LC 926. Flip String to Monotone Increasing

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. 1 <= S.length <= 20000
2. S only consists of '0' and '1' characters.

Runtime: 19 ms, faster than 17.65% of Java online submissions for Flip String to Monotone Increasing.

package date20190116;

public class flipstringtomonoincreasing926 {

  public static int minFlipsMonoIncr(String S) {
    int[][] lefttoright = new int[S.length()][];
    lefttoright[][] = S.charAt() == '' ?  : ;
    for(int i=; i<S.length(); i++){
      lefttoright[i][] = (S.charAt(i) == '' ?  : ) + lefttoright[i-][];
    }
    lefttoright[S.length()-][] = S.charAt(S.length()-) == '' ?  : ;
    for(int i=S.length()-; i>=; i--){
      lefttoright[i][] = (S.charAt(i) == '' ?  : ) + lefttoright[i+][];
    }
//    for(int[] x : lefttoright){
//      System.out.print(x[0] + " ");
//    }
    int ret = S.length();
    for(int i=; i<S.length()-; i++){
      ret = Math.min(ret, lefttoright[i][]+lefttoright[i+][]);
    }
    ret = Math.min(ret, lefttoright[][]);
    ret = Math.min(ret, lefttoright[S.length()-][]);
    return ret;
  }
  public static void main(String[] args){
    String s = "";
    minFlipsMonoIncr(s);

  }
}

another solution

Runtime: 12 ms, faster than 68.98% of Java online submissions for Flip String to Monotone Increasing.

class Solution {
    public int minFlipsMonoIncr(String S) {
      int[] dp = new int[S.length()+];
      int N = S.length();
      for(int i=; i < N; i++){
        dp[i+] = dp[i] + (S.charAt(i) == '' ?  : );
      }
      int ans = Integer.MAX_VALUE;
      for(int i=; i<N+; i++){
        ans = Math.min(ans, dp[i] + N-i - (dp[N] - dp[i]));
      }
      return ans;
    }

}