How do you solve Take K of Each Character From Left and Right on LeetCode?

Take K of Each Character From Left and Right (LeetCode #2516) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: You must check if it's possible to meet the requirement before proceeding.

What is the time complexity of Take K of Each Character From Left and Right?

The optimal solution for Take K of Each Character From Left and Right runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only traverse the string once with two pointers, making it much more efficient than the brute force approach.

What is the brute force solution for Take K of Each Character From Left and Right?

The brute force approach for Take K of Each Character From Left and Right is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible combination of characters taken from the left and right sides of the string. This means we will iterate through the string and count characters until we reach at least k of each character. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Take K of Each Character From Left and Right?

Take K of Each Character From Left and Right uses the following concepts: Hash Table, String, Sliding Window. The recommended patterns to study are: Hash Map, Sliding Window.

What are the interview tips for Take K of Each Character From Left and Right?

Always clarify edge cases, such as when k is 0 or greater than the length of the string. Explain your thought process clearly as you work through the problem. Practice optimizing your brute force solution to a more efficient approach.

What are common mistakes when solving Take K of Each Character From Left and Right?

Not checking if k is achievable before processing the string. Overlooking the need to count characters accurately from both ends.

#2516

Take K of Each Character From Left and Right

Medium

Hash TableStringSliding WindowHash MapSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently find the minimum number of characters needed to satisfy the k requirement for each character. This reduces the need for nested loops and improves performance.

⚙️

Algorithm

5 steps

1Step 1: Count the frequency of 'a', 'b', and 'c' in the string s.
2Step 2: If any character's count is less than k, return -1 immediately.
3Step 3: Use two pointers to represent the left and right ends of the string. Start with both pointers at the ends and move them inward while counting characters.
4Step 4: Maintain a count of characters taken from both ends and check if the counts meet the requirement of at least k for each character.
5Step 5: Update the minimum time taken whenever the requirement is met.

solution.py24 lines

1def minMinutes(s, k):
2    count = {'a': 0, 'b': 0, 'c': 0}
3    for char in s:
4        count[char] += 1
5    if count['a'] < k or count['b'] < k or count['c'] < k:
6        return -1
7    left, right = 0, len(s) - 1
8    total = 0
9    while left <= right:
10        total += 1
11        if s[left] == 'a': count['a'] -= 1
12        elif s[left] == 'b': count['b'] -= 1
13        else: count['c'] -= 1
14        left += 1
15        if all(v <= 0 for v in (count['a'] - k, count['b'] - k, count['c'] - k)):
16            return total
17        total += 1
18        if s[right] == 'a': count['a'] -= 1
19        elif s[right] == 'b': count['b'] -= 1
20        else: count['c'] -= 1
21        right -= 1
22        if all(v <= 0 for v in (count['a'] - k, count['b'] - k, count['c'] - k)):
23            return total
24    return -1

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once with two pointers, making it much more efficient than the brute force approach.

1You must check if it's possible to meet the requirement before proceeding.
2Using a sliding window can significantly reduce the time complexity.

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