LeetCode Study Lists: Blind 75, NeetCode 150 and Grind 75

The five lists

None of the five below originate here. We reproduce each one by problem number, in the order the list itself uses, and link every entry to its page in this library. 525 of the entries they name resolve to a problem here; a problem two lists share is counted once. Nothing is padded to make a title's number come out right — each page states what it actually holds.

Blind 75A 75-problem list assembled from the patterns that keep reappearing in technical interviews, grouped by topic. Every entry links to a Python solution with complexity analysis.75 problems · 19 Easy · 49 Medium · 7 Hard75 problems
NeetCode 150An expansion of the Blind 75 into eighteen topic groups, from arrays and hashing through advanced graphs and two-dimensional dynamic programming.150 problems · 28 Easy · 101 Medium · 21 Hard150 problems
Grind 75The Blind 75 reworked into a week-by-week schedule that starts with the easiest high-yield problems and ends on the hard ones. Ordered for study, not by topic.75 problems · 23 Easy · 43 Medium · 9 Hard75 problems
LeetCode 75The official study plan, seventy-five problems across twenty-two topic groups from array and string work up to monotonic stacks. Skewed easier than the Blind 75 and better as a first pass.75 problems · 22 Easy · 53 Medium · 0 Hard75 problems
Top Interview 150The official interview-preparation list, a hundred and fifty problems covering every data structure and technique a technical screen is likely to reach for.150 problems · 40 Easy · 92 Medium · 18 Hard150 problems

Eyes only

Stuck on the one they actually ask?

In the live round, Stealth Interview reads the problem off your screen, returns a working solution with a step-by-step explanation and its time and space complexity, and transcribes what the interviewer says. It's a desktop app for macOS and Windows, invisible to screen sharing.

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python

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

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

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python

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

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 TerraClear chat

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.

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 TerraClear chat

Primary display

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

LeetCode Study Lists: Blind 75,
NeetCode 150 and Three More

The five lists

Stuck on the one they actually ask?

Approach

Approach

Approach

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