Finding the Minimum Window in S which Contains All Elements from T | LeetCode

Given a set T of characters and a string S, find the minimum window in S which will contain all the characters in T in complexity O(n).
eg,
S = “ADOBECODEBANC”
T = “ABC”
Minimum window is “BANC”.
From http://www.darrensunny.me/leetcode-minimum-window-substring/
 we can use a HashMap-like data structure (with name stat) to count the number of appearance of each character in T. Each time we encounter a character in T, we decrease its appearance in stat. We also maintain the number of matches between the characters in T and the window, by increasing it as long as the expected appearance of the encountered character is nonnegative. When the number of matches equals to the length of T, we know the current window meets the requirement.
e advance the right side of the working window if the current character is not included in T. Once it is included, we update stat and the number of matches (if necessary). And when the current window meets the requirement as we go, we advance the left side of the working window if it is safe to do, i.e. the character at the left does not appear in T, or it is redundant (observed by its expected appearance). When the left side of the window cannot be advanced, we have got a potential minimum such window.

public String minWindow(String S, String T) {

        if (S == null || S.length() == 0 || T == null || T.length() == 0)

            return "";

        int m = T.length(), n = S.length();

 

        // Preprocessing: count the number of appearance of each unique character in T

        HashMap<Character, Integer> stat = new HashMap<Character, Integer>();

        for (int i = 0; i < m; i++) {

            char c = T.charAt(i);

            if (stat.containsKey(c))

                stat.put(c, stat.get(c)+1);

            else

                stat.put(c, 1);

        }

 

        int matches = 0;        // Current number of matches

        int minimumBegin = -1, minimumEnd = n;  // Indices of the minimum window

        int begin = 0, end = 0;     // Indices of the working window

        for (; end < n; end++) {       // Move the right side of the working window

            // Move end to the next position where S[end] appears in T

            char endChar = S.charAt(end);

            if (!stat.containsKey(endChar))     // Ignore characters not in T

                continue;

            // Update stat, and matches if necessary

            stat.put(endChar, stat.get(endChar)-1); // Decrement the corresponding desired appearance

            if (stat.get(endChar) >= 0)     // The current matched character is not redundant

                matches++;

            // S[begin...end] meets the requirement

            while (matches == m) {

                // Move begin to the next position where S[begin] appears in T

                char beginChar = S.charAt(begin);

                while (!stat.containsKey(beginChar))

                    beginChar = S.charAt(++begin);

                // Update stat

                stat.put(beginChar, stat.get(beginChar)+1);

                // S[begin+1...end] lacks a character same as S[begin]

                // A possible minimum window is found

                if (stat.get(beginChar) > 0) {

                    if (end-begin < minimumEnd-minimumBegin) {

                        minimumBegin = begin;

                        minimumEnd = end;

                    }

                    matches--;

                }

                // Move the left side of the working window

                begin++;

            }

        }

 

        if (minimumBegin >= 0)      // A minimum such window was found

            return S.substring(minimumBegin, minimumEnd+1);

        else        // No such window was found

            return "";

    }

Code from http://leetcode.com/2010/11/finding-minimum-window-in-s-which.html

bool minWindow(const char* S, const char *T,

               int &minWindowBegin, int &minWindowEnd) {

  int sLen = strlen(S);

  int tLen = strlen(T);

  int needToFind[256] = {0};

 

  for (int i = 0; i < tLen; i++)

    needToFind[T[i]]++;

 

  int hasFound[256] = {0};

  int minWindowLen = INT_MAX;

  int count = 0;

  for (int begin = 0, end = 0; end < sLen; end++) {

    // skip characters not in T

    if (needToFind[S[end]] == 0) continue;

    hasFound[S[end]]++;

    if (hasFound[S[end]] <= needToFind[S[end]])

      count++;

 

    // if window constraint is satisfied

    if (count == tLen) {

      // advance begin index as far right as possible,

      // stop when advancing breaks window constraint.

      while (needToFind[S[begin]] == 0 ||

            hasFound[S[begin]] > needToFind[S[begin]]) {

        if (hasFound[S[begin]] > needToFind[S[begin]])

          hasFound[S[begin]]--;

        begin++;

      }

 

      // update minWindow if a minimum length is met

      int windowLen = end - begin + 1;

      if (windowLen < minWindowLen) {

        minWindowBegin = begin;

        minWindowEnd = end;

        minWindowLen = windowLen;

      } // end if

    } // end if

  } // end for

 

  return (count == tLen) ? true : false;

}

Read full article from Finding the Minimum Window in S which Contains All Elements from T | LeetCode