LeetCode #3748 — HARD

Count Stable Subarrays

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

You are given an integer array nums.

A subarray of nums is called stable if it contains no inversions, i.e., there is no pair of indices i < j such that nums[i] > nums[j].

You are also given a 2D integer array queries of length q, where each queries[i] = [l_i, r_i] represents a query. For each query [l_i, r_i], compute the number of stable subarrays that lie entirely within the segment nums[l_i..r_i].

Return an integer array ans of length q, where ans[i] is the answer to the i^th query.

Note:

A single element subarray is considered stable.

Example 1:

Input: nums = [3,1,2], queries = [[0,1],[1,2],[0,2]]

Output: [2,3,4]

Explanation:

For queries[0] = [0, 1], the subarray is [nums[0], nums[1]] = [3, 1].
- The stable subarrays are [3] and [1]. The total number of stable subarrays is 2.
For queries[1] = [1, 2], the subarray is [nums[1], nums[2]] = [1, 2].
- The stable subarrays are [1], [2], and [1, 2]. The total number of stable subarrays is 3.
For queries[2] = [0, 2], the subarray is [nums[0], nums[1], nums[2]] = [3, 1, 2].
- The stable subarrays are [3], [1], [2], and [1, 2]. The total number of stable subarrays is 4.

Thus, ans = [2, 3, 4].

Example 2:

Input: nums = [2,2], queries = [[0,1],[0,0]]

Output: [3,1]

Explanation:

For queries[0] = [0, 1], the subarray is [nums[0], nums[1]] = [2, 2].
- The stable subarrays are [2], [2], and [2, 2]. The total number of stable subarrays is 3.
For queries[1] = [0, 0], the subarray is [nums[0]] = [2].
- The stable subarray is [2]. The total number of stable subarrays is 1.

Thus, ans = [3, 1].

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁵
1 <= queries.length <= 10⁵
queries[i] = [l_i, r_i]
0 <= l_i <= r_i <= nums.length - 1

Step 01

Brute Force Baseline

Problem summary: You are given an integer array nums. A subarray of nums is called stable if it contains no inversions, i.e., there is no pair of indices i < j such that nums[i] > nums[j]. You are also given a 2D integer array queries of length q, where each queries[i] = [li, ri] represents a query. For each query [li, ri], compute the number of stable subarrays that lie entirely within the segment nums[li..ri]. Return an integer array ans of length q, where ans[i] is the answer to the ith query. Note: A single element subarray is considered stable.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Binary Search

Example 1

[3,1,2]
[[0,1],[1,2],[0,2]]

Example 2

[2,2]
[[0,1],[0,0]]

Step 02

Core Insight

What unlocks the optimal approach

Identify maximal non-decreasing segments. Each segment of length <code>L</code> contributes <code>L * (L + 1) / 2</code> stable subarrays.
Build a prefix array of total stable subarrays ending at each index.
For query <code>[l, r]</code>, compute the prefix sum in the range and adjust for the left segment crossing <code>l</code>.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3748: Count Stable Subarrays
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;
    }
}

// Accepted solution for LeetCode #3748: Count Stable Subarrays
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
}

# Accepted solution for LeetCode #3748: Count Stable Subarrays
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

// Accepted solution for LeetCode #3748: Count Stable Subarrays
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3748: Count Stable Subarrays
// 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;
//     }
// }

// Accepted solution for LeetCode #3748: Count Stable Subarrays
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.