Welcome to Subscribe On Youtube

Formatted question description: https://leetcode.ca/all/2342.html

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

  • Difficulty: Medium.
  • Related Topics: Array, Hash Table, Sorting, Heap (Priority Queue).
  • Similar Questions: .

Problem

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.**

  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

Solution (Java, C++, Python)

  • class Solution {
    public int maximumSum(int[] nums) {
        Map<Integer, Integer> map = new HashMap<>();
        int res = -1;
        for (int num : nums) {
            int s = 0;
            for (char digit : String.valueOf(num).toCharArray()) {
                s += Integer.valueOf(digit - '0');
            }
            if (!map.containsKey(s)) {
                map.put(s, num);
            } else {
                res = Math.max(res, map.get(s) + num);
                map.put(s, Math.max(map.get(s), num));
            }
        }
        return res;
    }
}

############

class Solution {
    public int maximumSum(int[] nums) {
        int ans = -1;
        int[] d = new int[100];
        for (int v : nums) {
            int y = 0;
            for (int x = v; x > 0; x /= 10) {
                y += x % 10;
            }
            if (d[y] > 0) {
                ans = Math.max(ans, d[y] + v);
            }
            d[y] = Math.max(d[y], v);
        }
        return ans;
    }
}

  • class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        ans = -1
        d = defaultdict(int)
        for v in nums:
            x, y = v, 0
            while x:
                y += x % 10
                x //= 10
            if y in d:
                ans = max(ans, d[y] + v)
            d[y] = max(d[y], v)
        return ans

############

# 2342. Max Sum of a Pair With Equal Sum of Digits
# https://leetcode.com/problems/max-sum-of-a-pair-with-equal-sum-of-digits/

class Solution:
    def maximumSum(self, nums: List[int]) -> int:
        n = len(nums)
        cnt = defaultdict(list)
        
        for x in nums:
            cnt[sum(int(k) for k in str(x))].append(x)
        
        res = -1
        
        for values in cnt.values():
            if len(values) >= 2:
                values.sort()
                res = max(res, values[-1] + values[-2])
            
        
        return res


  • class Solution {
public:
    int maximumSum(vector<int>& nums) {
        vector<vector<int>> d(100);
        for (int& v : nums) {
            int y = 0;
            for (int x = v; x > 0; x /= 10) {
                y += x % 10;
            }
            d[y].emplace_back(v);
        }
        int ans = -1;
        for (auto& vs : d) {
            if (vs.size() > 1) {
                sort(vs.rbegin(), vs.rend());
                ans = max(ans, vs[0] + vs[1]);
            }
        }
        return ans;
    }
};

  • func maximumSum(nums []int) int {
	ans := -1
	d := [100]int{}
	for _, v := range nums {
		y := 0
		for x := v; x > 0; x /= 10 {
			y += x % 10
		}
		if d[y] > 0 {
			ans = max(ans, d[y]+v)
		}
		d[y] = max(d[y], v)
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

  • function maximumSum(nums: number[]): number {
    const d: number[] = Array(100).fill(0);
    let ans = -1;
    for (const v of nums) {
        let x = 0;
        for (let y = v; y; y = (y / 10) | 0) {
            x += y % 10;
        }
        if (d[x]) {
            ans = Math.max(ans, d[x] + v);
        }
        d[x] = Math.max(d[x], v);
    }
    return ans;
}


  • impl Solution {
    pub fn maximum_sum(nums: Vec<i32>) -> i32 {
        let mut d = vec![0; 100];
        let mut ans = -1;

        for &v in nums.iter() {
            let mut x: usize = 0;
            let mut y = v;
            while y > 0 {
                x += (y % 10) as usize;
                y /= 10;
            }
            if d[x] > 0 {
                ans = ans.max(d[x] + v);
            }
            d[x] = d[x].max(v);
        }

        ans
    }
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

All Problems

All Solutions