How do you solve Minimum Suffix Flips on LeetCode?

Minimum Suffix Flips (LeetCode #1529) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Flipping from the left affects all bits to the right.

What is the brute force solution for Minimum Suffix Flips?

The brute force approach for Minimum Suffix Flips is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible index to flip and observing the changes in the string. This method is straightforward but inefficient, as it may require multiple passes over the string to achieve the desired result. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Suffix Flips?

Minimum Suffix Flips uses the following concepts: String, Greedy. The recommended patterns to study are: Greedy, Two Pointers.

What are the interview tips for Minimum Suffix Flips?

Start with a simple approach to demonstrate understanding. Explain your thought process clearly as you code. Consider edge cases, like all zeros or all ones.

What are common mistakes when solving Minimum Suffix Flips?

Not considering the initial state of the string. Overcomplicating the flipping logic.

#1529

Minimum Suffix Flips

Medium

StringGreedyGreedyTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal solution leverages the observation that we only need to count transitions between '0' and '1' in the target string. Each transition indicates a necessary flip operation, allowing us to minimize the number of operations.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to count the number of flips.
2Step 2: Iterate through the target string from the first character to the last.
3Step 3: Count a flip whenever the current character differs from the previous character.
4Step 4: If the first character is '1', increment the flip count by one.
5Step 5: Return the total flip count.

solution.py11 lines

1# Full working Python code
2
3def min_flips(target: str) -> int:
4    flips = 0
5    for i in range(len(target)):
6        if i == 0 and target[i] == '1':
7            flips += 1
8        elif i > 0 and target[i] != target[i - 1]:
9            flips += 1
10    return flips
11

ℹ

Complexity note: This is efficient because it only requires a single pass through the target string, counting transitions, leading to linear time complexity.

1Flipping from the left affects all bits to the right.
2Count transitions between '0' and '1' to minimize operations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.