flip string to monotone increasing medium

Author: weiy

DP/FlipStringToMonotoneIncreasing.py

@@ -0,0 +1,76 @@
+"""
+A string of '0's and '1's is monotone increasing if it consists of some number of '0's (possibly 0), followed by some number of '1's (also possibly 0.)
+
+We are given a string S of '0's and '1's, and we may flip any '0' to a '1' or a '1' to a '0'.
+
+Return the minimum number of flips to make S monotone increasing.
+
+ 
+
+Example 1:
+
+Input: "00110"
+Output: 1
+Explanation: We flip the last digit to get 00111.
+Example 2:
+
+Input: "010110"
+Output: 2
+Explanation: We flip to get 011111, or alternatively 000111.
+Example 3:
+
+Input: "00011000"
+Output: 2
+Explanation: We flip to get 00000000.
+ 
+
+Note:
+
+1 <= S.length <= 20000
+S only consists of '0' and '1' characters.
+
+
+给定一个由 '0' 和 '1' 组成的字符串，把它变成单调递增的字符串。返回最少的改变次数。
+
+思路：
+从左向右走过一遍，把找到的 1 变成 0。
+从右向左过一遍，把找到的 0 变成 1。
+
+最后过一遍，找到相加最少的一个点即可（可能不止一个）。
+
+测试地址：
+https://leetcode.com/contest/weekly-contest-107/problems/flip-string-to-monotone-increasing/
+
+"""
+
+class Solution(object):
+    def minFlipsMonoIncr(self, S):
+        """
+        :type S: str
+        :rtype: int
+        """
+        
+        x = [0] if S[0] == '0' else [1]
+        
+        # left to right
+        # 1 -> 0
+        for i in range(1, len(S)):
+            if S[i] == '1':
+                x.append(x[-1]+1)
+            else:
+                x.append(x[-1])
+        
+        # right to left
+        # 0 -> 1
+        S = S[::-1]
+        y = [0] if S[0] == '1' else [1]
+        for i in range(1, len(S)):
+            if S[i] == '0':
+                y.append(y[-1]+1)
+            else:
+                y.append(y[-1])
+        
+        y.reverse()
+        
+        return min([i+j for i,j in zip(x,y)]) - 1
+