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Leetcode 2578. Split With Minimum Sum

Problem Statement

1. Split With Minimum Sum

You are given a positive integer num. Split it into two non-negative integers num1 and num2 such that:

• The concatenation of num1 and num2 equals the original integer num.
• num1 and num2 should not contain any leading zeroes unless they are zero themselves.

Your task is to minimize the sum of the two numbers num1 and num2.

Write a function splitMinSum such that:

public int splitMinSum(int num)

• num is a positive integer.
• The function returns the minimum possible sum of num1 and num2 after splitting num as described.

Clarifying Questions

1. Input Constraints: What is the maximum value of num?
   • Assume typical integer constraints: 1 ≤ num ≤ 10^9.
2. Output Requirements: Are num1 and num2 required to be non-zero?
   • They can both be zero if required, otherwise non-negative integers with no leading zeroes.
3. Edge Cases:
   • What should the function return if num contains only one digit?
     • Since num can’t be split into two non-zero digits meaningfully, we can consider one of the splits as zero. For example, if num = 5, the splits can be num1 = 0 and num2 = 5 or vice-versa.

Strategy

To solve this problem, observe that for minimizing the sum of the two numbers num1 and num2, we need to balance their digit values as closely as possible. The following steps will guide our approach:

1. Convert num to a String: This helps in easily splitting the number at any position.
2. Generate all Possible Splits: For a number represented as a string s, generate all possible splits at each position i where 1 ≤ i ≤ length of s - 1.
3. Calculate Sum of the Splits: For each split, convert the substrings to integers and calculate their sums.
4. Find the Minimum Sum: Track and return the minimum sum encountered among all possible splits.

Code

public class Solution {
    public int splitMinSum(int num) {
        String numStr = String.valueOf(num);
        int minSum = Integer.MAX_VALUE;
        
        for (int i = 1; i < numStr.length(); i++) {
            int num1 = Integer.parseInt(numStr.substring(0, i));
            int num2 = Integer.parseInt(numStr.substring(i));
            
            int currentSum = num1 + num2;
            minSum = Math.min(minSum, currentSum);
        }
        
        return minSum;
    }

    public static void main(String[] args) {
        Solution solution = new Solution();
        System.out.println(solution.splitMinSum(4325));  // Test case
    }
}

Time Complexity

• String Conversion: O(N), where N is the number of digits in num.
• Substring and Integer Parsing: Each split operation involves substring and parsing operations which take O(N) for each split since the maximum length of a number (given the constraints) can be up to 10 digits.
• Total Complexity: O(N^2) where N is the number of digits in num.

This approach effectively balances the simplicity of implementation and efficient computation for typical constraints on num.

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