Leetcode: Flip String to Monotone Increasing

Flip String to Monotone Increasing



Similar Problems:


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.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


  • Solution:
// Blog link: https://code.dennyzhang.com/flip-string-to-monotone-increasing
// Basic Ideas: dynamic programming
// dp[i][0]: how many changes to the make the sequence ends with 0
// dp[i][1]: how many changes to the make the sequence ends with 1
// Complexity: Time O(n), Space O(n)
func minFlipsMonoIncr(S string) int {
    len_s := len(S)
    dp := make([][]int, len_s)
    for i := range dp { dp[i] = []int{0, 0} }
    for i, ch := range S {
        if i == 0 {
            if ch == '1' { dp[i][0] = 1 }
            if ch == '0' { dp[i][1] = 1 }
            continue
        }
        min := dp[i-1][0]
        if dp[i-1][1] < min { min = dp[i-1][1] }
        if ch == '0' {
            dp[i][0] = dp[i-1][0]
            dp[i][1] = min + 1
        } else {
            dp[i][0] = dp[i-1][0] + 1
            dp[i][1] = min
        }
    }    
    res := dp[len_s-1][1]
    if dp[len_s-1][0] < dp[len_s-1][1] {
        res = dp[len_s-1][0]
    }
    return res
}
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