1063MediumArrayHash TableSortingHeap (Priority Queue)

The Heaviest Pair With Matching Digit Totals

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Trains the technique from

LeetCode 2342Max Sum of a Pair With Equal Sum of Digits

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A list of positive whole numbers reads values. The digit total of a number is what its decimal digits add up to.

Return the largest amount any two values at different positions add up to, given that their digit totals match, or -1 when no two positions have matching digit totals.

Examples

Example 1

Input: values = [51, 71, 17, 42]
Output: 93

The digit totals are six, eight, eight and six. The pair on six adds to ninety-three and the pair on eight to eighty-eight, so ninety-three wins.

Example 2

Input: values = [1, 2, 3]
Output: -1

The three digit totals are all different, so no pair qualifies.

Example 3

Input: values = [9, 90, 900]
Output: 990

All three values have a digit total of nine, so the two largest pair up.

Constraints

1 <= values.length <= 10^5
1 <= values[i] <= 10^9

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def maximum_sum(values: list[int]) -> int:

Java

public int maximumSum(int[] values)

The intended approach

Time O(n * d) for n values of at most d digits.Space O(1), since the groups number at most eighty-two.

Values can only pair with others sharing their digit total, so group them by that total. Digit totals of numbers under a billion never exceed eighty-one, so there are few groups however long the list is. Within a group the best pair is plainly the two largest values, so nothing needs storing but the largest seen so far per group. Walk the list once: work out the value's digit total, and if that group already holds a value, their sum is a candidate; then keep whichever of the two is larger as the group's standing value. That is right because a later value pairs best with the largest earlier one, and every pair is considered when its later member is reached. Ending with no candidate at all is the case answering `-1`.

Pythonentry point: maximum_sum

def maximum_sum(values):
    best_in = {}
    answer = -1
    for value in values:
        total = 0
        left = value
        while left:
            total += left % 10
            left //= 10
        if total in best_in:
            here = best_in[total] + value
            if here > answer:
                answer = here
            if value > best_in[total]:
                best_in[total] = value
        else:
            best_in[total] = value
    return answer

The obvious version

Take every pair of positions, work out both digit totals from scratch, and where they match measure the two values added together against the best so far. Costs the list's length squared rather than one pass with a standing value per group.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_maximum_sum(values):
    def digits(value):
        total = 0
        left = value
        while left:
            total += left % 10
            left //= 10
        return total
    n = len(values)
    answer = -1
    for i in range(n):
        for j in range(i + 1, n):
            if digits(values[i]) != digits(values[j]):
                continue
            here = values[i] + values[j]
            if here > answer:
                answer = here
    return answer

How this problem was checked

Reference solution executed against 15 test cases — 15 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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