What is the time complexity of Minimum Changes To Make Alternating Binary String?

The optimal solution for Minimum Changes To Make Alternating Binary String runs in O(n) time and uses O(1) space. The time complexity is O(n) because we only make a single pass through the string, and the space complexity is O(1) since we use a fixed number of counters.

What is the brute force solution for Minimum Changes To Make Alternating Binary String?

The brute force approach for Minimum Changes To Make Alternating Binary String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each possible alternating string and counting how many changes are needed to convert the given string into one of these patterns. 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 Changes To Make Alternating Binary String?

Minimum Changes To Make Alternating Binary String uses the following concepts: String. The recommended patterns to study are: Two Pointers, Sliding Window.

What are the interview tips for Minimum Changes To Make Alternating Binary String?

Always think about edge cases, such as very short strings. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice writing clean, efficient code, as readability is important in interviews.

What are common mistakes when solving Minimum Changes To Make Alternating Binary String?

Not considering both possible starting characters for the alternating pattern. Overcomplicating the problem by trying to generate all possible strings.

#1758

Minimum Changes To Make Alternating Binary String

Easy

StringTwo PointersSliding Window

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)

Instead of generating two strings, we can simply count how many changes are needed while traversing the string once. This reduces the time complexity significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize two counters for changes needed for both patterns (starting with '0' and starting with '1').
2Step 2: Traverse the string character by character.
3Step 3: For each character, check if it matches the expected character for both patterns and increment the respective counters if it doesn't.
4Step 4: Return the minimum of the two counters.

solution.py13 lines

1# Full working Python code
2
3def min_operations(s):
4    count0 = 0
5    count1 = 0
6    for i in range(len(s)):
7        expected_char_for_0 = '0' if i % 2 == 0 else '1'
8        expected_char_for_1 = '1' if i % 2 == 0 else '0'
9        if s[i] != expected_char_for_0:
10            count0 += 1
11        if s[i] != expected_char_for_1:
12            count1 += 1
13    return min(count0, count1)

ℹ

Complexity note: The time complexity is O(n) because we only make a single pass through the string, and the space complexity is O(1) since we use a fixed number of counters.

1The string can only be made alternating by changing adjacent characters.
2There are only two valid alternating patterns: starting with '0' or starting with '1'.

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