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

The optimal solution for Minimum Number of Swaps to Make the Binary String Alternating runs in Unknown time and uses Unknown space.

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

Minimum Number of Swaps to Make the Binary String Alternating uses the following concepts: String, Greedy. The recommended patterns to study are: .

Minimum Number of Swaps to Make the Binary String Alternating — LeetCode #1864 (Medium)

Tags: String, Greedy

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach involves checking every possible pair of characters to see if swapping them can lead to an alternating string. This is simple but inefficient as it requires many comparisons.

The time complexity is O(n²) because we are checking every pair of characters, leading to n(n-1)/2 comparisons. The space complexity is O(1) as we are using a constant amount of extra space.

Step 1: Generate all possible strings by swapping any two characters in the input string.
Step 2: Check each generated string to see if it is alternating.
Step 3: Count the number of swaps needed to reach an alternating string and return the minimum.

For s = '111000': 1. Initial string: '111000' 2. Check swap (0,1): '111000' (not alternating) 3. Check swap (0,2): '111000' (not alternating) 4. Check swap (1,2): '111000' (not alternating) 5. Check swap (1,3): '110100' (not alternating) 6. Check swap (1,4): '101100' (not alternating) 7. Check swap (1,5): '100110' (not alternating) 8. Check swap (2,3): '110010' (not alternating) 9. Check swap (2,4): '101010' (alternating, 1 swap needed) 10. Result: 1 swap needed.

Optimal Solution approach

Time complexity: Unknown. Space complexity: Unknown.

The optimal solution involves counting the number of '0's and '1's in the string. Based on their counts, we can determine if it's possible to form an alternating string and calculate the minimum swaps needed.

Step 1: Count the number of '0's and '1's in the string.
Step 2: Check if the absolute difference between the counts is greater than 1. If so, return -1 (impossible to alternate).
Step 3: Calculate the expected positions for '0's and '1's based on the counts.
Step 4: Count mismatches for both expected patterns and return the minimum swaps needed.

#1864

Minimum Number of Swaps to Make the Binary String Alternating

Medium

String↗Greedy↗

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time UnknownSpace Unknown

⚙️

Algorithm

4 steps

1Step 1: Count the number of '0's and '1's in the string.
2Step 2: Check if the absolute difference between the counts is greater than 1. If so, return -1 (impossible to alternate).
3Step 3: Calculate the expected positions for '0's and '1's based on the counts.
4Step 4: Count mismatches for both expected patterns and return the minimum swaps needed.

solution.py14 lines

1def min_swaps(s):
2    count0 = s.count('0')
3    count1 = s.count('1')
4    if abs(count0 - count1) > 1:
5        return -1
6    swaps1 = swaps2 = 0
7    for i in range(len(s)):
8        expected = '01' if count0 >= count1 else '10'
9        if s[i] != expected[i % 2]:
10            swaps1 += 1
11        expected = '10' if count0 >= count1 else '01'
12        if s[i] != expected[i % 2]:
13            swaps2 += 1
14    return min(swaps1 // 2, swaps2 // 2)

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