How do you solve Minimum Consecutive Cards to Pick Up on LeetCode?

Minimum Consecutive Cards to Pick Up (LeetCode #2260) 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 HashMap allows for efficient lookups and updates, reducing the need for nested loops.

What is the brute force solution for Minimum Consecutive Cards to Pick Up?

The brute force approach for Minimum Consecutive Cards to Pick Up is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray of cards to find the minimum length that contains a pair of matching cards. This is straightforward but inefficient as it examines all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Consecutive Cards to Pick Up?

Minimum Consecutive Cards to Pick Up uses the following concepts: Array, Hash Table, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Consecutive Cards to Pick Up?

Always consider the simplest approach first, then think about optimizations. Explain your thought process clearly; it shows your understanding of the problem. Practice identifying patterns in problems, as they can help you recognize efficient solutions.

What are common mistakes when solving Minimum Consecutive Cards to Pick Up?

Not using a HashMap to track indices, leading to unnecessary nested loops. Failing to handle cases where no matching cards exist, returning incorrect results.

#2260

Minimum Consecutive Cards to Pick Up

Medium

ArrayHash TableSliding WindowHash 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)

The optimal approach uses a HashMap to track the last index of each card value. This allows us to efficiently find the distance between matching cards in a single pass through the array.

⚙️

Algorithm

6 steps

1Step 1: Initialize a HashMap to store the last occurrence of each card value.
2Step 2: Initialize a variable to store the minimum length found, set to infinity.
3Step 3: Iterate through the cards array, for each card, check if it exists in the HashMap.
4Step 4: If it exists, calculate the distance from the last occurrence to the current index and update the minimum length.
5Step 5: Update the HashMap with the current index of the card.
6Step 6: After the loop, return the minimum length found or -1 if no pairs exist.

solution.py8 lines

1def minimumCardPickup(cards):
2    last_index = {}
3    min_length = float('inf')
4    for i, card in enumerate(cards):
5        if card in last_index:
6            min_length = min(min_length, i - last_index[card] + 1)
7        last_index[card] = i
8    return min_length if min_length != float('inf') else -1

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the cards array. The space complexity is O(n) due to the HashMap storing the last indices of card values.

1Using a HashMap allows for efficient lookups and updates, reducing the need for nested loops.
2The problem can be solved in a single pass, making it more efficient than checking all pairs.

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