Sunday, October 16, 2016

LeetCode 424 - Longest Repeating Character Replacement


https://leetcode.com/problems/longest-repeating-character-replacement/
Given a string that consists of only uppercase English letters, you can replace any letter in the string with another letter at most k times. Find the length of a longest substring containing all repeating letters you can get after performing the above operations.
Note:
Both the string's length and k will not exceed 104.
Example 1:
Input:
s = "ABAB", k = 2

Output:
4

Explanation:
Replace the two 'A's with two 'B's or vice versa.
Example 2:
Input:
s = "AABABBA", k = 1

Output:
4

Explanation:
Replace the one 'A' in the middle with 'B' and form "AABBBBA".
The substring "BBBB" has the longest repeating letters, which is 4.

https://discuss.leetcode.com/topic/63471/java-sliding-window-easy-to-understand
The problem is similar to longest substring with most K distinct characte. But this time, the constraint is we can only have most K characters that is different with the most frequent character in the substring. For example in the sliding window:
"ABBBAC" most frequent character is B with count 3, all other character is count as different to B, 
    which is A and C, and the result is 2 + 1 = 3. 
Each time we count the different characters. If it is not bigger than k we extend the sliding window.
Since we only have 26 characters, keep the count in a integer array is good enough.
    public int characterReplacement(String s, int k) {
        if(s == null || s.length() == 0){
            return 0;
        }
        int max = 0;
        int[] ch = new int[26];
        char[] str = s.toCharArray();
        for(int i=0, j=0; i<s.length(); i++){
            while(j < s.length()){
                ch[str[j] - 'A']++;
                if(count(ch) > k){  //If exceed k, break
                    ch[str[j] - 'A']--;
                    break;
                }
                j++;
            }
            max = Math.max(max, j-i);
            ch[str[i] - 'A']--;
        }
        return max;
    }
    //Count the number of character that is different to the longest character
    public int count(int[] ch){
        int max = 0;
        int sum = 0;
        for(int val:ch){
            sum += val;
            max = Math.max(max, val);
        }
        return sum - max;
    }
https://discuss.leetcode.com/topic/63416/sliding-window-similar-to-finding-longest-substring-with-k-distinct-characters/5
The problem says that we can make at most k changes to the string (any character can be replaced with any other character). So, let's say there were no constraints like the k. Given a string convert it to a string with all same characters with minimal changes. The answer to this is
length of the entire string - number of times of the maximum occurring character in the string
Given this, we can apply the at most k changes constraint and maintain a sliding window such that
(length of substring - number of times of the maximum occurring character in the substring) <= k
    int characterReplacement(string s, int k) {
        vector<int> counts(26, 0);
        int start = 0;
        int maxCharCount = 0;
        int n = s.length();
        int result = 0;
        for(int end = 0; end < n; end++){
            counts[s[end]-'A']++;
            if(maxCharCount < counts[s[end]-'A']){
                maxCharCount = counts[s[end]-'A'];
            }
            while(end-start-maxCharCount+1 > k){
                counts[s[start]-'A']--;
                start++;
                for(int i = 0; i < 26; i++){
                    if(maxCharCount < counts[i]){
                        maxCharCount = counts[i];
                    }
                }
            }
            result = max(result, end-start+1);
        }
        return result;
    }
https://discuss.leetcode.com/topic/63494/java-12-lines-o-n-sliding-window-solution-with-explanation
There's no edge case for this question. The initial step is to extend the window to its limit, that is, the longest we can get to with maximum number of modifications. Until then the variable start will remain at 0.
Then as end increase, the whole substring from 0 to end will violate the rule, so we need to update start accordingly (slide the window). We move start to the right until the whole string satisfy the constraint again. Then each time we reach such situation, we update our max length.
The solution is great, but I have a small question here :
when you move the start pointer, why we the max count may change:
for example :
s = "bbcdef", k = 2.
when move end to 'e', we need to remove first 'b' so that maxcount will change to 1.
But for s = "bbcdd", it wont change when we move end to second 'd' and remove first 'b'.
In this question we may don't need to worry about that because we just need to get maxlength of substring. But what if the question is to find all the strategy, can we try to use similar method ?

when move the left begin of bmaxcount will not be change to 1.It will always the biggest one unless another one is more than it. That is why explain that
our sliding windows need not shrink, even if a window may cover an invalid substring
this situation it will shift the whole window to the right by one
    public int characterReplacement(String s, int k) {
        int len = s.length();
        int[] count = new int[26];
        int start = 0, maxCount = 0, maxLength = 0;
        for (int end = 0; end < len; end++) {
            maxCount = Math.max(maxCount, ++count[s.charAt(end) - 'A']);
            while (end - start + 1 - maxCount > k) {
                count[s.charAt(start) - 'A']--;
                start++;
            }
            maxLength = Math.max(maxLength, end - start + 1);
        }
        return maxLength;
    }






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