76. Minimum Window Substring

When t's elements are not all present: move right pointer right
When all of t's elements are present: move left pointer right; if right pointer hasn't reached the end, continue moving right pointer
If t is found: break and restart

Optimization

The code finds the minimum length window in string s that contains all characters in string t. It uses two pointers, left and right, to traverse s and track the window. The isAll function checks if all characters in t are present in the current window. Improvements:

Use a dict_keys dictionary instead of isAll to track character counts in the current window, updating directly instead of calling Counter repeatedly.

Remove unused variables ans, hash_1, dict_t.

Initialize right outside the loop, and use while right < len(s) and not isAll(dict_keys, dict_t) to terminate when all characters in t are found.

Track the minimum window length and the start/end indices, return s[start:end+1] at the end.

get() method syntax:

dict.get(key[, value])

Parameters:

key — the key to look up.
value — optional, the default to return if the key doesn't exist.

Return value: Returns the value for the key, or None (or the default) if not found.

Code

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        ans = 1000001
        hash_1 = {}

        def isAll(dict_keys, target):
            for i in target:
                if i not in dict_keys:
                    return False
                else:
                    if dict_keys[i] < target[i]:
                        return False
            return True

        dict_t = Counter(t)
        left = 0
        right = 0
        while right < len(s):
            dict1 = Counter(s[left:right + 1])
            if isAll(dict1, dict_t):
                ans = min(ans, right - left + 1)
                hash_1[right - left + 1] = s[left:right+1]
                print(ans, hash_1)
                left += 1
            else:
                right += 1
            print(ans, hash_1, dict1.keys())
        return hash_1[ans] if len(hash_1) != 0 else ""

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        # store character counts in t
        dict_t = Counter(t)
        # store characters in current sliding window
        dict_keys = {}
        left = 0
        right = 0
        min_len = float('inf')
        start = 0
        end = 0
        while right < len(s):
            # if the current character is in t, add it to the window dict
            if s[right] in dict_t:
                dict_keys[s[right]] = dict_keys.get(s[right], 0) + 1
            # if the current window contains all characters in t
            while right < len(s) and isAll(dict_keys, dict_t):
                if right - left + 1 < min_len:
                    min_len = right - left + 1
                    start = left
                    end = right
                # move left pointer right, remove character from window dict
                if s[left] in dict_keys:
                    dict_keys[s[left]] -= 1
                    if dict_keys[s[left]] == 0:
                        del dict_keys[s[left]]
                left += 1
            right += 1
        return s[start:end + 1] if min_len != float('inf') else ""

def isAll(dict_keys, target):
    for key in target:
        if key not in dict_keys or dict_keys[key] < target[key]:
            return False
    return True

贡献者

longsizhuo贡献 5 次 · 最近 2026/05/06

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76. Minimum Window Substring

Problem

Approach

Optimization

Code

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