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

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