Solved! Leetcode 1529. Minimum Suffix Flips

Description: Minimum Suffix Flips

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

<strong>Input:</strong> target = "10111"
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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.

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

<strong>Input:</strong> target = "101"
<strong>Output:</strong> 3
<strong>Explanation:</strong> 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.

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

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

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'.

Solution

class Solution {

    public int minFlips(String target) {
        char prev = '0';
        int count = 0;
        for(int i=0;i<target.length();i++){
            char c = target.charAt(i);
            //check if the previous character is same as the current character
            //if not, increment operations count and update prev character
            if(c != prev){
                count++;
                prev = c;
            }
        }

        return count;
    }
}

class Solution {

public int minFlips(String target) {

char prev = '0';

int count = 0;

for(int i=0;i<target.length();i++){

char c = target.charAt(i);

//check if the previous character is same as the current character

//if not, increment operations count and update prev character

if(c != prev){

count++;

prev = c;

}

return count;

}

Time Complexity

O(n), where n is the number of characters in a string

Space Complexity

O(1)

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