LeetCode-in-Ruby.github.io
76. Minimum Window Substring
Hard
Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring__, return the empty string "".
The testcases will be generated such that the answer is unique.
A substring is a contiguous sequence of characters within the string.
Example 1:
Input: s = “ADOBECODEBANC”, t = “ABC”
Output: “BANC”
Explanation: The minimum window substring “BANC” includes ‘A’, ‘B’, and ‘C’ from string t.
Example 2:
Input: s = “a”, t = “a”
Output: “a”
Explanation: The entire string s is the minimum window.
Example 3:
Input: s = “a”, t = “aa”
Output: “”
Explanation: Both ‘a’s from t must be included in the window. Since the largest window of s only has one ‘a’, return empty string.
Constraints:
m == s.lengthn == t.length1 <= m, n <= 105sandtconsist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n) time?
Solution
# @param {String} s
# @param {String} t
# @return {String}
def min_window(s, t)
return '' if t.size > s.size
fixnum_max = (2 ** (0.size * 8 - 2) - 1)
mp = Hash.new(0)
t.each_char do |c|
mp[c] += 1
end
begin_ptr = 0
end_ptr = 0
counter = mp.size
len = fixnum_max
head = 0
while end_ptr < s.size
ch = s[end_ptr].chr
if mp.include?(ch)
mp[ch] -= 1
counter -= 1 if mp[ch].zero?
end
end_ptr += 1
while counter.zero?
tmp = s[begin_ptr].chr
if mp.include?(tmp)
mp[tmp] += 1
counter += 1 if mp[tmp].positive?
end
if end_ptr - begin_ptr < len
len = end_ptr - begin_ptr
head = begin_ptr
end
begin_ptr += 1
end
end
len == fixnum_max ? '' : s[head..head + len - 1]
end