lintcode-medium-Minimum Window Substring

Given a string source and a string target, find the minimum window in source which will contain all the characters in target.

Notice

If there is no such window in source that covers all characters in target, return the emtpy string "".

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

Clarification

Should the characters in minimum window has the same order in target?

• Not necessary.

Example

For source = "ADOBECODEBANC", target = "ABC", the minimum window is"BANC"

Challenge

Can you do it in time complexity O(n) ?

这题思路和minimum sum subarray差不多

public class Solution {
    /**
     * @param source: A string
     * @param target: A string
     * @return: A string denote the minimum window
     *          Return "" if there is no such a string
     */
    public String minWindow(String source, String target) {
        // write your code

        if(source == null || source.length() == 0 || target == null || target.length() == 0 || source.length() < target.length())
            return "";

        int[] tar = new int[128];
        int[] sou = new int[128];

        for(int i = 0; i < target.length(); i++)
            tar[target.charAt(i)]++;

        int left = 0;
        int right = 0;
        int start = -1;
        int end = source.length();
        int win = 0;

        int len = source.length();

        for(; right < len; right++){
            if(tar[source.charAt(right)] != 0){
                sou[source.charAt(right)]++;

                if(sou[source.charAt(right)] <= tar[source.charAt(right)]){
                    win++;
                }

                if(win == target.length()){
                    while(left < right){
                        if(tar[source.charAt(left)] == 0){
                            left++;
                        }
                        else if(sou[source.charAt(left)] > tar[source.charAt(left)]){
                            sou[source.charAt(left)]--;
                            left++;
                        }
                        else{
                            break;
                        }
                    }

                    win--;

                    if(right - left < end - start){
                        start = left;
                        end = right;
                    }
                }

            }
        }

        if(start == -1)
            return "";
        else
            return source.substring(start, end + 1);
    }
}