#1529

Medium

auto_awesomeGreedy choice plus invariant validation

LeetCode Problem Workspace

Minimum Suffix Flips

Find the minimum number of operations to convert a binary string to a target string using bit flips.

Problem Statement

You are given a binary string target of length n and another binary string s of length n that starts as all zeros. Your task is to transform s into target through the minimum number of operations. In each operation, you can choose an index i where 0 <= i < n and flip all bits from index i to the end of the string. A flip means changing '0' to '1' and '1' to '0'.

Return the minimum number of operations required to transform s into target. The problem can be efficiently solved using a greedy strategy with invariant validation, where flips are performed only when mismatches occur in certain positions of the string.

Examples

Example 1

Input: target = "10111"

Output: 3

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

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

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 Approach

Greedy Strategy

The key to solving this problem lies in identifying when to flip the bits. We can iterate over the target string from left to right. If a mismatch occurs at position i, we perform a flip starting from i. This guarantees that all the bits from position i to the end of the string are flipped, and the operation is recorded. This method minimizes the number of flips required.

Tracking Flips

We need to track the state of the string after each flip. Instead of flipping the entire suffix physically, we can simply track whether the number of flips is odd or even. If the number of flips up to a given point is odd, it means the bit has been flipped, and if even, the bit remains unchanged.

Invariant Validation

The greedy approach ensures that each flip operation is chosen at an optimal point, where the mismatch occurs. After each flip, we maintain the invariant that the mismatch is resolved or addressed, reducing unnecessary operations. The solution ensures no backtracking is needed once a flip is performed.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of this approach is O(n), where n is the length of the target string. We iterate through the string once, checking for mismatches and performing flips only when needed. The space complexity is O(1), as we only need a few variables to track the number of flips and the current state of the string, without requiring extra space for additional data structures.

What Interviewers Usually Probe

Candidate demonstrates an understanding of greedy algorithms by applying it to string manipulation.
Candidate proposes a solution with O(n) time complexity, which is optimal for this problem.
Candidate identifies and implements the invariant validation effectively, minimizing the number of operations.

Common Pitfalls or Variants

Common pitfalls

Failing to track the flip count correctly, leading to unnecessary operations or missed flips.
Overcomplicating the solution by trying to optimize beyond the greedy approach, which is already optimal.
Misunderstanding the effect of each flip and failing to track its impact correctly on the string.

Follow-up variants

Consider extending the problem to include different types of strings, such as strings with characters other than '0' and '1'.
Explore the impact of reverse operations, where flips are performed from right to left instead of left to right.
Modify the problem to consider minimizing the number of flips, but with constraints on the number of flips allowed per operation.

FAQ

How do you approach solving Minimum Suffix Flips?

The problem can be solved using a greedy strategy, flipping bits only when mismatches occur. This minimizes the number of operations required.

What is the time complexity of solving Minimum Suffix Flips?

The time complexity is O(n), where n is the length of the target string. This is because we only need to iterate through the string once.

What does the greedy choice in Minimum Suffix Flips involve?

The greedy choice involves flipping at positions where mismatches occur, ensuring each flip operation reduces the number of mismatches optimally.

Can you optimize the solution further for Minimum Suffix Flips?

The greedy approach is already optimal for this problem, and no further optimization is needed as it achieves the best time complexity.

What are the common mistakes in solving Minimum Suffix Flips?

Common mistakes include incorrectly tracking the flip count, overcomplicating the solution, and failing to account for the impact of each flip.

terminal

Solution

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 $1$.

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

Continue Practicing

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