How do you solve Find Valid Pair of Adjacent Digits in String on LeetCode?

Find Valid Pair of Adjacent Digits in String (LeetCode #3438) 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: Using a frequency map allows for efficient counting and checking of conditions in a single pass.

What is the brute force solution for Find Valid Pair of Adjacent Digits in String?

The brute force approach for Find Valid Pair of Adjacent Digits in String is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks every pair of adjacent digits in the string to see if they meet the criteria. It's straightforward but inefficient for larger strings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Valid Pair of Adjacent Digits in String?

Find Valid Pair of Adjacent Digits in String uses the following concepts: Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Valid Pair of Adjacent Digits in String?

Always clarify the problem requirements before jumping into coding. Think about edge cases, such as strings with only one type of digit. Discuss your thought process and approach with the interviewer before coding.

What are common mistakes when solving Find Valid Pair of Adjacent Digits in String?

Forgetting to check if the digits in the pair are different. Not correctly counting the occurrences of each digit.

#3438

Find Valid Pair of Adjacent Digits in String

Easy

Hash TableStringCountingHash MapArray

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)

This approach uses a HashMap to count the occurrences of each digit first, allowing us to check pairs in a single pass, which is much more efficient.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each digit in the string.
2Step 2: Iterate through the string and for each pair of adjacent digits, check if they are different.
3Step 3: For each valid pair, check their counts in the frequency map against their numeric values. If valid, return the pair.

solution.py8 lines

1def find_valid_pair(s):
2    from collections import Counter
3    freq = Counter(s)
4    for i in range(len(s) - 1):
5        if s[i] != s[i + 1]:
6            if freq[s[i]] == int(s[i]) and freq[s[i + 1]] == int(s[i + 1]):
7                return s[i] + s[i + 1]
8    return ''

ℹ

Complexity note: The time complexity is O(n) because we traverse the string twice: once for counting and once for checking pairs. The space complexity is O(n) due to the frequency map.

1Using a frequency map allows for efficient counting and checking of conditions in a single pass.
2Adjacent pairs can be checked in linear time after counting, significantly improving performance.

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