1529. Minimum Suffix Flips

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Description

You are given a 0-indexed binary string target of length n. You have another binary string s of length n that is initially set to all zeros. You want to make s equal to target.

In one operation, you can pick an index i where 0 <= i < n and flip all bits in the inclusive range [i, n - 1]. Flip means changing '0' to '1' and '1' to '0'.

Return the minimum number of operations needed to make s equal to target.

 

Example 1:

Input: target = "10111"
Output: 3
Explanation: Initially, s = "00000".
Choose index i = 2: "00000" -> "00111"
Choose index i = 0: "00111" -> "11000"
Choose index i = 1: "11000" -> "10111"
We need at least 3 flip operations to form target.

Example 2:

Input: target = "101"
Output: 3
Explanation: Initially, s = "000".
Choose index i = 0: "000" -> "111"
Choose index i = 1: "111" -> "100"
Choose index i = 2: "100" -> "101"
We need at least 3 flip operations to form target.

Example 3:

Input: target = "00000"
Output: 0
Explanation: We do not need any operations since the initial s already equals target.

 

Constraints:

• n == target.length
• 1 <= n <= 105
• target[i] is either '0' or '1'.

Solutions

Solution 1: Greedy

We traverse the string $\textit{target}$ from left to right, using a variable $\textit{ans}$ to record the number of flips. When we reach index $i$, if the parity of the current flip count $\textit{ans}$ is different from $\textit{target}[i]$, we need to perform a flip operation at index $i$ and increment $\textit{ans}$ by $1$.

The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.

Python3

class Solution:
    def minFlips(self, target: str) -> int:
        ans = 0
        for v in target:
            if (ans & 1) ^ int(v):
                ans += 1
        return ans

Java

class Solution {
    public int minFlips(String target) {
        int ans = 0;
        for (int i = 0; i < target.length(); ++i) {
            int v = target.charAt(i) - '0';
            if (((ans & 1) ^ v) != 0) {
                ++ans;
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int minFlips(string target) {
        int ans = 0;
        for (char c : target) {
            int v = c - '0';
            if ((ans & 1) ^ v) {
                ++ans;
            }
        }
        return ans;
    }
};

Go

func minFlips(target string) int {
	ans := 0
	for _, c := range target {
		v := int(c - '0')
		if ((ans & 1) ^ v) != 0 {
			ans++
		}
	}
	return ans
}

TypeScript

function minFlips(target: string): number {
    let ans = 0;
    for (const c of target) {
        if (ans % 2 !== +c) {
            ++ans;
        }
    }
    return ans;
}

Rust

impl Solution {
    pub fn min_flips(target: String) -> i32 {
        let mut ans = 0;
        for c in target.chars() {
            let bit = (c as u8 - b'0') as i32;
            if ans % 2 != bit {
                ans += 1;
            }
        }
        ans
    }
}