How hard is LeetCode 1944. Number of Visible People in a Queue?

LeetCode 1944. Number of Visible People in a Queue is rated Hard on LeetCode.

What is the time complexity of LeetCode 1944. Number of Visible People in a Queue?

The Python solution on this page runs in O(n).

What is the space complexity of LeetCode 1944. Number of Visible People in a Queue?

The Python solution on this page uses O(n) auxiliary space.

What topics does LeetCode 1944. Number of Visible People in a Queue cover?

LeetCode 1944. Number of Visible People in a Queue is tagged Stack, Array and Monotonic Stack on LeetCode.

HardNo. 1944

Number of Visible People in a Queue — LeetCode 1944 Python Solution

Pattern: Stack
Topics: Stack, Array, Monotonic Stack
Reading time: 3 min
Source: leetcode.com

The problem

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.

Example

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.

Python solution

PythonSolution

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

Complexity

Time: O(n)
Space · auxiliary: O(n)

Pattern: Stack

When the most recent unresolved thing is the one that matters, use a stack. LeetCode 1944. Number of Visible People in a Queue is filed here because LeetCode tags it Stack, which is the vocabulary this hub collects.

The stack guide has the Python template for the pattern and the 194 LeetCode problems that use it.

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Frequently asked questions

How hard is LeetCode 1944. Number of Visible People in a Queue?: LeetCode 1944. Number of Visible People in a Queue is rated Hard on LeetCode.
What is the time complexity of LeetCode 1944. Number of Visible People in a Queue?: The Python solution on this page runs in O(n).
What is the space complexity of LeetCode 1944. Number of Visible People in a Queue?: The Python solution on this page uses O(n) auxiliary space.
What topics does LeetCode 1944. Number of Visible People in a Queue cover?: LeetCode 1944. Number of Visible People in a Queue is tagged Stack, Array and Monotonic Stack on LeetCode.

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python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

Technical interview · 45 minsolution.pyShared editor · Python 3Longest substring without repeatsMediumGiven a string s, return the length of the longest contiguous substring in which every character is unique.s = "abcabcbb"
→ 31def length_of_longest_substring(s: str) -> int:2 last_seen = {}3 left = best = 04 for right, ch in enumerate(s):5 if ch in last_seen and last_seen[ch] >= left:

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 Terra

Transcribing...

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

90%Subscription

python

Please analyze this context and help me answer the question or solve the problem.

Transcript

"Okay, next one. Given a string, return the length of the longest substring that has no repeating characters. Walk me through your approach before you code it."

python

def length_of_longest_substring(s: str) -> int:    last_seen = {}                    # char -> index    left = best = 0    for right, ch in enumerate(s):        if ch in last_seen and last_seen[ch] >= left:            left = last_seen[ch] + 1  # jump past it        last_seen[ch] = right        best = max(best, right - left + 1)    return best    last_seen = {}                    # char -> index

Approach

At each index r, the window's left edge is:

left r = max (left r -1, last[s r] + 1)

Time O(n): each character enters and leaves the window once.
Space O(min(n, k)): k is the size of the character set.

Type your message...

GPT-6.1 TerraClear chat

Transcribing...

Primary display

QuitEscHide+QCapture+PCrop+OSubmit+EnterPress to get answerLeft/Right+Up/Down+

The problem

Example

Python solution

Complexity

Pattern: Stack

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Frequently asked questions

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