Leetcode 2516: Take K of Each Character From Left and Right
You are given a string
s consisting of the characters ‘a’, ‘b’, and ‘c’, and a non-negative integer k. Each minute, you can choose to take either the leftmost or the rightmost character from s. Your task is to determine the minimum number of minutes required to collect at least k occurrences of each character (‘a’, ‘b’, and ‘c’). If it is not possible to collect k of each character, return -1.Problem
Approach
Steps
Complexity
Input: You are given a string `s` consisting of only the letters 'a', 'b', and 'c', and a non-negative integer `k`.
Example: s = "abcabcabc", k = 2
Constraints:
• 1 <= s.length <= 10^5
• s consists of only the letters 'a', 'b', and 'c'.
• 0 <= k <= s.length
Output: Return the minimum number of minutes required to take at least `k` of each character, or -1 if it is not possible.
Example: Output: 6
Constraints:
• If it's not possible to take `k` of each character, return -1.
Goal: We need to compute the minimum number of minutes required to collect at least `k` occurrences of each character from the string `s`.
Steps:
• Count the occurrences of each character in `s`.
• If there are less than `k` occurrences of any character ('a', 'b', or 'c'), return -1.
• Use a sliding window technique to find the minimum subarray that contains at least `k` occurrences of each character.
• Return the size of this subarray, or -1 if no such subarray exists.
Goal: The string length and values for `k` are bounded by the problem constraints.
Steps:
• The string `s` will contain only 'a', 'b', and 'c'.
• The value of `k` must be less than or equal to the length of the string.
Assumptions:
• The input string `s` contains only the characters 'a', 'b', and 'c'.
• The integer `k` is non-negative and less than or equal to the length of the string.
• Input: s = "abcabcabc", k = 2
• Explanation: We need to find the shortest time to collect at least 2 occurrences of each character. By taking 3 characters from the left and 3 characters from the right, we can collect 2 'a's, 2 'b's, and 2 'c's in a total of 6 minutes.
• Input: s = "aaa", k = 2
• Explanation: Since there are no 'b' or 'c' characters in the string, it is impossible to collect at least 2 of each character, so the answer is -1.
Approach: The goal is to find the minimum number of minutes required to collect at least `k` of each character ('a', 'b', and 'c').
Observations:
• We can use a sliding window approach to find the shortest subarray containing the required characters.
• If we can't find a subarray with the required characters, return -1.
• The problem is essentially about finding the minimum window containing all the required characters.
Steps:
• Count the frequency of 'a', 'b', and 'c' in the string.
• Check if any of the characters have fewer than `k` occurrences, and return -1 if so.
• Use a sliding window to find the shortest window that contains at least `k` occurrences of each character.
• Return the size of the window, or -1 if no such window exists.
Empty Inputs:
• If `s` is empty, return -1.
Large Inputs:
• Ensure the solution works efficiently for large inputs (up to 10^5 characters).
Special Values:
• If `k` is 0, the result should be 0 because no characters need to be collected.
Constraints:
• Ensure that the sliding window approach handles all edge cases.
int takeCharacters(string s, int k) {
map<char, int> mp;
for(int i = 0; i < s.size(); i++)
mp[s[i]]++;
if(mp['a'] < k || mp['b'] < k || mp['c'] < k) return -1;
int j = 0, mx = 0;
for(int i = 0; i < s.size(); i++) {
// select max window which does not decrements frq below k;
mp[s[i]]--;
while(j <= i && mp[s[i]] < k) {
mp[s[j]]++;
j++;
}
mx = max(mx, i - j + 1);
}
return s.size() - mx;
}
1 : Function Definition
int takeCharacters(string s, int k) {
Define the function `takeCharacters`, which takes a string `s` and an integer `k` as parameters, and returns the minimum number of characters to remove to make the frequency of all characters at least `k`.
2 : Map Initialization
map<char, int> mp;
Initialize a map `mp` to store the frequency of each character in the string.
3 : Loop
for(int i = 0; i < s.size(); i++)
Iterate through each character in the string `s`.
4 : Frequency Counting
mp[s[i]]++;
Increment the frequency count of the current character in the map.
5 : Condition Check
if(mp['a'] < k || mp['b'] < k || mp['c'] < k) return -1;
Check if any of the characters 'a', 'b', or 'c' appear fewer than `k` times. If true, return -1 as it's not possible to achieve the condition.
6 : Variable Initialization
int j = 0, mx = 0;
Initialize two variables: `j` to represent the left boundary of the sliding window, and `mx` to store the maximum valid window size.
7 : Sliding Window
for(int i = 0; i < s.size(); i++) {
Start a second loop to iterate over the string, using a sliding window approach to find the largest window where all characters meet the frequency requirement.
8 : Decrement Frequency
mp[s[i]]--;
Decrement the frequency of the current character as it's included in the window.
9 : Sliding Window Adjustment
while(j <= i && mp[s[i]] < k) {
Start a while loop to move the left boundary `j` forward to maintain the frequency condition for all characters in the window.
10 : Increment Frequency
mp[s[j]]++;
Increment the frequency of the character at the left boundary `j` since it's being excluded from the window.
11 : Window Adjustment
j++;
Move the left boundary `j` forward to shrink the window.
12 : Max Window Size
mx = max(mx, i - j + 1);
Update the maximum valid window size if the current window is larger than the previously found windows.
13 : Return Statement
return s.size() - mx;
Return the number of characters to be removed, which is the total length of the string minus the size of the largest valid window.
Best Case: O(n)
Average Case: O(n)
Worst Case: O(n)
Description: We use a sliding window technique, which processes each character once.
Best Case: O(1)
Worst Case: O(1)
Description: We only need a fixed amount of space for the character counts.
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