Problem statement

accounts[i][j] is the money the i-th customer holds in the j-th bank. A customer's wealth is the sum of their accounts.

Return the wealth of the richest customer.

Example 1

Input: accounts = [[1,2,3],[3,2,1]]
Output: 6

Example 2

Input: accounts = [[1,5],[7,3],[3,5]]
Output: 10
Explanation: Customer 2: 7 + 3 = 10.

Constraints

• 1 <= customers, banks <= 8
• 1 <= accounts[i][j] <= 100

Hints for Richest Customer Wealth

1. Hint 1

   Sum each row.

2. Hint 2

   Track the running maximum.

How to solve Richest Customer Wealth

Each customer's wealth is just the sum of their row, so the answer is the largest row sum. One pass over the matrix, keeping a running maximum — no sorting, no extra storage.

Approach

1. Start the best wealth at zero.
2. For each row, add up its entries.
3. If that total beats the current best, keep it.
4. Return the best.

Why it works

There is no structure to exploit here — every account must be read to know its customer's total, so O(rows · cols) is optimal. Tracking a running maximum rather than collecting all the sums keeps the extra space constant.

Complexity

• Time — O(rows · cols)
• Space — O(1)

Pitfalls

• Reset the row sum for each customer; carrying it over accumulates across rows.
• The answer is the maximum row sum, not the total of the whole matrix.
• All values are positive here, so seeding the best at zero is safe — with possible negatives you would seed from the first row instead.

Reference solution

Python

from typing import List

def maximumWealth(accounts: List[List[int]]) -> int:
    return max(sum(a) for a in accounts)

Java

public static int maximumWealth(int[][] accounts) {
    int best = 0;
    for (int[] row : accounts) {
        int sum = 0;
        for (int v : row) sum += v;
        if (sum > best) best = sum;
    }
    return best;
}

C++

int maximumWealth(vector<vector<int>>& accounts) {
    int best = 0;
    for (const auto& row : accounts) {
        int sum = 0;
        for (int v : row) sum += v;
        if (sum > best) best = sum;
    }
    return best;
}

JavaScript

var maximumWealth = function(accounts) {
    let best = 0;
    for (const a of accounts) {
        let sum = 0;
        for (const x of a) sum += x;
        best = Math.max(best, sum);
    }
    return best;
};

Also on the editorial tab: C, C#, Go, Kotlin, PHP, Ruby, Rust, Swift, TypeScript.

More arrays problems

• Brightest String of Diyas Easy
• Design Neighbor Sum Service Easy
• Design an Ordered Stream Easy
• Design HashMap Easy
• Design HashSet Easy
• Range Sum Query - Immutable Easy