How do you solve Minimum Operations to Make a Special Number on LeetCode?

Minimum Operations to Make a Special Number (LeetCode #2844) 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: A number is special if it ends with '00', '25', '50', or '75'.

What is the brute force solution for Minimum Operations to Make a Special Number?

The brute force approach for Minimum Operations to Make a Special Number is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible subsequences of the string and checking if any of them form a special number (divisible by 25). This is simple but inefficient, as it explores many unnecessary combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Operations to Make a Special Number?

Minimum Operations to Make a Special Number uses the following concepts: Math, String, Greedy, Enumeration. The recommended patterns to study are: Greedy, Enumeration.

What are the interview tips for Minimum Operations to Make a Special Number?

Always clarify the problem constraints and requirements before diving into coding. Discuss your thought process and approach with the interviewer. Test your solution with edge cases to ensure robustness.

What are common mistakes when solving Minimum Operations to Make a Special Number?

Not considering all valid pairs that can form special numbers. Overlooking the need to count deletions correctly.

#2844

Minimum Operations to Make a Special Number

Medium

MathStringGreedyEnumerationGreedyEnumeration

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 focuses on finding the last two digits that can form a special number directly by scanning the string from the end. This avoids unnecessary checks and directly counts the deletions needed.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to count deletions and set it to 0.
2Step 2: Traverse the string from the end to the beginning, looking for pairs of digits that can form '00', '25', '50', or '75'.
3Step 3: For each valid pair found, calculate the number of deletions needed and update the minimum operations.

solution.py17 lines

1# Full working Python code
2
3def min_operations_optimal(num):
4    n = len(num)
5    min_ops = float('inf')
6    for i in range(n - 1, -1, -1):
7        if num[i] == '0':
8            for j in range(i - 1, -1, -1):
9                if num[j] == '0':
10                    min_ops = min(min_ops, (i - j - 1))
11                elif num[j] == '5':
12                    min_ops = min(min_ops, (i - j - 1) + 1)
13                elif num[j] == '2':
14                    min_ops = min(min_ops, (i - j - 1) + 2)
15                elif num[j] == '7':
16                    min_ops = min(min_ops, (i - j - 1) + 2)
17    return min_ops if min_ops != float('inf') else n

ℹ

Complexity note: The time complexity is O(n) because we only traverse the string once, checking pairs of digits in a single pass. This is efficient compared to the brute force method.

1A number is special if it ends with '00', '25', '50', or '75'.
2The optimal solution leverages the structure of the problem by focusing on the last two digits.

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