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chris · chris · commit fb73ea05ee7f · 2019-01-15T14:11:28.000+08:00
diff --git a/fruit-into-baskets.py b/fruit-into-baskets.py
@@ -0,0 +1,46 @@
+#https://leetcode.com/problems/fruit-into-type_baskets/
+
+class Solution(object):
+    #Time Efficiency: O(N), where N is the fruit count in the trees.
+    #Space Efficiency: O(K), where K is the total types of fruit in the trees.
+    def totalFruit(self, tree):
+        counter = 0 #max fruit count
+        start = 0 #the point where from start to i has only two types of fruit
+        mark = {} #we keep track of each type of fruit's index last seen
+        type_basket = [] #types of fruit in the basket now
+
+        for i in range(len(tree)):
+            t = tree[i]
+            if t not in type_basket:
+                if len(type_basket)<2:
+                    """
+                    if this type of fruit not in type_basket
+                    and the type_basket is not full yet
+                    add the type to the type_basket
+                    """
+                    type_basket.append(t)
+
+                elif len(type_basket)==2:
+                    """
+                    if this type of fruit not in type_basket
+                    and the type_basket has two types already
+                    we get rid of the type with smaller last seen index
+                    by moving the start to its index+1
+                    so now between start and i has only two types of fruit
+                    """
+                    t1 = type_basket[0]
+                    t2 = type_basket[1]
+                    if mark[t1]<mark[t2]:
+                        start = mark[t1]+1
+                        type_basket[0] = t
+                    else:
+                        start = mark[t2]+1
+                        type_basket[1] = t
+                else:
+                    #basket should not have more than two types
+                    print('ERROR')
+                    return 0
+            
+            counter = max(counter, i-start+1)
+            mark[t] = i
+        return counter
diff --git a/longest-substring-without-repeating-characters.py b/longest-substring-without-repeating-characters.py
@@ -0,0 +1,37 @@
+class Solution(object):
+
+    """
+    everytime we iterate to the next char (on index i)
+    we move the start to the place, where all the char between the start and i are all unique
+    and keep undating counter til the end
+    
+    how do we move the start?
+    we write down the index of the char we last seen
+    if char is seen, we set the start to (the index we seen char last time)+1
+    so we don't repeat char, BUT
+
+    another rule of start is not moving leftward
+    bc this will cause more repeating char between start and i
+    so we have to keep start incremental
+    
+    we only iterate through the string once so
+    Time Efficiency is O(N)
+
+    we use a dict to keep track of the index of char so
+    Space Efficiency is O(N)
+    """
+
+    def lengthOfLongestSubstring(self, s):
+        if s is None or s=='': return 0
+        counter = 0 #max length of substring
+        start = 0 #index where all the char between the start and i are all unique
+        mark = {} #all the char index we last seen
+
+        for i in range(len(s)):
+            char_now = s[i]
+            if char_now in mark:
+                start = max(start, mark[char_now]+1)
+            counter = max(counter, i-start+1)
+            mark[char_now] = i
+                
+        return counter
