Is Minimum Suffix Flips easy or hard?

Minimum Suffix Flips is classified as a Medium problem on LeetCode. The implementation is short, but the challenge lies in recognizing the greedy insight that flips correspond to transitions in the target string.

How to solve Minimum Suffix Flips in O(n)?

Iterate through the target string while maintaining the current effective bit after previous flips. If the effective value differs from the target character, increment the flip counter and toggle the state. This greedy strategy ensures each index is processed once, giving O(n) time complexity.

What is the best approach for Minimum Suffix Flips?

The optimal approach is a greedy transition-count method. Traverse the target string and track the current effective bit after previous flips. Whenever the desired bit differs from the current state, perform a flip and toggle the state. This solves the problem in O(n) time and O(1) space.

What data structure is used in Minimum Suffix Flips?

The problem mainly uses string traversal with a greedy strategy. No complex data structures are required. A simple counter and a variable tracking the current flip state are enough to implement the optimal solution.

What is the time complexity of Minimum Suffix Flips?

The optimal solution runs in O(n) time because it scans the target string once. Each character is processed with constant-time checks. The space complexity is O(1) since only a few variables are used to track the current flip state.

Minimum Suffix Flips Python or Java solution approach?

Both Python and Java implementations follow the same logic: iterate through the string, compare the effective bit with the target bit, and increment the flip counter when they differ. The algorithm runs in O(n) time and uses constant extra space.

Is Minimum Suffix Flips asked at Google, Amazon, or Meta interviews?

Greedy string problems like Minimum Suffix Flips appear in interviews at large tech companies including Amazon and Google. The problem tests your ability to recognize patterns in state transitions and avoid unnecessary simulation of operations.

Description Hints 1 Solution Video 8:23

1529. Minimum Suffix Flips

Medium73.9% acceptance5 companies

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:

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:

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:

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'.

Asked at 5 companies

Visa JPMorgan Chase Morgan Stanley Amazon Microsoft

Topics

Greedy String

Hints open one at a time

Hint 1

Consider a strategy where the choice of bulb with number i is increasing. In such a strategy, you no longer need to worry about bulbs that have been set to the left.

8:23

Still stuck?Read 3 approaches, or watch a walkthrough.

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