function countStableSubarrays(nums: number[], queries: number[][]): number[] {
+    const n = nums.length;
+    const seg: number[] = [];
+    const s: number[] = [0];
+
+    let l = 0;
+    for (let r = 0; r < n; r++) {
+        if (r === n - 1 || nums[r] > nums[r + 1]) {
+            seg.push(l);
+            const k = r - l + 1;
+            s.push(s[s.length - 1] + (k * (k + 1)) / 2);
+            l = r + 1;
+        }
+    }
+
+    const ans: number[] = [];
+    for (const [left, right] of queries) {
+        const i = _.sortedIndex(seg, left + 1);
+        const j = _.sortedIndex(seg, right + 1) - 1;
+
+        if (i > j) {
+            const k = right - left + 1;
+            ans.push((k * (k + 1)) / 2);
+        } else {
+            const a = seg[i] - left;
+            const b = right - seg[j] + 1;
+            ans.push((a * (a + 1)) / 2 + s[j] - s[i] + (b * (b + 1)) / 2);
+        }
+    }
+
+    return ans;
+}
 ```
 
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