76. Minimum Window Substring

[tech-cow]

'''
start[6:12]

Problem Solving:
    [?] Sliding window algorithm
        * Potentially scan through with 2 pointers
            - left being the window shrinker
            - right being the window extender
            
            extend right point until meeting the condition of S contains all T
                take GlobalMin
                then shrink the size of left
        
        2 pointer algorithm takes O(N), not sure yet for checking S contains all T part, I am a bit stuck here.
        Stuck:
            One way to verify Target in S, is the making a set of every window
                but that takes O(len(window)) time complexity for convertion
                    
                    Can we make a dictionary of S before hand
                    
                    "ADOBECODEBANC"
                    
                    {
                        char  :  count
                        A     :     1    
                    }
                    
                I can think of a way in which having Target becomes a hashset
                Having a fixed hashtable to store S's state initially
                Having a second hashtable to change count frequency
                    If visted, count of associated element decrement by 1
                
                
                Each time, for loop -> Target hashset:
                    then find comparision of inital hashtable & frequently changed hastable to get the diff
                        If diff more than 1 for every element in hashset
                            Meet the condition
                
                Time Complexity: 
                    O(N) + O(size of Target)
                    
        [6:26] Ok, having brainfart , solution time?
        [7:06] Done, need to re-do
'''

from collections import Counter
class Solution:
    def minWindow(self, nums , target):
        if nums is None: return ""
        targetHash = Counter(target)
        curHash = {}
        
        targetCount = len(targetHash)
        matched = 0
        
        globalMin = float("inf")
        size = len(nums)
        res = ""
        
        j = 0
        for i in range(size):
            # 不断增大窗口大小，知道满足所有Match需求
            while j < size and matched < targetCount:
                if nums[j] in targetHash:
                    curHash[nums[j]] = curHash.get(nums[j], 0) + 1
                    if curHash[nums[j]] == targetHash[nums[j]]:
                        matched += 1
                j += 1
                
            # 更新最小值
            if j - i < globalMin and matched == targetCount:
                globalMin = j - i
                res = nums[i:j]

            # 删除left的一个元素，使得窗口继续滑动
            if nums[i] in targetHash:
                if curHash[nums[i]] == targetHash[nums[i]]:
                    matched -= 1
                curHash[nums[i]] -= 1
        return res