最小視窗子字串
“最小視窗子串”問題的詳細指南和 Python 實作。
1. 學習
「最小視窗子字串」問題是滑動視窗部分的關鍵挑戰。
此實作著重於 Python 中的簡單層級邏輯。
在我們提供的解決方案中,我們優先考慮技術準確性和程式碼可讀性。
2. Real-World Applications
3. Visual Intuition
可視化最小視窗子字串的邏輯流程。
4. Prerequisites
5. Step-by-Step Thinking
1. Understand the problem
仔細閱讀最小視窗子字串的問題陳述。
2. Formulate brute force
起草一個簡單的迭代解決方案。
3. Identify inefficiency
尋找冗餘計算。
4. Optimize search path
使用散列或排序來加速該過程。
5. Final Implementation
清理生產標準代碼。
問題陳述
給定長度分別為 m 和 n 的兩個字串 s 和 t,傳回 s 的最小視窗子字串,使得 t 中的每個字元(包括重複字元)都包含在視窗中。如果不存在這樣的子字串,則傳回空字串 ""。
答案保證是唯一的。
寫一個函數 minWindow(s: str, t: str) -> str。
- •m == len(s), n == len(t)
- •1 <= m, n <= 10^5
- •s and t consist of uppercase and lowercase English letters
範例
s = "ADOBECODEBANC", t = "ABC"
"BANC"
The minimum window substring "BANC" (indices 9-12) contains 'A', 'B', and 'C' from t.
s = "a", t = "a"
"a"
The entire string s is the minimum window.
s = "a", t = "aa"
""
Both 'a's from t must be included. Since s only has one 'a', return empty string.
Need a Hint?
Edge Cases to Watch
- 空輸入結構
- 單元素輸入
- 大數值範圍
準備好解決了嗎?
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面試見解和變化
複雜度分析分解
為什麼時間: Directly evaluates all possibilities.
為什麼選擇太空: Uses standard local memory.
為什麼時間: Optimized paths reduce total operations.
為什麼選擇太空: May trade memory for speed.
最佳化解決方案Python程式碼
最佳化解決方案Python程式碼
def min_window_opt(s, t):
if t == "": return ""
countT, window = {}, {}
for c in t: countT[c] = 1 + countT.get(c, 0)
have, need = 0, len(countT)
res, resLen = [-1, -1], float("infinity")
l = 0
for r in range(len(s)):
c = s[r]
window[c] = 1 + window.get(c, 0)
if c in countT and window[c] == countT[c]:
have += 1
while have == need:
if (r - l + 1) < resLen:
resLen = r - l + 1
res = [l, r]
window[s[l]] -= 1
if s[l] in countT and window[s[l]] < countT[s[l]]:
have -= 1
l += 1
l, r = res
return s[l : r + 1] if resLen != float("infinity") else ""暴力破解代碼(劇透保護)
暴力破解代碼(劇透保護)
def min_window_brute(s, t):
if not t: return ""
def contains_all(counts_s, counts_t):
for char in counts_t:
if counts_s.get(char, 0) < counts_t[char]:
return False
return True
t_counts = {}
for c in t: t_counts[c] = 1 + t_counts.get(c, 0)
res, res_len = "", float("inf")
for i in range(len(s)):
for j in range(i, len(s)):
sub = s[i : j + 1]
sub_counts = {}
for c in sub: sub_counts[c] = 1 + sub_counts.get(c, 0)
if contains_all(sub_counts, t_counts):
if (j - i + 1) < res_len:
res_len = j - i + 1
res = sub
return resAlgorithm Pattern Checklist
When dealing with Sliding Window data patterns.
- Are constraints clear?
- Is there a linear or logarithmic optimization possible?
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