Solution For Minimum Suffix Flips

Problem Statement:

Given a binary string s, we can flip every consecutive group of k characters. For example, if we have s = “00110”, and k = 3, we can flip the first and second group of characters to get “10010”, or the second and third group of characters to get “00101”.

Return the minimum number of flips required to convert s into a string with no consecutive groups of the same character.

Example 1:

Input: s = “010110”, k = 2
Output: 2
Explanation: We can flip the first and second, or third and fourth segments.

Example 2:

Input: s = “0010111”, k = 2
Output: 1

Approach:

The problem can be solved using a greedy approach. We need to start from the beginning of the string and check if the current segment of k characters needs to be flipped. If yes, we can flip it and count it as a flip. We need to continue this process till the end of the string. We can optimize this approach by keeping track of the number of flips done so far, and the current segment that needs to be flipped.

Algorithm:

1. Initialize count variable to 0.
2. Start from the beginning of the string, and for each segment of k characters:
   a. Check if the segment needs to be flipped.
   b. If yes, flip the segment and increment the count variable by 1.
   c. Move to the next segment.
3. Return the count variable.

Code:

Python solution:

class Solution:
def minFlips(self, s: str, k: int) -> int:
n = len(s)
count = 0
for i in range(n-k+1):
if s[i:i+k].count(‘0’) == k//2:
continue
count += 1
temp = s[i:i+k][::-1] for j in range(k):
s[i+j] = ‘1’ if temp[j] == ‘0’ else ‘0’
return count

Time Complexity: O(N*K)

Space Complexity: O(N)

Step by Step Implementation For Minimum Suffix Flips

class Solution {
    public int minFlips(String target) {
        int flips = 0;
        for (int i = 0; i < target.length(); i++) {
            if (target.charAt(i) == '1') {
                flips++;
            }
        }
        return flips;
    }
}

class Solution:
    def minFlips(self, target: str) -> int:
        # Initialize count and flip to 0
        count = 0
        flip = 0
        
        # Iterate through target
        for t in target:
            # If the current value is not equal to the flip value
            if t != str(flip):
                # Increment count
                count += 1
                # Flip the value
                flip = 1 - flip
                
        # Return count
        return count

This problem can be solved using a simple greedy algorithm. The idea is to keep track of the number of flips needed to make the suffixes of the given string equal. We can do this by iterating over the string from right to left. For each character, we compare it to the character at the same index from the end of the string. If they are different, we increment the number of flips needed. If they are the same, we do nothing.

At the end, we return the minimum number of flips needed.

function minSuffixFlips(s) {
    let flips = 0;
    
    for (let i = 0; i < s.length; i++) {
        if (s[i] !== s[s.length - 1 - i]) {
            flips++;
        }
    }
    
    return flips;
}

class Solution {
public:
    int minFlips(string target) {
        int count = 0;
        for (int i = 0; i < target.size(); i++) {
            if (target[i] == '1') {
                count++;
            }
        }
        return count;
    }
};

using System; 

public class Solution { 

// Function to find minimum flips required to 

// make all suffixes of given binary string 

// alternate 

static int minFlips(string str) 

{ 

int n = str.Length, flips = 0; 

// Iterate over all substrings starting 

// from 2nd character 

for (int i = 1; i < n; i++) 

{ 

// If i-th character is same as i-1 character 

// but i-th character is different from 

// (i+1)-th character, then we need a flip at 

// i 

if (str[i] == str[i - 1] && str[i] != str[(i + 1) % n]) 

flips++; 

} 

return flips; 

} 

// Driver code 

public static void Main() 

{ 

string str = "001"; 

Console.Write(minFlips(str)); 

} 

}