Related: LeetCode 67 - Minimum Window Substring
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.
http://tech-queries.blogspot.com/2010/12/finding-minimum-window-in-array-which.html
Read full article from Finding the Minimum Window in S which Contains All Elements from T | LeetCode
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 "";
}
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;
}
http://tech-queries.blogspot.com/2010/12/finding-minimum-window-in-array-which.html
Read full article from Finding the Minimum Window in S which Contains All Elements from T | LeetCode
