Leetcode 2342 Max Sum of a Pair With Equal Sum of Digits Solution in Java | Hindi Coding Community

 

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

Java Code :

class Solution {

public int maximumSum(int[] nums) {

ArrayList<Integer> table[] = new ArrayList[82];

for(int n: nums){

int sum = 0, N = n;

for(; n != 0; n/=10) sum += n%10;

if(table[sum] == null) table[sum] = new ArrayList<Integer>();

table[sum].add(N);

}

int ans = -1;

for(ArrayList<Integer> v: table)

if(v != null && v.size() > 1){

int a = 0, b = 0;

for(Integer x : v)

if(x >= a){b = a; a = x;}

else if(x > b) b = x;

ans = Math.max(ans, a + b);

}

return ans;

}

}