Richest Customer Wealth - Practice Coding Problems

Richest Customer Wealth - Problem

Array Matrix Easy

Imagine you're a bank analyst tasked with identifying the wealthiest customer across multiple banking institutions. You have access to a comprehensive financial report represented as a matrix where each row represents a customer and each column represents a different bank.

Given an m x n integer matrix accounts where accounts[i][j] represents the amount of money the i^th customer has in the j^th bank, your goal is to find the maximum total wealth among all customers.

A customer's wealth is calculated as the sum of all money they have across all their bank accounts. Return the wealth of the richest customer.

Example: If customer 0 has [1, 2, 3] in three banks, their total wealth is 1 + 2 + 3 = 6.

Input & Output

example_1.py — Basic Case

$ Input: accounts = [[1,2,3],[3,2,1]]

› Output: 6

💡 Note: Customer 1 has wealth = 1 + 2 + 3 = 6. Customer 2 has wealth = 3 + 2 + 1 = 6. Both have maximum wealth of 6.

example_2.py — Unequal Wealth

$ Input: accounts = [[1,5],[7,3],[3,5]]

› Output: 10

💡 Note: Customer 1 has wealth = 1 + 5 = 6. Customer 2 has wealth = 7 + 3 = 10. Customer 3 has wealth = 3 + 5 = 8. Maximum wealth is 10.

example_3.py — Single Customer

$ Input: accounts = [[2,8,7]]

› Output: 17

💡 Note: There's only one customer with wealth = 2 + 8 + 7 = 17.

Visualization

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Understanding the Visualization

Start Analysis

Begin with the first customer's portfolio

Sum Accounts

Add up all account balances for this customer

Track Maximum

Compare with the highest wealth seen so far

Next Customer

Move to the next customer and repeat

Final Result

Return the highest total wealth found

Key Takeaway

🎯 Key Insight: We only need to track the running maximum while calculating each customer's total wealth, making this an optimal O(1) space solution.

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

We visit each element in the matrix exactly once, where m is customers and n is banks

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track current and maximum wealth

✓ Linear Space

Constraints

m == accounts.length
n == accounts[i].length
1 ≤ m, n ≤ 50
1 ≤ accounts[i][j] ≤ 100

Asked in

a Amazon 25 G Google 18 ⊞ Microsoft 15 Apple 12

This problem is solved optimally with a single pass through the matrix using nested loops. For each customer (row), we sum all their bank accounts (columns) and track the maximum wealth seen. Time complexity is O(m × n) where we must visit each element once, and space complexity is O(1) as we only need variables to track current and maximum wealth.

Common Approaches

Approach	Time	Space	Notes
✓ Nested Loop Approach	O(m × n)	O(1)	Iterate through each customer and sum their wealth across all banks

Nested Loop Approach — Algorithm Steps

Initialize maxWealth to 0
For each customer (outer loop)
Calculate current customer's wealth by summing all their accounts (inner loop)
Update maxWealth if current wealth is greater
Return maxWealth

Visualization

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Step-by-Step Walkthrough

Initialize

Start with maxWealth = 0

Process Row 1

Sum all accounts for customer 1

Update Maximum

Compare with current maximum

Continue

Repeat for all customers

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

int maximumWealth(int** accounts, int accountsSize, int* accountsColSize) {
    int maxWealth = 0;
    for (int i = 0; i < accountsSize; i++) {
        int currentWealth = 0;
        for (int j = 0; j < accountsColSize[i]; j++) {
            currentWealth += accounts[i][j];
        }
        if (currentWealth > maxWealth) {
            maxWealth = currentWealth;
        }
    }
    return maxWealth;
}

int main() {
    // Example usage
    int row1[] = {1, 2, 3};
    int row2[] = {3, 2, 1};
    int* accounts[] = {row1, row2};
    int colSizes[] = {3, 3};
    printf("%d\n", maximumWealth(accounts, 2, colSizes));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m × n)

We visit each element in the matrix exactly once, where m is customers and n is banks

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables to track current and maximum wealth

✓ Linear Space

Constraints

m == accounts.length
n == accounts[i].length
1 ≤ m, n ≤ 50
1 ≤ accounts[i][j] ≤ 100

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Ln 1, Col 1

Smart Actions

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Input & Output

Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Nested Loop Approach — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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