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LeetCode 2844: Minimum Operations to Make a Special Number Solution

Master LeetCode problem 2844 (Minimum Operations to Make a Special Number), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2844. Minimum Operations to Make a Special Number

Medium
MathStringGreedyEnumeration

Problem Explanation

Explanation

To solve this problem, we can iterate through the given number and consider all possible positions to split the number into two parts. We then calculate the number of operations required to make each part divisible by 25. The minimum number of operations needed to make the entire number divisible by 25 is the sum of operations required for the two parts.

Algorithm:

  1. Iterate through the given number and consider all possible split positions.
  2. Calculate the number of operations required to make each part divisible by 25.
  3. Keep track of the minimum number of operations found so far.

Time Complexity: O(n^2) where n is the length of the input number. Space Complexity: O(1)

Solution Code

class Solution {
    public int minOperations(String num) {
        int n = num.length();
        int minOps = Integer.MAX_VALUE;
        
        for (int i = 0; i < n; i++) {
            for (int j = i + 1; j < n; j++) {
                int ops = 0;
                String part1 = num.substring(0, i) + num.substring(j);
                String part2 = num.substring(i, j);
                
                ops += countOps(part1);
                ops += countOps(part2);
                
                minOps = Math.min(minOps, ops);
            }
        }
        
        return minOps;
    }
    
    private int countOps(String part) {
        int n = part.length();
        int ops1 = helper(part, "00");
        int ops2 = helper(part, "25");
        int ops3 = helper(part, "50");
        int ops4 = helper(part, "75");
        
        return Math.min(ops1, Math.min(ops2, Math.min(ops3, ops4)));
    }
    
    private int helper(String part, String target) {
        int i = part.length() - 1;
        int j = target.length() - 1;
        int ops = 0;
        
        while (i >= 0 && j >= 0) {
            if (part.charAt(i) == target.charAt(j)) {
                i--;
                j--;
            } else {
                ops++;
                i--;
            }
        }
        
        ops += j + 1;
        return ops;
    }
}

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