查找包含另一个字符串作为子序列的最小长度 substring

Question

Let's say I have a string s1 = "eabegcghdefgh" and another string s2 = "egh" .

The code should return answer 4 because of the substring "efgh" of s1 (since it's the smallest substring which contains s2 as a subsequence).

Note: There can be a couple of substrings possible, but "efgh" is the smallest such substring.

In other words, find the length of smallest substring of s1 which contains all characters of another string s2 , but in order.

Please note: I want to ask how to do it in O(n) time complexity.

Answer 1

public static String smallestWindow(String S, String T) {
    if (S.equals(T))  //If S and T are equal, then simply return S.
        return S;
    /**
     * Use sliding window. 

If a substring W of S exists such that T is a subsequence of W, then a substring of S that is longer than or have the same length as W must also exist. Therefore, find such a substring of S starting from index 0 in S and index 0 in T. If the last character of T is reached, then the current index in S is the end index of the substring, and find the maximum possible start index of the substring in S. After the start index and the end index are found, the minimum window size can be updated, and store the start index and the end index as well.

Then set the index in T to 0 and try to find another substring in S. Repeat the process until the end of S is reached. After all the characters in S are visited, the minimum substring length can be obtained, and return the shortest substring.

     */
    int sLength = S.length(), tLength = T.length();
    int start = 0, end = sLength - 1;
    int sIndex = 0, tIndex = 0;
    while (sIndex < sLength) {
        if (S.charAt(sIndex) == T.charAt(tIndex))
            tIndex++;
        if (tIndex == tLength) {
            int rightIndex = sIndex;
            tIndex--;
            while (tIndex >= 0) {
                if (S.charAt(sIndex) == T.charAt(tIndex))
                    tIndex--;
                sIndex--;
            }
            sIndex++;
            if (rightIndex - sIndex < end - start) {
                start = sIndex;
                end = rightIndex;
            }
            tIndex = 0;
        }
        sIndex++;
    }
    int windowSize = end - start + 1;
        return S.substring(start, start + windowSize);
}

Answer 2

This is a basic Sliding window problem often known as "Smallest window in a string containing all the characters of another string ". We approach this problem by having two pointers "low" and "high" starting at the first index of the string to be searched and we increment the high pointer until all the characters of the pattern is matched and then we try to shrink the substring by incrementing the low pointer.

Here is a java snippet for the problem.

 public static String smallestWindow(String s, String pat){
    if(s == null || s.length() == 0)return "-1";
    int map[] = new int[128];
for(char k : pat.toCharArray())
map[k]++;
int count =0,start=-1,end=-1,lo=0,hi=0,min= Integer.MAX_VALUE;

for(hi = 0;hi<s.length();hi++)
{
    if(map[s.charAt(hi)] >0)
     count++;

     map[s.charAt(hi)]--;

     if(count == pat.length())
     {
         while(lo<hi && map[s.charAt(lo)] <0)
         { map[s.charAt(lo)]++;
             lo++;
         }
         if(min > hi - lo +1)
         {
             min = hi - lo+1;
             start = lo;
             end = hi+1;
         }
         map[s.charAt(lo)]++;

         lo++;
         count--; 
     }
}
return start == -1? "-1" : s.substring(start,end);
}

i recommend you to watch this video to understand it well. Smallest window in a string containing all the characters of another string