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Minimum Suffix Flips

Number: 1652

Difficulty: Medium

Paid? No

Companies: J.P. Morgan, IBM, Amazon

Problem Description

Given a binary string target and an initial string s of the same length filled with zeros, determine the minimum number of operations needed to transform s into target. In one operation, you can select an index i (0 ≤ i < n) and flip all bits from i to the end (n - 1).

Key Insights

  • The operation flips a suffix, affecting all bits from the chosen index to the end.
  • Observe that a flip is required only when the current bit in target does not match the expected state (starting with '0').
  • Count transitions: each time the target bit differs from the expected value, an operation is performed and the expected state toggles.
  • The solution greedily counts these transitions, which directly equals the minimum number of operations needed.

Space and Time Complexity

Time Complexity: O(n) where n is the length of target. Space Complexity: O(1).

Solution

The algorithm iterates through the target string with an expected bit initially set to '0' (to match the all-zero initial string s). For every bit in target that differs from the expected bit, an operation (flip) is counted, and the expected bit toggles. This approach works because flipping a suffix affects the state of all subsequent bits, so correcting each deviation in order is optimal. Key data structures include just simple variables for counting operations and tracking the expected state.

Code Solutions

# Python implementation of Minimum Suffix Flips

class Solution:
    def minFlips(self, target: str) -> int:
        flips = 0
        expected = '0'  # The expected value starts as '0'
        # Iterate through each character in the target string
        for ch in target:
            # If the current bit doesn't match the expected bit, flip the segment
            if ch != expected:
                flips += 1
                # Toggle the expected bit after a flip
                expected = '1' if expected == '0' else '0'
        return flips

# Example usage:
# solution = Solution()
# print(solution.minFlips("10111"))
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