2815. Max Pair Sum in an Array 

Description

You are given an integer array nums. You have to find the maximum sum of a pair of numbers from nums such that the largest digit in both numbers is equal.

For example, 2373 is made up of three distinct digits: 2, 3, and 7, where 7 is the largest among them.

Return the maximum sum or -1 if no such pair exists.

Example 1:

Input: nums = [112,131,411]

Output: -1

Explanation:

Each numbers largest digit in order is [2,3,4].

Example 2:

Input: nums = [2536,1613,3366,162]

Output: 5902

Explanation:

All the numbers have 6 as their largest digit, so the answer is 2536 + 3366 = 5902.

Example 3:

Input: nums = [51,71,17,24,42]

Output: 88

Explanation:

Each number's largest digit in order is [5,7,7,4,4].

So we have only two possible pairs, 71 + 17 = 88 and 24 + 42 = 66.

Constraints:

2 <= nums.length <= 100
1 <= nums[i] <= 10⁴

Solutions

Solution 1: Enumeration

First, we initialize the answer variable $ans=-1$. Next, we directly enumerate all pairs $(nums[i], nums[j])$ where $i \lt j$, and calculate their sum $v=nums[i] + nums[j]$. If $v$ is greater than $ans$ and the largest digit of $nums[i]$ and $nums[j]$ are the same, then we update $ans$ with $v$.

The time complexity is $O(n^2 \times \log M)$, where $n$ is the length of the array and $M$ is the maximum value in the array.

Python3

class Solution:
    def maxSum(self, nums: List[int]) -> int:
        ans = -1
        for i, x in enumerate(nums):
            for y in nums[i + 1 :]:
                v = x + y
                if ans < v and max(str(x)) == max(str(y)):
                    ans = v
        return ans

Java

class Solution {
    public int maxSum(int[] nums) {
        int ans = -1;
        int n = nums.length;
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                int v = nums[i] + nums[j];
                if (ans < v && f(nums[i]) == f(nums[j])) {
                    ans = v;
                }
            }
        }
        return ans;
    }

    private int f(int x) {
        int y = 0;
        for (; x > 0; x /= 10) {
            y = Math.max(y, x % 10);
        }
        return y;
    }
}

C++

class Solution {
public:
    int maxSum(vector<int>& nums) {
        int ans = -1;
        int n = nums.size();
        auto f = [](int x) {
            int y = 0;
            for (; x; x /= 10) {
                y = max(y, x % 10);
            }
            return y;
        };
        for (int i = 0; i < n; ++i) {
            for (int j = i + 1; j < n; ++j) {
                int v = nums[i] + nums[j];
                if (ans < v && f(nums[i]) == f(nums[j])) {
                    ans = v;
                }
            }
        }
        return ans;
    }
};

Go

func maxSum(nums []int) int {
	ans := -1
	f := func(x int) int {
		y := 0
		for ; x > 0; x /= 10 {
			y = max(y, x%10)
		}
		return y
	}
	for i, x := range nums {
		for _, y := range nums[i+1:] {
			if v := x + y; ans < v && f(x) == f(y) {
				ans = v
			}
		}
	}
	return ans
}

TypeScript

function maxSum(nums: number[]): number {
    const n = nums.length;
    let ans = -1;
    const f = (x: number): number => {
        let y = 0;
        for (; x > 0; x = Math.floor(x / 10)) {
            y = Math.max(y, x % 10);
        }
        return y;
    };
    for (let i = 0; i < n; ++i) {
        for (let j = i + 1; j < n; ++j) {
            const v = nums[i] + nums[j];
            if (ans < v && f(nums[i]) === f(nums[j])) {
                ans = v;
            }
        }
    }
    return ans;
}