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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.
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.length
n == t.length
1 <= m, n <= 105
s
andt
consist of uppercase and lowercase English letters.
Follow up: Could you find an algorithm that runs in O(m + n)
time?
Python
# time complexity: O(len(s) + len(t))
# space complexity: O(len(s) + len(t))
from collections import Counter, defaultdict
class Solution:
def minWindow(self, s: str, t: str) -> str:
reqCount = defaultdict(int)
window = defaultdict(int)
result = [-1, -1]
resultLen = float('inf')
current = 0
for char in t:
reqCount[char] += 1
required = len(reqCount)
left = 0
for right in range(len(s)):
char = s[right]
if char in reqCount:
window[char] += 1
if window[char] == reqCount[char]:
current += 1
while current == required:
if (right - left + 1) < resultLen:
resultLen = right - left + 1
result = [left, right]
leftChar = s[left]
if leftChar in window:
window[leftChar] -= 1
if window[leftChar] < reqCount[leftChar]:
current -= 1
left += 1
return s[result[0]:result[1] + 1] if resultLen != float('inf') else ""
S = "ADOBECODEBANC"
T = "ABC"
print(Solution().minWindow(S, T))