Max Sum of a Pair With Equal Sum of Digits

Problem Description

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.

Key Insights

• The problem requires finding pairs of indices with equal digit sums.
• A hash table can be used to store the maximum number associated with each digit sum.
• The solution involves iterating through the array, calculating digit sums, and updating the maximum sums in the hash table.
• The final result is derived from the maximum sums of matching digit sums.

Space and Time Complexity

Time Complexity: O(n * d), where n is the number of elements in the array and d is the number of digits in the largest number. Space Complexity: O(d), where d is the number of unique digit sums stored in the hash table.

Solution

To solve this problem, we will use a hash table (dictionary) to keep track of the maximum number corresponding to each digit sum. The algorithm proceeds as follows:

1. Define a helper function to calculate the sum of digits of a number.
2. Iterate through the array and compute the digit sum for each number.
3. For each unique digit sum, maintain the maximum number encountered so far.
4. If a digit sum has been seen before, calculate the potential maximum sum using the current number and the maximum number associated with that digit sum.
5. Return the largest sum found or -1 if no valid pairs exist.

Code Solutions