0926. Flip String to Monotone Increasing - kumaeki/LeetCode GitHub Wiki

0926. Flip String to Monotone Increasing


A binary string is monotone increasing if it consists of some number of 0's (possibly none), followed by some number of 1's (also possibly none).

You are given a binary string s. You can flip s[i] changing it from 0 to 1 or from 1 to 0.

Return the minimum number of flips to make s monotone increasing.

Example 1:

Input: s = "00110"

Output: 1

Explanation: We flip the last digit to get 00111.

Example 2:

Input: s = "010110"

Output: 2

Explanation: We flip to get 011111, or alternatively 000111.

Example 3:

Input: s = "00011000"

Output: 2

Explanation: We flip to get 00000000.

Constraints:

  • 1 <= s.length <= 10^5
  • s[i] is either '0' or '1'.

解法1

class Solution {
    public int minFlipsMonoIncr(String s) {
        int size = s.length();
        
        // 从左往右, 把所有碰到的0都变为1, 用的次数
        int[] left = new int[size + 2];
        for(int i = 1; i <= size; i++)
            left[i] = left[i - 1] + (s.charAt(i - 1) == '1' ? 1 : 0);
        
        
        // 从右往左, 把所有碰到的1都变为0, 用的次数
        int[] right = new int[size + 2];
        for(int i = size; i > 0; i--)
            right[i] = right[i + 1] + (s.charAt(i - 1) == '0' ? 1 : 0);
        
        // 对于每个位置,左边都变为0, 右边都变为1的总次数,找最小值
        int result = Integer.MAX_VALUE;
        for(int i = 0; i <= size; i++)
            result = Math.min(result, left[i] + right[i + 1]);
        
        return result;
    }
}
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