What is the time complexity of Minimum Number of Flips to Make the Binary String Alternating?

The optimal solution for Minimum Number of Flips to Make the Binary String Alternating runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only iterate through the string once to count the mismatches. The space complexity is O(1) since we use a fixed number of variables.

What is the brute force solution for Minimum Number of Flips to Make the Binary String Alternating?

The brute force approach for Minimum Number of Flips to Make the Binary String Alternating is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible cyclic shifts of the string and counting the number of flips needed for each shift to make it alternating. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Flips to Make the Binary String Alternating?

Minimum Number of Flips to Make the Binary String Alternating uses the following concepts: String, Dynamic Programming, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Minimum Number of Flips to Make the Binary String Alternating?

Always analyze the problem for patterns before diving into coding. Discuss your thought process with the interviewer; they may provide hints. Practice problems that involve string manipulations and counting mismatches.

What are common mistakes when solving Minimum Number of Flips to Make the Binary String Alternating?

Not considering the two possible alternating patterns. Overcomplicating the problem by trying to generate all shifts.

#1888

Minimum Number of Flips to Make the Binary String Alternating

Medium

StringDynamic ProgrammingSliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal solution leverages the fact that we only need to count mismatches at even and odd indices for two possible alternating patterns. This avoids the need for generating all shifts, significantly improving efficiency.

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Algorithm

3 steps

1Step 1: Count the number of '0's and '1's at even and odd indices for the original string.
2Step 2: Calculate the flips needed for both alternating patterns: starting with '0' and starting with '1'.
3Step 3: Return the minimum of the two calculated values.

solution.py10 lines

1def minFlips(s):
2    even0, even1, odd0, odd1 = 0, 0, 0, 0
3    for i, char in enumerate(s):
4        if i % 2 == 0:
5            if char == '0': even0 += 1
6            else: even1 += 1
7        else:
8            if char == '0': odd0 += 1
9            else: odd1 += 1
10    return min(even1 + odd0, even0 + odd1)

ℹ

Complexity note: The time complexity is O(n) because we only iterate through the string once to count the mismatches. The space complexity is O(1) since we use a fixed number of variables.

1Understanding the structure of the string is crucial; we only need to focus on even and odd indices.
2Cyclic shifts can be avoided by counting mismatches directly.

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