README_EN.md

1672. Richest Customer Wealth

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Description

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

Solutions

Python3

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

Java

class Solution {
    public int maximumWealth(int[][] accounts) {
        int ans = 0;
        for (var e : accounts) {
            int s = 0;
            for (int v : e) {
                s += v;
            }
            ans = Math.max(ans, s);
        }
        return ans;
    }
}

C++

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

Go

func maximumWealth(accounts [][]int) int {
    ans := 0
    for _, e := range accounts {
        s := 0
        for _, v := range e {
            s += v
        }
        if ans < s {
            ans = s
        }
    }
    return ans
}

TypeScript

function maximumWealth(accounts: number[][]): number {
    return accounts.reduce(
        (res, account) =>
            Math.max(
                res,
                account.reduce((p, v) => p + v),
            ),
        0,
    );
}

Rust

impl Solution {
    pub fn maximum_wealth(accounts: Vec<Vec<i32>>) -> i32 {
        accounts
            .iter()
            .map(|account| account.iter().sum())
            .max()
            .unwrap()
    }
}

Kotlin

class Solution {
    fun maximumWealth(accounts: Array<IntArray>): Int {
        var max = 0
        for (account in accounts) {
            val sum = account.sum()
            if (sum > max) {
                max = sum
            }
        }
        return max
    }
}

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