How do you solve Check If Word Is Valid After Substitutions on LeetCode?

Check If Word Is Valid After Substitutions (LeetCode #1003) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: The string must be able to form complete 'abc' sequences.

What is the brute force solution for Check If Word Is Valid After Substitutions?

The brute force approach for Check If Word Is Valid After Substitutions is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach tries to generate all possible strings by inserting 'abc' in various positions and checks if the generated string matches the input. This is not efficient but helps us understand the problem. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check If Word Is Valid After Substitutions?

Check If Word Is Valid After Substitutions uses the following concepts: String, Stack. The recommended patterns to study are: Stack, Greedy.

What are the interview tips for Check If Word Is Valid After Substitutions?

Always consider edge cases, like strings that are too short or do not contain 'abc'. Explain your thought process clearly while coding to demonstrate your understanding. Practice writing code on a whiteboard or in a shared document to simulate the interview environment.

What are common mistakes when solving Check If Word Is Valid After Substitutions?

Not checking for the complete sequence 'abc' before popping from the stack. Assuming the order of characters can be arbitrary without considering the sequence.

#1003

Check If Word Is Valid After Substitutions

Medium

StringStackStackGreedy

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a stack to keep track of the characters and ensures that every 'abc' sequence is properly formed. This is efficient and directly checks for the validity of the string.

⚙️

Algorithm

4 steps

1Step 1: Initialize an empty stack.
2Step 2: Traverse each character in the string s.
3Step 3: Push each character onto the stack. If the top three characters of the stack are 'abc', pop them off.
4Step 4: After processing all characters, check if the stack is empty; if it is, return true; otherwise, return false.

solution.py9 lines

1def is_valid(s):
2    stack = []
3    for char in s:
4        stack.append(char)
5        if len(stack) >= 3 and stack[-3:] == ['a', 'b', 'c']:
6            stack.pop()
7            stack.pop()
8            stack.pop()
9    return len(stack) == 0

ℹ

Complexity note: The time complexity is O(n) because we traverse the string once, and the stack operations (push/pop) are O(1). The space complexity is O(n) in the worst case when no 'abc' sequences are formed.

1The string must be able to form complete 'abc' sequences.
2Using a stack helps efficiently manage the formation and removal of 'abc' sequences.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.