LeetCode #3113 — HARD

Find the Number of Subarrays Where Boundary Elements Are Maximum

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

The Problem

Problem Statement

You are given an array of positive integers nums.

Return the number of subarrays of nums, where the first and the last elements of the subarray are equal to the largest element in the subarray.

Example 1:

Input: nums = [1,4,3,3,2]

Output: 6

Explanation:

There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:

subarray [1,4,3,3,2], with its largest element 1. The first element is 1 and the last element is also 1.
subarray [1,4,3,3,2], with its largest element 4. The first element is 4 and the last element is also 4.
subarray [1,4,3,3,2], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [1,4,3,3,2], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [1,4,3,3,2], with its largest element 2. The first element is 2 and the last element is also 2.
subarray [1,4,3,3,2], with its largest element 3. The first element is 3 and the last element is also 3.

Hence, we return 6.

Example 2:

Input: nums = [3,3,3]

Output: 6

Explanation:

There are 6 subarrays which have the first and the last elements equal to the largest element of the subarray:

subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.
subarray [3,3,3], with its largest element 3. The first element is 3 and the last element is also 3.

Hence, we return 6.

Example 3:

Input: nums = [1]

Output: 1

Explanation:

There is a single subarray of nums which is [1], with its largest element 1. The first element is 1 and the last element is also 1.

Hence, we return 1.

Constraints:

1 <= nums.length <= 10⁵
1 <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: You are given an array of positive integers nums. Return the number of subarrays of nums, where the first and the last elements of the subarray are equal to the largest element in the subarray.

Baseline thinking

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

Pattern signal: Array · Binary Search · Stack

Example 1

[1,4,3,3,2]

Example 2

[3,3,3]

Example 3

[1]

Core Insight

What unlocks the optimal approach

For each element <code>nums[i]</code>, we can count the number of valid subarrays ending with it.
For each index <code>i</code>, find the nearest index <code>j</code> on its left <code>(j < i)</code> such that <code>nums[j] < nums[i]</code>. This can be done via a monotonic stack.
For each index <code>i</code>, find the number of indices <code>k</code> in the window <code>[j + 1, i]</code> such that <code>nums[k] == nums[i]</code>, this is the number of the valid subarrays ending with <code>nums[i]</code>. This can be done by sliding window.
Sum the answer of all the indices <code>i</code> to get the final result.
Is it possible to use DSU as an alternate solution?

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 #3113: Find the Number of Subarrays Where Boundary Elements Are Maximum
class Solution {
    public long numberOfSubarrays(int[] nums) {
        Deque<int[]> stk = new ArrayDeque<>();
        long ans = 0;
        for (int x : nums) {
            while (!stk.isEmpty() && stk.peek()[0] < x) {
                stk.pop();
            }
            if (stk.isEmpty() || stk.peek()[0] > x) {
                stk.push(new int[] {x, 1});
            } else {
                stk.peek()[1]++;
            }
            ans += stk.peek()[1];
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3113: Find the Number of Subarrays Where Boundary Elements Are Maximum
func numberOfSubarrays(nums []int) (ans int64) {
	stk := [][2]int{}
	for _, x := range nums {
		for len(stk) > 0 && stk[len(stk)-1][0] < x {
			stk = stk[:len(stk)-1]
		}
		if len(stk) == 0 || stk[len(stk)-1][0] > x {
			stk = append(stk, [2]int{x, 1})
		} else {
			stk[len(stk)-1][1]++
		}
		ans += int64(stk[len(stk)-1][1])
	}
	return
}

# Accepted solution for LeetCode #3113: Find the Number of Subarrays Where Boundary Elements Are Maximum
class Solution:
    def numberOfSubarrays(self, nums: List[int]) -> int:
        stk = []
        ans = 0
        for x in nums:
            while stk and stk[-1][0] < x:
                stk.pop()
            if not stk or stk[-1][0] > x:
                stk.append([x, 1])
            else:
                stk[-1][1] += 1
            ans += stk[-1][1]
        return ans

// Accepted solution for LeetCode #3113: Find the Number of Subarrays Where Boundary Elements Are Maximum
// 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 #3113: Find the Number of Subarrays Where Boundary Elements Are Maximum
// class Solution {
//     public long numberOfSubarrays(int[] nums) {
//         Deque<int[]> stk = new ArrayDeque<>();
//         long ans = 0;
//         for (int x : nums) {
//             while (!stk.isEmpty() && stk.peek()[0] < x) {
//                 stk.pop();
//             }
//             if (stk.isEmpty() || stk.peek()[0] > x) {
//                 stk.push(new int[] {x, 1});
//             } else {
//                 stk.peek()[1]++;
//             }
//             ans += stk.peek()[1];
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #3113: Find the Number of Subarrays Where Boundary Elements Are Maximum
function numberOfSubarrays(nums: number[]): number {
    const stk: number[][] = [];
    let ans = 0;
    for (const x of nums) {
        while (stk.length > 0 && stk.at(-1)![0] < x) {
            stk.pop();
        }
        if (stk.length === 0 || stk.at(-1)![0] > x) {
            stk.push([x, 1]);
        } else {
            stk.at(-1)![1]++;
        }
        ans += stk.at(-1)![1];
    }
    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(n)

Space

O(n)

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.

Breaking monotonic invariant

Wrong move: Pushing without popping stale elements invalidates next-greater/next-smaller logic.

Usually fails on: Indices point to blocked elements and outputs shift.

Fix: Pop while invariant is violated before pushing current element.

Find the Number of Subarrays Where Boundary Elements Are Maximum

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Off-by-one on range boundaries

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

Breaking monotonic invariant