Randomized LeetCode Problems: 12 Python Solutions

Randomized problems by difficulty

Problems with a complete Python solution are listed first, then by ascending problem number.

Medium (9)

Medium problems
No.	Problem	Topics
0380	Insert Delete GetRandom O(1)	Design, Array, Hash Table +2
0382	Linked List Random Node	Reservoir Sampling, Linked List, Math +1
0384	Shuffle an Array	Design, Array, Math +1
0398	Random Pick Index	Reservoir Sampling, Hash Table, Math +1
0470	Implement Rand10() Using Rand7()	Math, Rejection Sampling, Probability and Statistics +1
0478	Generate Random Point in a Circle	Geometry, Math, Rejection Sampling +1
0497	Random Point in Non-overlapping Rectangles	Reservoir Sampling, Array, Math +4
0519	Random Flip Matrix	Reservoir Sampling, Hash Table, Math +1
0528	Random Pick with Weight	Array, Math, Binary Search +2

Hard (3)

Hard problems
No.	Problem	Topics
0381	Insert Delete GetRandom O(1) - Duplicates allowed	Design, Array, Hash Table +2
0710	Random Pick with Blacklist	Array, Hash Table, Math +3
1515	Best Position for a Service Centre	Geometry, Array, Math +1

Eyes only

When the Randomized problem arrives live.

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.

Get Stealth Interview See pricing

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 Terra

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

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+

Randomized LeetCode Problems: All 12,
With Python Solutions

Randomized problems by difficulty

Medium (9)

Hard (3)

When the Randomized problem arrives live.

Approach

Approach

Approach

Keep exploring

Randomized problems by difficulty

Medium (9)9

Hard (3)3

When the Randomized problem arrives live.

Approach

Approach

Approach

Keep exploring

Medium (9)

Hard (3)