LeetCode #2347 — EASY

Best Poker Hand

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

You are given an integer array ranks and a character array suits. You have 5 cards where the i^th card has a rank of ranks[i] and a suit of suits[i].

The following are the types of poker hands you can make from best to worst:

"Flush": Five cards of the same suit.
"Three of a Kind": Three cards of the same rank.
"Pair": Two cards of the same rank.
"High Card": Any single card.

Return a string representing the best type of poker hand you can make with the given cards.

Note that the return values are case-sensitive.

Example 1:

Input: ranks = [13,2,3,1,9], suits = ["a","a","a","a","a"]
Output: "Flush"
Explanation: The hand with all the cards consists of 5 cards with the same suit, so we have a "Flush".

Example 2:

Input: ranks = [4,4,2,4,4], suits = ["d","a","a","b","c"]
Output: "Three of a Kind"
Explanation: The hand with the first, second, and fourth card consists of 3 cards with the same rank, so we have a "Three of a Kind".
Note that we could also make a "Pair" hand but "Three of a Kind" is a better hand.
Also note that other cards could be used to make the "Three of a Kind" hand.

Example 3:

Input: ranks = [10,10,2,12,9], suits = ["a","b","c","a","d"]
Output: "Pair"
Explanation: The hand with the first and second card consists of 2 cards with the same rank, so we have a "Pair".
Note that we cannot make a "Flush" or a "Three of a Kind".

Constraints:

ranks.length == suits.length == 5
1 <= ranks[i] <= 13
'a' <= suits[i] <= 'd'
No two cards have the same rank and suit.

Step 01

Brute Force Baseline

Problem summary: You are given an integer array ranks and a character array suits. You have 5 cards where the ith card has a rank of ranks[i] and a suit of suits[i]. The following are the types of poker hands you can make from best to worst: "Flush": Five cards of the same suit. "Three of a Kind": Three cards of the same rank. "Pair": Two cards of the same rank. "High Card": Any single card. Return a string representing the best type of poker hand you can make with the given cards. Note that the return values are case-sensitive.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map

Example 1

[13,2,3,1,9]
["a","a","a","a","a"]

Example 2

[4,4,2,4,4]
["d","a","a","b","c"]

Example 3

[10,10,2,12,9]
["a","b","c","a","d"]

Core Insight

What unlocks the optimal approach

Sequentially check the conditions 1 through 4, and return the outcome corresponding to the first met condition.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2347: Best Poker Hand
class Solution {
    public String bestHand(int[] ranks, char[] suits) {
        boolean flush = true;
        for (int i = 1; i < 5 && flush; ++i) {
            flush = suits[i] == suits[i - 1];
        }
        if (flush) {
            return "Flush";
        }
        int[] cnt = new int[14];
        boolean pair = false;
        for (int x : ranks) {
            if (++cnt[x] == 3) {
                return "Three of a Kind";
            }
            pair = pair || cnt[x] == 2;
        }
        return pair ? "Pair" : "High Card";
    }
}

// Accepted solution for LeetCode #2347: Best Poker Hand
func bestHand(ranks []int, suits []byte) string {
	flush := true
	for i := 1; i < 5 && flush; i++ {
		flush = suits[i] == suits[i-1]
	}
	if flush {
		return "Flush"
	}
	cnt := [14]int{}
	pair := false
	for _, x := range ranks {
		cnt[x]++
		if cnt[x] == 3 {
			return "Three of a Kind"
		}
		pair = pair || cnt[x] == 2
	}
	if pair {
		return "Pair"
	}
	return "High Card"
}

# Accepted solution for LeetCode #2347: Best Poker Hand
class Solution:
    def bestHand(self, ranks: List[int], suits: List[str]) -> str:
        # if len(set(suits)) == 1:
        if all(a == b for a, b in pairwise(suits)):
            return 'Flush'
        cnt = Counter(ranks)
        if any(v >= 3 for v in cnt.values()):
            return 'Three of a Kind'
        if any(v == 2 for v in cnt.values()):
            return 'Pair'
        return 'High Card'

// Accepted solution for LeetCode #2347: Best Poker Hand
impl Solution {
    pub fn best_hand(ranks: Vec<i32>, suits: Vec<char>) -> String {
        if suits.iter().all(|v| *v == suits[0]) {
            return "Flush".to_string();
        }
        let mut count = [0; 14];
        let mut is_pair = false;
        for &v in ranks.iter() {
            let i = v as usize;
            count[i] += 1;
            if count[i] == 3 {
                return "Three of a Kind".to_string();
            }
            is_pair = is_pair || count[i] == 2;
        }
        (if is_pair { "Pair" } else { "High Card" }).to_string()
    }
}

// Accepted solution for LeetCode #2347: Best Poker Hand
function bestHand(ranks: number[], suits: string[]): string {
    if (suits.every(v => v === suits[0])) {
        return 'Flush';
    }
    const count = new Array(14).fill(0);
    let isPair = false;
    for (const v of ranks) {
        if (++count[v] === 3) {
            return 'Three of a Kind';
        }
        isPair = isPair || count[v] === 2;
    }
    if (isPair) {
        return 'Pair';
    }
    return 'High Card';
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.