Write a function: class Solution { public int solution(int[] A); } that, given an array A consisting of N integers, returns the maximum sum of two numbers whose digits add up to an equal sum. If there are no two numbers whose digits have an equal sum, the function should return -1. Examples: 1. Given A = [51, 71, 17,42], the function should return 93. There are two pairs of numbers whose digits add up to an equal sum: (51, 42) and (17, 71). The first pair sums up to 93. 2. Given A = [42, 33, 60), the function should return 102. The digits of all numbers in A add up to the same sum, and choosing to add 42 and 60 gives the result 102. 3. Given A = [51, 32, 43], the function should return -1, since all numbers in A have digits that add up to different, unique sums. Write an efficient algorithm for the following assumptions: • N is an integer within the range [1..200,000); • each element of array A is an integer within the range [1..1,000,000,000).

The idea that, for each pair of devices v and w, there’s a strict dichotomy between being “in range” or “out of range” is a simplified abstraction. more accurately, there’s a power decay function f (·) that specifies, for a pair of devices at distance δ, the signal strength f(δ) that they’ll be able to achieve on their wireless connection. (we’ll assume that f (δ) decreases with increasing δ.) we might want to build this into our notion of back-up sets as follows: among the k devices in the back-up set of v, there should be at least one that can be reached with very high signal strength, at least one other that can be reached with moderately high signal strength, and so forth. more concretely, we have values p1 ≥ p2 ≥ . . ≥ pk, so that if the back-up set for v consists of devices at distances d1≤d2≤≤dk,thenweshouldhavef(dj)≥pj foreachj. give an algorithm that determines whether it is possible to choose a back-up set for each device subject to this more detailed condition, still requiring that no device should appear in the back-up set of more than b other devices. again, the algorithm should output the back-up sets themselves, provided they can be found.\