Template : Two Pointers & Hash -> String process

Reference : https://leetcode.com/discuss/72701/here-10-line-template-that-can-solve-most-substring-problems

/* For most substring problem, we are given a string and need to find a substring of it which satisfy some restrictions. A general way is to use a hashmap assisted with two pointers. */

int findSubstring(string s){
        int[] map = new int[256];
        int counter; // check whether the substring is valid
        int begin=0, end=0; //two pointers, one point to tail and one  head
        int d; //the length of substring

        for() { /* initialize the hash map here */ }

        while(end<s.size()){

            if(map[s[end++]]-- ?){  /* modify counter here */ }

            while(/* counter condition */){ 

                 /* update d here if finding minimum*/

                //increase begin to make it invalid/valid again

                if(map[s[begin++]]++ ?){ /*modify counter here*/ }
            }  

            /* update d here if finding maximum*/
        }
        return d;
  }

3. Longest Substring Without Repeating Characters

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        int[] map = new int[256];
        
        int begin = 0, end = 0; 
        int count = 0, max = 0;
        while (end < s.length()) {
            if (map[s.charAt(end++)]++ > 0) count++;
            while (count > 0) if (map[s.charAt(begin++)]-- > 1) count--;
            max = max > (end - begin) ? max : (end - begin);
        }
        
        return max;
    }
}

159. Longest Substring with At Most Two Distinct Characters

Given a string, find the length of the longest substring T that contains at most 2 distinct characters.

For example, Given s = “eceba”,

T is "ece" which its length is 3.

public class Solution {
    public int lengthOfLongestSubstringTwoDistinct(String s) {
        if (s == null || s.isEmpty()) return 0;
        if (s.length() <= 2) return s.length();
        
        int[] map = new int[256];
        
        int begin = 0, end = 0; // two pointers
        int count = 0, max = 0;
        
        while (end < s.length()) {
            if (map[s.charAt(end++)]++ == 0) count++; // new diff character
            while (count > 2) if (map[s.charAt(begin++)]-- == 1) count--;
            max = max > (end - begin) ? max : (end - begin);
        }
        
        return max;
    }
}

340. Longest Substring with At Most K Distinct Characters

Given a string, find the length of the longest substring T that contains at most k distinct characters.

For example, Given s = “eceba” and k = 2,

T is "ece" which its length is 3.

public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        if (k == 0 || s == null || s.isEmpty()) return 0;
        if (s.length() <= k) return s.length();
        
        int[] map = new int[256];
        int begin = 0, end = 0;
        int count = 0, max = 0;
        
        while (end < s.length()) {
            if (map[s.charAt(end++)]++ == 0) count++;
            while (count > k) if (map[s.charAt(begin++)]-- == 1) count--;
            max = max > (end - begin) ? max : (end - begin);
        }
        
        return max;
    }
}

76. Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

public class Solution {
    public String minWindow(String s, String t) {
        if (s == null || s.isEmpty() || t == null || t.isEmpty()) return "";
        int[] map =new int[256];
        
        for (int i = 0; i < t.length(); i++) {
            map[t.charAt(i)]++;
        }
        
        int begin = 0, head = 0, end = 0;
        int min = Integer.MAX_VALUE;
        int count = t.length();
        
        while (end < s.length()) {
            if (map[s.charAt(end++)]-- > 0) count--;
            while (count == 0) {
                min = min > end - begin ? (end - (head = begin)) : min; 
                if (map[s.charAt(begin++)]++ >= 0) count++;
            }
        }
        
        return min == Integer.MAX_VALUE ? "" : s.substring(head, head + min);
    }
}