Richest Customer Wealth

Richest Customer Wealth - Problem

Array Matrix Easy

You are given an m x n integer matrix accounts where accounts[i][j] is the amount of money the i^th customer has in the j^th bank.

Return the wealth of the richest customer.

A customer's wealth is the sum of money they have in all their bank accounts. The richest customer is the customer that has the maximum wealth.

Input & Output

Example 1 — Equal Wealth

$ 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 customers have equal wealth of 6.

Example 2 — Different Wealth

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

› Output: 10

💡 Note: Customer 1: 1+5 = 6, Customer 2: 7+3 = 10, Customer 3: 3+5 = 8. The richest customer has wealth = 10.

Example 3 — Single Account

$ Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]

› Output: 17

💡 Note: Customer 1: 2+8+7 = 17, Customer 2: 7+1+3 = 11, Customer 3: 1+9+5 = 15. Maximum wealth is 17.

Constraints

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

Visualization

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Asked in

LC LeetCode 25 A Amazon 15

The key insight is to iterate through each customer and sum their bank accounts to find total wealth. The optimal approach uses built-in sum functions for cleaner code. Time: O(m×n), Space: O(1)

Common Approaches

✓ Brute Force - Nested Loop

⏱️ Time: O(m×n) Space: O(1)

For each customer, iterate through all their bank accounts to calculate total wealth, then keep track of the maximum wealth seen so far. This is straightforward but involves nested loops.

Optimized Single Pass

⏱️ Time: O(m×n) Space: O(1)

Instead of manually looping through each account, use language built-in sum functions to calculate each customer's wealth in one operation. This makes the code cleaner and more readable while maintaining the same time complexity.

Brute Force - Nested Loop — Algorithm Steps

Iterate through each customer (row)
For each customer, sum all their bank accounts (columns)
Track the maximum wealth encountered

Visualization

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Step-by-Step Walkthrough

Customer 1

Sum accounts: 1+2+3 = 6

Customer 2

Sum accounts: 3+2+1 = 6

Result

Maximum wealth is 6

Code -

solution.c — C

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

int solution(int** accounts, int m, int* colSizes) {
    int maxWealth = 0;
    for (int i = 0; i < m; i++) {
        int currentWealth = 0;
        for (int j = 0; j < colSizes[i]; j++) {
            currentWealth += accounts[i][j];
        }
        if (currentWealth > maxWealth) {
            maxWealth = currentWealth;
        }
    }
    return maxWealth;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Simple parsing for test cases like [[1,2,3],[3,2,1]]
    int** accounts = malloc(100 * sizeof(int*));
    int* colSizes = malloc(100 * sizeof(int));
    int m = 0;
    
    char* ptr = line + 1; // skip first '['
    while (*ptr && *ptr != ']') {
        if (*ptr == '[') {
            accounts[m] = malloc(100 * sizeof(int));
            colSizes[m] = 0;
            ptr++;
            
            while (*ptr && *ptr != ']') {
                if (*ptr >= '0' && *ptr <= '9') {
                    int num = 0;
                    while (*ptr >= '0' && *ptr <= '9') {
                        num = num * 10 + (*ptr - '0');
                        ptr++;
                    }
                    accounts[m][colSizes[m]++] = num;
                } else {
                    ptr++;
                }
            }
            m++;
        }
        ptr++;
    }
    
    int result = solution(accounts, m, colSizes);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(m×n)

Visit each cell in the m×n matrix exactly once

✓ Linear Growth

Space Complexity

O(1)

Only use variables to track current sum and maximum wealth

✓ Linear Space

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Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force - Nested Loop — Algorithm Steps

Visualization

Code -

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