How do you solve Best Poker Hand on LeetCode?

Best Poker Hand (LeetCode #2347) 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: Understanding the hierarchy of poker hands is crucial.

What is the brute force solution for Best Poker Hand?

The brute force approach for Best Poker Hand is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will check each condition for the poker hand in a straightforward manner. We will first check if all cards have the same suit, then check for three of a kind, a pair, and finally default to high card. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Best Poker Hand?

Best Poker Hand uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Best Poker Hand?

Clarify the rules of poker hands if unsure. Think about edge cases, like all cards being the same rank or suit. Explain your thought process as you implement the solution.

What are common mistakes when solving Best Poker Hand?

Not checking all suits before counting ranks. Overlooking the need to check for the best hand hierarchy.

#2347

Best Poker Hand

Easy

ArrayHash TableCountingHash 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)

In the optimal approach, we will use a single pass to count the ranks and check the suits simultaneously. This reduces the number of checks and improves efficiency.

⚙️

Algorithm

3 steps

1Step 1: Initialize a counter for ranks and a flag for suits.
2Step 2: Traverse through the ranks and suits to populate the rank counter and check if all suits are the same.
3Step 3: Determine the best hand based on the counts.

solution.py10 lines

1def bestHand(ranks, suits):
2    suit_check = len(set(suits)) == 1
3    count = Counter(ranks)
4    if suit_check:
5        return 'Flush'
6    if any(v >= 3 for v in count.values()):
7        return 'Three of a Kind'
8    if any(v == 2 for v in count.values()):
9        return 'Pair'
10    return 'High Card'

ℹ

Complexity note: The complexity is O(n) because we are only making a single pass through the ranks and suits, counting occurrences and checking conditions simultaneously.

1Understanding the hierarchy of poker hands is crucial.
2Efficient counting and checking can significantly reduce complexity.

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