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

Approach 1: Max Row Sum

Given a m x n grid, we need to find the maximum row sum. We iterate each row and define wealth = 0, then iterate each column, add each accounts[i][j] to wealth. After that, update the answer if the wealth for the current row is greater than the current maximum sum.

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

Actually we don't need to define n, m, i and j if we write in the following style.

class Solution {
public:
    int maximumWealth(vector<vector<int>>& accounts) {
        int ans = 0;
        for (auto x : accounts) {
            int wealth = 0;
            for (auto y : x) wealth += y;
            ans = max(ans, wealth);
        }
        return ans;
    }
};

Approach 2: Max Row Sum (STL)

Same idea as approach 1 but it is using STL.

class Solution {
public:
    int maximumWealth(vector<vector<int>>& accounts) {
        int ans = 0;
        for (auto x : accounts) {
            ans = max(ans, accumulate(x.begin(), x.end(), 0));
        }
        return ans;
    }
}

class Solution:
    def maximumWealth(self, accounts: List[List[int]]) -> int:
        res = 0
        for i in accounts:
            res = max(res, sum(i))
        return res

/**
 * @param {number[][]} accounts
 * @return {number}
 */
var maximumWealth = function (accounts) {
  let ans = 0;
  for (i of accounts) {
    ans = Math.max(ans, sum(i));
  }
  return ans;
};

function sum(arr) {
  let res = 0;
  for (i of arr) {
    res += i;
  }
  return res;
}

1672 - Richest Customer Wealth (Easy)

Problem Link

Problem Statement

Approach 1: Max Row Sum

Approach 2: Max Row Sum (STL)

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