How hard is LeetCode 2575. Find the Divisibility Array of a String?

LeetCode 2575. Find the Divisibility Array of a String is rated Medium on LeetCode.

What is the time complexity of LeetCode 2575. Find the Divisibility Array of a String?

The Python solution on this page runs in O(n), where n is the length of the string `word`.

What is the space complexity of LeetCode 2575. Find the Divisibility Array of a String?

The Python solution on this page uses O(1) auxiliary space.

What topics does LeetCode 2575. Find the Divisibility Array of a String cover?

LeetCode 2575. Find the Divisibility Array of a String is tagged Array, Math and String on LeetCode.

MediumNo. 2575

Find the Divisibility Array of a String — LeetCode 2575 Python Solution

Pattern: Math and Number Theory
Topics: Array, Math, String
Reading time: 2 min
Source: leetcode.com

The problem

You are given a 0-indexed string word of length n consisting of digits, and a positive integer m. The divisibility array div of word is an integer array of length n such that: div[i] = 1 if the numeric value of word[0,...,i] is divisible by m, or div[i] = 0 otherwise.

Example

Input: word = "998244353", m = 3
Output: [1,1,0,0,0,1,1,0,0]
Explanation: There are only 4 prefixes that are divisible by 3: "9", "99", "998244", and "9982443".

Python solution

PythonSolution

class Solution:
    def divisibilityArray(self, word: str, m: int) -> List[int]:
        ans = []
        x = 0
        for c in word:
            x = (x * 10 + int(c)) % m
            ans.append(1 if x == 0 else 0)
        return ans

Complexity

Time: O(n)where n is the length of the string `word`
Space · auxiliary: O(1)

Pattern: Math and Number Theory

Find the closed form, the invariant, or the modular identity — and skip the loop entirely. LeetCode 2575. Find the Divisibility Array of a String is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Math.

The math and number theory guide has the Python template for the pattern and the 485 LeetCode problems that use it.

Related problems
No.	Problem	Difficulty
2125	Number of Laser Beams in a Bank	Medium
0043	Multiply Strings	Medium
0048	Rotate Image	Medium
0066	Plus One	Easy
0168	Excel Sheet Column Title	Easy
0171	Excel Sheet Column Number	Easy

Frequently asked questions

How hard is LeetCode 2575. Find the Divisibility Array of a String?: LeetCode 2575. Find the Divisibility Array of a String is rated Medium on LeetCode.
What is the time complexity of LeetCode 2575. Find the Divisibility Array of a String?: The Python solution on this page runs in O(n), where n is the length of the string `word`.
What is the space complexity of LeetCode 2575. Find the Divisibility Array of a String?: The Python solution on this page uses O(1) auxiliary space.
What topics does LeetCode 2575. Find the Divisibility Array of a String cover?: LeetCode 2575. Find the Divisibility Array of a String is tagged Array, Math and String on LeetCode.

Stuck on problems like this in a live interview?

Stealth Interview is a desktop app for macOS and Windows. It reads the problem off your screen and returns a working solution with a step-by-step explanation and its time and space complexity, invisible to screen sharing.

Download for Mac Download for Windows

Free to download. Check it runs on your machine, then pick a plan.

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."

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: Math and Number Theory

Related problems

Frequently asked questions

Stuck on problems like this in a live interview?

Approach

Approach

Approach