What is the time complexity of Minimum Number of Operations to Make String Sorted?

The optimal solution for Minimum Number of Operations to Make String Sorted runs in O(n) time and uses O(n) space. The complexity is O(n) because we only traverse the string a couple of times and use a fixed-size array for counting characters.

What is the brute force solution for Minimum Number of Operations to Make String Sorted?

The brute force approach for Minimum Number of Operations to Make String Sorted is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach simulates the operations described in the problem. We repeatedly find the largest index where the string is not sorted and perform the necessary swaps and reversals until the string is sorted. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Operations to Make String Sorted?

Minimum Number of Operations to Make String Sorted uses the following concepts: Hash Table, Math, String, Combinatorics, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Operations to Make String Sorted?

Clarify the operations and constraints before coding. Think about the implications of each operation on the string's order. Practice similar problems to get comfortable with combinatorial counting.

What are common mistakes when solving Minimum Number of Operations to Make String Sorted?

Not considering edge cases like already sorted strings. Confusing the operations with simple swaps without reversals.

#1830

Minimum Number of Operations to Make String Sorted

Hard

Hash TableMathStringCombinatoricsCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(n)

The optimal solution leverages combinatorial counting to determine the number of operations needed to sort the string without simulating each step. This approach is faster and avoids unnecessary iterations.

⚙️

Algorithm

4 steps

1Step 1: Count the frequency of each character in the string.
2Step 2: For each character, calculate how many characters are smaller than it that come after it in the string.
3Step 3: Use the counts to determine the number of swaps needed to sort the string.
4Step 4: Return the total number of operations modulo 10^9 + 7.

solution.py13 lines

1def min_operations(s):
2    MOD = 10**9 + 7
3    count = [0] * 26
4    for char in s:
5        count[ord(char) - ord('a')] += 1
6    operations = 0
7    total = 0
8    for char in s:
9        index = ord(char) - ord('a')
10        total += sum(count[:index])
11        operations += total
12        count[index] -= 1
13    return operations % MOD

ℹ

Complexity note: The complexity is O(n) because we only traverse the string a couple of times and use a fixed-size array for counting characters.

1The problem can be viewed as counting inversions in the string.
2Understanding character frequency can help optimize the solution.

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