1672. Richest Customer Wealth

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

Solution 1: Summation

We traverse accounts and find the maximum sum of each row.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the grid, respectively. The space complexity is $O(1)$.

Python3JavaC++GoTypeScriptRustPHPCKotlin

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

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

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;
    }
};

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
}

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

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

class Solution {
    /**
     * @param Integer[][] $accounts
     * @return Integer
     */
    function maximumWealth($accounts) {
        $rs = 0;
        for ($i = 0; $i < count($accounts); $i++) {
            $sum = 0;
            for ($j = 0; $j < count($accounts[$i]); $j++) {
                $sum += $accounts[$i][$j];
            }
            if ($sum > $rs) {
                $rs = $sum;
            }
        }
        return $rs;
    }
}

#define max(a, b) (((a) > (b)) ? (a) : (b))

int maximumWealth(int** accounts, int accountsSize, int* accountsColSize) {
    int ans = INT_MIN;
    for (int i = 0; i < accountsSize; i++) {
        int sum = 0;
        for (int j = 0; j < accountsColSize[i]; j++) {
            sum += accounts[i][j];
        }
        ans = max(ans, sum);
    }
    return ans;
}

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
    }
}

1672. Richest Customer Wealth

Description

Solutions

Solution 1: Summation

Comments