What is the brute force approach and complexity for Hand of Straights?

Backtrack: try every card value as the start of the next group, and undo when a choice fails with time complexity O(N^(N/G)) and space complexity O(N).

How do you identify the pattern for Hand of Straights?

Look for key signals in the prompt: "groups of consecutive cards" indicates A group is a run of values, so its start value decides the whole group.; "rearrange into groups of groupSize" indicates Every card must be used, so the copies of each value matter.; "return true or false" indicates Any one valid split is enough. We never need to report the split itself..

Hand of Straights

Medium·LeetCode #846 Pro OnlyBasic Reinforcement

LeetCode

Problem Statement

Alice has a hand of cards. Each card shows a number. She wants to split all of her cards into groups of exactly groupSize cards, where the numbers in each group are consecutive, like 3, 4, 5.

Return true if she can do this, and false if not.

Input

An array hand of card values, and an integer groupSize.

Output

true if the cards can be split into groups of consecutive values, otherwise false.

Example 1

Input: hand = [1,2,3,6,2,3,4,7,8], groupSize = 3

Output: true

Split into [1,2,3], [2,3,4] and [6,7,8].

Example 2

Input: hand = [1,2,3,4,5], groupSize = 4

Output: false

Five cards can't be split into groups of four.

Example 3

Input: hand = [1,2,3,3,4,5], groupSize = 3

Output: true

Two copies of 3 are fine: [1,2,3] and [3,4,5].

Constraints

• 1 ≤ hand.length ≤ 10^4
• 0 ≤ hand[i] ≤ 10^9
• 1 ≤ groupSize ≤ hand.length

Pattern Recognition Clues

Recognising these key signals in the problem description immediately identifies the pattern:

“groups of consecutive cards”

→ A group is a run of values, so its start value decides the whole group.

“rearrange into groups of groupSize”

→ Every card must be used, so the copies of each value matter.

“return true or false”

→ Any one valid split is enough. We never need to report the split itself.

Approach Complexity Ladder

Brute Force — try every choice of group startO(N^(N/G))O(N)

Backtrack: try every card value as the start of the next group, and undo when a choice fails

Optimal — ordered map of counts, always start from the smallest cardO(N log N)O(N)

Ordered map of counts; the smallest remaining card starts each group, which takes groupSize consecutive values

Brute Force Intuition

Pick any unused card as the start of the next group. The group is that value and the next groupSize − 1 values. Take those cards out, and recurse on what is left. If the rest fails, put the cards back and try another start.

Recognition rule for this sub-pattern:

The best choice changes as you go — keep candidates in a heap and always pop the current best.

Remember this · Greedy

Trigger

Maximise count / minimise cost with a natural "best next step", sorting makes the choice obvious, reachability.

Think

Is there a choice I can always make safely without looking ahead?

Key question

Why can't the greedy choice be worse than the optimal one? (Try to break it with a small example.)

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