medium Minimum Suffix Flips

Problem Statement

You are given a target binary string of length n. Initially, you also have another string str of length n containing all 0s. You need to make str equal to the target by performing some number of operations.

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

Return the minimum number of operations required to make string str equal to target.

Examples

• Example 1:

  • Input: target = "001"
  • Expected Output: 1
  • Justification: Flipping the last character in the str string from 0 to 1 will make the string "001", achieving the goal in one operation.

• Example 2:

  • Input: target = "11001"
  • Expected Output: 3
  • Justification: Starting with s as "00000", the process to make s equal to "11001" is as follows:
    1. Flip the last three digits to get "00111".
    2. Flip all five digits to get "11000".
    3. Flip the last digit to get "11001". This process requires three operations, thus the expected output is 3.

• Example 3:

  • Input: target = "100010"
  • Expected Output: 4
  • Justification: Beginning with s as "000000", to transform s into "100010", the sequence of operations could be:
    1. Flip the last two digits to get "000011".
    2. Flip the last digit to get "000010".
    3. Flip all six digits to get "111101".
    4. Flip the last five digits to get "100010". This sequence involves four operations, making the expected output 4.

Solution

To solve this problem, we'll focus on counting the transitions between 0s and 1s in the given string. The core idea is that to minimize the number of flips, we should only consider flipping when there's a change in the character from 0 to 1 or from 1 to 0. This approach works because flipping at each transition point ensures that we are making the most out of each flip, gradually converting the string into a uniform one with either all 0s or all 1s. By only focusing on these transition points, we effectively reduce the problem to counting how many times the character changes in the string, which directly correlates to the minimum number of flips required. This strategy is effective because it directly targets the sources of inconsistency in the string, ensuring a minimal number of operations to achieve uniformity.

Step-by-step Algorithm

1. Initialize a variable flips to 0. This will keep track of the number of flips required to make the string uniform.
2. Initialize a variable prev to '0' to represent the initial expected character in the string, considering the string initially contains all '0's.
3. Iterate through each character in the target string using a for loop. In each iteration, compare the current character with prev:
   • If the current character is different from prev, this indicates a transition point where a flip is needed.
   • Increment the flips counter by 1 to account for this flip.
   • Update prev to the current character, as this character is now the last flipped state in our operation.
4. After completing the iteration through the string, return the flips counter as the total number of flips required to make the string uniform.

Algorithm Walkthrough

Let's consider the Input "100010"

1. Initialization

   • flips = 0
   • prev = '0'

2. First Character: '1'

   • Current character ('1') is different from prev ('0').
   • Increment flips to 1.
   • Update prev to '1'.

3. Second Character: '0'

   • Current character ('0') is different from prev ('1').
   • Increment flips to 2.
   • Update prev to '0'.

4. Third Character: '0'

   • Current character ('0') is the same as prev ('0'), no flip needed.
   • flips remains 2.

5. Fourth Character: '0'

   • Current character ('0') is the same as prev ('0'), no flip needed.
   • flips remains 2.

6. Fifth Character: '1'

   • Current character ('1') is different from prev ('0').
   • Increment flips to 3.
   • Update prev to '1'.

7. Sixth Character: '0'

   • Current character ('0') is different from prev ('1').
   • Increment flips to 4.
   • Update prev to '0'.

8. Final Result

   • After iterating through the entire string, the total flips required to make the string "100010" uniform is 4.

Code

public class Solution {

  public int minFlips(String target) {
    int flips = 0; // Initialize flips counter
    char prev = '0'; // Initialize previous character as '0'
    // Iterate through the string
    for (int i = 0; i < target.length(); i++) {
      if (target.charAt(i) != prev) {
        flips++; // Increment flips if current char is different from prev
        prev = target.charAt(i); // Update prev to current char
      }
    }
    return flips; // Return the total flips required
  }

  public static void main(String[] args) {
    Solution solution = new Solution();
    System.out.println(solution.minFlips("001")); // Expected output: 1
    System.out.println(solution.minFlips("11001")); // Expected output: 3
    System.out.println(solution.minFlips("100010")); // Expected output: 4
  }
}

Complexity Analysis

• Time Complexity: , where n is the length of the input string. The algorithm iterates through the string once to count the transitions.
• Space Complexity: , as the space used does not depend on the input size and only a fixed number of variables are used.