1672. Richest Customer Wealth leetcode solution in c++

 

You are given an m x n integer grid accounts where accounts[i][j] is the amount of money the i​​​​​​​​​​​th​​​​ customer has in the j​​​​​​​​​​​th​​​​ bank. Return the wealth that the richest customer has.

A customer's wealth is the amount of money they have in all their bank accounts. The richest customer is the customer that has the maximum wealth.

 

Example 1:

Input: accounts = [[1,2,3],[3,2,1]]
Output: 6
Explanation:
1st customer has wealth = 1 + 2 + 3 = 6
2nd customer has wealth = 3 + 2 + 1 = 6
Both customers are considered the richest with a wealth of 6 each, so return 6.

Example 2:

Input: accounts = [[1,5],[7,3],[3,5]]
Output: 10
Explanation: 
1st customer has wealth = 6
2nd customer has wealth = 10 
3rd customer has wealth = 8
The 2nd customer is the richest with a wealth of 10.

Example 3:

Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
Output: 17

 

Constraints:

• m == accounts.length
• n == accounts[i].length
• 1 <= m, n <= 50
• 1 <= accounts[i][j] <= 100

solution<1>

Runtime: 18 ms, faster than 5.58% of C++ online submissions for Richest Customer Wealth.

Memory Usage: 7.9 MB, less than 16.36% of C++ online submissions for Richest Customer Wealth.

class Solution {
public:
    int maximumWealth(vector<vector<int>>& accounts) {
        int max=0;
        for(int i=0;i<accounts.size();i++)
        {
            int sum=0;
            for(int j=0;j<accounts[i].size();j++)
            {
               sum+=accounts[i][j];
               if(sum>max)
                 {
                      max=sum;
                 }
            }
        }
        return max;
    }
};

solution<2>

Runtime: 4 ms, faster than 86.84% of C++ online submissions for Richest Customer Wealth.

Memory Usage: 7.9 MB, less than 53.67% of C++ online submissions for Richest Customer Wealth.

class Solution {
public:
    int maximumWealth(vector<vector<int>>& accounts) {
        int ans=0;
        int sum=0;
        for(int i=0;i<accounts.size();i++)
        {
            for(int j=0;j<accounts[i].size();j++)
            {
               sum+=accounts[i][j];
            }
            ans=max(ans,sum);
            sum=0;
        }
        return ans;
    }
};