Leetcode Longest Repeating Character Replacement problem solution

In this Leetcode Longest Repeating Character Replacement problem solution, You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times.

Return the length of the longest substring containing the same letter you can get after performing the above operations.

Problem solution in Python.

class Solution(object):
	def characterReplacement(self, s, k):
		"""
		:type s: str
		:type k: int
		:rtype: int
		using the sliding window technique
		"""
		window_start = 0
		longest_substring, most_repeated = 0 , 0
		freq_map = {}

		for window_end in range(len(s)):
			curr_char = s[window_end]
			if curr_char not in freq_map:
				freq_map[curr_char] = 0
			freq_map[curr_char] += 1

			most_repeated = max(most_repeated, freq_map[curr_char])

			if window_end - window_start + 1 - most_repeated > k:
				left_char = s[window_start]
				freq_map[left_char] -= 1
				window_start += 1
			longest_substring = max(longest_substring, window_end - window_start + 1)

		return longest_substring

Problem solution in Java.

public int characterReplacement(String s, int k) {
int start=0;
int longCharCount=0;
Map<Character, Integer> map = new HashMap<>();
int maxLength=0;

    for(int end=0;end<s.length();end++){
        char endchar = s.charAt(end);
        map.put(endchar, map.getOrDefault(endchar,0)+1);
        longCharCount = Math.max(longCharCount, map.get(endchar));
        
        if(end-start+1-longCharCount >k){
            char startChar=s.charAt(start);
            map.put(startChar, map.getOrDefault(startChar,0)-1);
            start++;
        }
        
        maxLength = Math.max(maxLength, end-start+1);
    }
    
    return maxLength;

}

Problem solution in C++.

class Solution {
public:
    int characterReplacement(string s, int k) {
        int size = s.size(); int ret = 0;
        vector<int> count(26, 0);
        int start = 0; int end = 0; int Freq = 0; 
        for(; end<size; end++){
            count[s[end]-'A'] += 1;
            Freq = max(Freq, count[s[end]-'A']);
            while((end-start+1)-Freq > k) {
               
                count[s[start]-'A'] -= 1;
                start += 1;
                
            }
            ret=max(ret,end-start+1);
        }
        return ret;
    }
};

Problem solution in C.

int characterReplacement(char * s, int k){
    if(s[0]==0)return 0;
    int n=strlen(s);
    int max=0;
    int i;
    int diff=0;
    int *record=calloc(26,4);
    int l;
    
    for(i=0,l=0;i<n;i++){
        if(++record[s[i]-'A']<=max){
            if(diff==k){
                record[s[l++]-'A']--;
            }
            else{
                diff++;
            }
        }
        else {
            max++;
        }   
    }    
    return max+diff;
}

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