Recursion LeetCode Problems: 45 Python Solutions

Recursion problems by difficulty

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

Easy (9)

Easy problems
No.	Problem	Topics
0021	Merge Two Sorted Lists	Recursion, Linked List
0203	Remove Linked List Elements	Recursion, Linked List
0206	Reverse Linked List	Recursion, Linked List
0231	Power of Two	Bit Manipulation, Recursion, Math
0234	Palindrome Linked List	Stack, Recursion, Linked List +1
0326	Power of Three	Recursion, Math
0342	Power of Four	Bit Manipulation, Recursion, Math
0509	Fibonacci Number	Recursion, Memoization, Math +1
2094	Finding 3-Digit Even Numbers	Recursion, Array, Hash Table +2

Medium (22)

Medium problems
No.	Problem	Topics
0002	Add Two Numbers	Recursion, Linked List, Math
0024	Swap Nodes in Pairs	Recursion, Linked List
0050	Pow(x, n)	Recursion, Math
0143	Reorder List	Stack, Recursion, Linked List +1
0241	Different Ways to Add Parentheses	Recursion, Memoization, Math +2
0390	Elimination Game	Recursion, Math
0394	Decode String	Stack, Recursion, String
0486	Predict the Winner	Recursion, Array, Math +2
0779	K-th Symbol in Grammar	Bit Manipulation, Recursion, Math
0894	All Possible Full Binary Trees	Tree, Recursion, Memoization +2
1545	Find Kth Bit in Nth Binary String	Recursion, String, Simulation
1823	Find the Winner of the Circular Game	Recursion, Queue, Array +2
1922	Count Good Numbers	Recursion, Math
1969	Minimum Non-Zero Product of the Array Elements	Greedy, Recursion, Math
2487	Remove Nodes From Linked List	Stack, Recursion, Linked List +1
2550	Count Collisions of Monkeys on a Polygon	Recursion, Math
0247	Strobogrammatic Number IIPremium	Recursion, Array, String
0255	Verify Preorder Sequence in Binary Search TreePremium	Stack, Tree, Binary Search Tree +4
0439	Ternary Expression ParserPremium	Stack, Recursion, String
0544	Output Contest MatchesPremium	Recursion, String, Simulation
0776	Split BSTPremium	Tree, Binary Search Tree, Recursion +1
1265	Print Immutable Linked List in ReversePremium	Stack, Recursion, Linked List +1

Hard (14)

Hard problems
No.	Problem	Topics
0010	Regular Expression Matching	Recursion, String, Dynamic Programming
0025	Reverse Nodes in k-Group	Recursion, Linked List
0044	Wildcard Matching	Greedy, Recursion, String +1
0060	Permutation Sequence	Recursion, Math
0224	Basic Calculator	Stack, Recursion, Math +1
0233	Number of Digit One	Recursion, Math, Dynamic Programming
0273	Integer to English Words	Recursion, Math, String
0736	Parse Lisp Expression	Stack, Recursion, Hash Table +1
0761	Special Binary String	Recursion, String
1106	Parsing A Boolean Expression	Stack, Recursion, String
1808	Maximize Number of Nice Divisors	Recursion, Math, Number Theory
0248	Strobogrammatic Number IIIPremium	Recursion, Array, String
0770	Basic Calculator IV	Stack, Recursion, Hash Table +2
0772	Basic Calculator IIIPremium	Stack, Recursion, Math +1

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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...

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

Recursion LeetCode Problems: All 45,
With Python Solutions

Recursion problems by difficulty

Easy (9)

Medium (22)

Hard (14)

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Recursion problems by difficulty

Easy (9)9

Medium (22)22

Hard (14)14

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Easy (9)

Medium (22)

Hard (14)