Introduction

In this blog post, we will explore how to solve the LeetCode problem "2342" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed array nums consisting of positive integers. You can choose two indices i and j, such that i != j, and the sum of digits of the number nums[i] is equal to that of nums[j]. Return the maximum value of nums[i] + nums[j] that you can obtain over all possible indices i and j that satisfy the conditions. If no such pair of indices exists, return -1. Example 1: Input: nums = [18,43,36,13,7] Output: 54 Explanation: The pairs (i, j) that satisfy the conditions are: - (0, 2), both numbers have a sum of digits equal to 9, and their sum is 18 + 36 = 54. - (1, 4), both numbers have a sum of digits equal to 7, and their sum is 43 + 7 = 50. So the maximum sum that we can obtain is 54. Example 2: Input: nums = [10,12,19,14] Output: -1 Explanation: There are no two numbers that satisfy the conditions, so we return -1. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109

Explanation

• Calculate Digit Sums: Iterate through the nums array and calculate the sum of digits for each number.
  • Store Maximums by Digit Sum: Use a dictionary to store the maximum number encountered for each digit sum. If a digit sum is already present in the dictionary, update the value with the larger of the current value and the number being processed.
  • Find Maximum Pair Sum: Iterate through the dictionary of digit sums and their maximum numbers. If a digit sum has at least two numbers associated with it (meaning it was encountered more than once), calculate the sum of the two largest numbers with that digit sum and keep track of the maximum sum found so far.

• Runtime Complexity: O(n), Storage Complexity: O(n)

Code

    def maximum_sum(nums):
    """
    Finds the maximum sum of two numbers in the array such that the sum of their digits is equal.

    Args:
        nums: A list of positive integers.

    Returns:
        The maximum sum of two numbers satisfying the condition, or -1 if no such pair exists.
    """

    digit_sum_map = {}
    max_sum = -1

    for num in nums:
        digit_sum = 0
        temp = num
        while temp > 0:
            digit_sum += temp % 10
            temp //= 10

        if digit_sum in digit_sum_map:
            max_sum = max(max_sum, digit_sum_map[digit_sum] + num)
            digit_sum_map[digit_sum] = max(digit_sum_map[digit_sum], num)
        else:
            digit_sum_map[digit_sum] = num

    return max_sum