LeetCode #1944 — HARD

Number of Visible People in a Queue

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

The Problem

Problem Statement

There are n people standing in a queue, and they numbered from 0 to n - 1 in left to right order. You are given an array heights of distinct integers where heights[i] represents the height of the i^th person.

A person can see another person to their right in the queue if everybody in between is shorter than both of them. More formally, the i^th person can see the j^th person if i < j and min(heights[i], heights[j]) > max(heights[i+1], heights[i+2], ..., heights[j-1]).

Return an array answer of length n where answer[i] is the number of people the i^th person can see to their right in the queue.

Example 1:

Input: heights = [10,6,8,5,11,9]
Output: [3,1,2,1,1,0]
Explanation:
Person 0 can see person 1, 2, and 4.
Person 1 can see person 2.
Person 2 can see person 3 and 4.
Person 3 can see person 4.
Person 4 can see person 5.
Person 5 can see no one since nobody is to the right of them.

Example 2:

Input: heights = [5,1,2,3,10]
Output: [4,1,1,1,0]

Constraints:

n == heights.length
1 <= n <= 10⁵
1 <= heights[i] <= 10⁵
All the values of heights are unique.

Step 01

Brute Force Baseline

Problem summary: There are n people standing in a queue, and they numbered from 0 to n - 1 in left to right order. You are given an array heights of distinct integers where heights[i] represents the height of the ith person. A person can see another person to their right in the queue if everybody in between is shorter than both of them. More formally, the ith person can see the jth person if i < j and min(heights[i], heights[j]) > max(heights[i+1], heights[i+2], ..., heights[j-1]). Return an array answer of length n where answer[i] is the number of people the ith person can see to their right in the queue.

Baseline thinking

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

Pattern signal: Array · Stack

Example 1

[10,6,8,5,11,9]

Example 2

[5,1,2,3,10]

Core Insight

What unlocks the optimal approach

How to solve this problem in quadratic complexity ?
For every subarray start at index i, keep finding new maximum values until a value larger than arr[i] is found.
Since the limits are high, you need a linear solution.
Use a stack to keep the values of the array sorted as you iterate the array from the end to the start.
Keep popping from the stack the elements in sorted order until a value larger than arr[i] is found, these are the ones that person i can see.

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 #1944: Number of Visible People in a Queue
class Solution {
    public int[] canSeePersonsCount(int[] heights) {
        int n = heights.length;
        int[] ans = new int[n];
        Deque<Integer> stk = new ArrayDeque<>();
        for (int i = n - 1; i >= 0; --i) {
            while (!stk.isEmpty() && stk.peek() < heights[i]) {
                stk.pop();
                ++ans[i];
            }
            if (!stk.isEmpty()) {
                ++ans[i];
            }
            stk.push(heights[i]);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1944: Number of Visible People in a Queue
func canSeePersonsCount(heights []int) []int {
	n := len(heights)
	ans := make([]int, n)
	stk := []int{}
	for i := n - 1; i >= 0; i-- {
		for len(stk) > 0 && stk[len(stk)-1] < heights[i] {
			ans[i]++
			stk = stk[:len(stk)-1]
		}
		if len(stk) > 0 {
			ans[i]++
		}
		stk = append(stk, heights[i])
	}
	return ans
}

# Accepted solution for LeetCode #1944: Number of Visible People in a Queue
class Solution:
    def canSeePersonsCount(self, heights: List[int]) -> List[int]:
        n = len(heights)
        ans = [0] * n
        stk = []
        for i in range(n - 1, -1, -1):
            while stk and stk[-1] < heights[i]:
                ans[i] += 1
                stk.pop()
            if stk:
                ans[i] += 1
            stk.append(heights[i])
        return ans

// Accepted solution for LeetCode #1944: Number of Visible People in a Queue
impl Solution {
    pub fn can_see_persons_count(heights: Vec<i32>) -> Vec<i32> {
        let n = heights.len();
        let mut ans = vec![0; n];
        let mut stack = Vec::new();
        for i in (0..n).rev() {
            while !stack.is_empty() {
                ans[i] += 1;
                if heights[i] <= heights[*stack.last().unwrap()] {
                    break;
                }
                stack.pop();
            }
            stack.push(i);
        }
        ans
    }
}

// Accepted solution for LeetCode #1944: Number of Visible People in a Queue
function canSeePersonsCount(heights: number[]): number[] {
    const n = heights.length;
    const ans: number[] = new Array(n).fill(0);
    const stk: number[] = [];
    for (let i = n - 1; ~i; --i) {
        while (stk.length && stk.at(-1) < heights[i]) {
            ++ans[i];
            stk.pop();
        }
        if (stk.length) {
            ++ans[i];
        }
        stk.push(heights[i]);
    }
    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

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan left (or right) to find the next greater/smaller element. The inner scan can visit up to n elements per outer iteration, giving O(n²) total comparisons. No extra space needed beyond loop variables.

MONOTONIC STACK

O(n) time

O(n) space

Each element is pushed onto the stack at most once and popped at most once, giving 2n total operations = O(n). The stack itself holds at most n elements in the worst case. The key insight: amortized O(1) per element despite the inner while-loop.

Shortcut: Each element pushed once + popped once → O(n) amortized. The inner while-loop does not make it O(n²).

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.

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.