You are given a string s consisting of the characters 'a', 'b', and 'c' and a non-negative integer k. Each minute, you may take either the leftmost character of s, or the rightmost character of s.
Return the minimum number of minutes needed for you to take at least k of each character, or return -1 if it is not possible to take k of each character.
Example 1:
Input: s = "aabaaaacaabc", k = 2 Output: 8 Explanation: Take three characters from the left of s. You now have two 'a' characters, and one 'b' character. Take five characters from the right of s. You now have four 'a' characters, two 'b' characters, and two 'c' characters. A total of 3 + 5 = 8 minutes is needed. It can be proven that 8 is the minimum number of minutes needed.
Example 2:
Input: s = "a", k = 1 Output: -1 Explanation: It is not possible to take one 'b' or 'c' so return -1.
Constraints:
1 <= s.length <= 105s consists of only the letters 'a', 'b', and 'c'.0 <= k <= s.lengthThis approach utilizes a two-pointer technique to simultaneously check from the left and from the right, aiming to fulfill the requirement of obtaining at least 'k' of each character ('a', 'b', 'c'). The key idea is to keep track of character counts from both ends and to update pointers strategically to achieve the minimum number of moves.
This Python solution first checks if it's possible to form 'k' occurrences of each character by counting their frequencies. The two-pointer approach is implemented to keep track of minimal steps needed by effectively sliding a window approach from both ends of the string. Adjust pointers and update current character counts to determine the smallest window, ensuring minimal moves.
C++
Time Complexity: O(n), where n is the length of the string, as every character is processed at most twice.
Space Complexity: O(1), as the storage requirement does not scale with input size beyond a fixed size.
This approach involves identifying the largest contiguous subarray that lacks at least 'k' of one or more characters. By calculating the complement of such subarrays, we aim to find the minimal total of minutes required.
The Java implementation calculates the shortest segment where all characters meet the conditions less than `k`. The required length is derived from the complement idea and uses a greedy sliding window strategy to find the maximum length of allowable segments, yielding the minimal minutes required for removal.
JavaScript
Time Complexity: O(n), stemming from a single pass to compute with sliding window technique.
Space Complexity: O(1) as the space needed is independent of the input size.
| Approach | Complexity |
|---|---|
| Two-Pointer Technique | Time Complexity: O(n), where n is the length of the string, as every character is processed at most twice. |
| Sliding Window Complement (Greedy) | Time Complexity: O(n), stemming from a single pass to compute with sliding window technique. |
Take K of Each Character From Left and Right - Leetcode - Python • NeetCodeIO • 14,779 views views
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