Leetcode 1672: Richest Customer Wealth

Input: The input consists of a 2D list `accounts`, where each element represents a customer's balance across multiple banks.

Example: accounts = [[2, 3, 4], [1, 5, 2]]

Constraints:

• 1 <= m, n <= 50

• 1 <= accounts[i][j] <= 100

Output: Return the wealth of the richest customer.

Example: Output: 9

Constraints:

Goal: Calculate the wealth of each customer and return the maximum.

Steps:

• Iterate through each row in the `accounts` list.

• For each customer, calculate the total wealth by summing the elements of their corresponding row.

• Track the maximum wealth during the iteration.

• Return the maximum wealth after processing all customers.

Goal: The input will always satisfy the following constraints:

Steps:

• The `accounts` list will always have at least one customer.

• Each customer will have at least one bank account.

Assumptions:

• There is no need to handle negative balances.

• Input: Input: [[3, 1], [6, 7], [4, 3]]

• Explanation: The wealth of each customer is calculated by summing their balances. The richest customer is the one with the maximum wealth.

Approach: We can solve this problem by iterating through the list of accounts and calculating the sum of each customer's wealth. The final result will be the maximum wealth found during the iteration.

Observations:

• We need to calculate the wealth of each customer.

• The solution involves summing the elements of each row and finding the maximum.

• A simple approach of iterating over the 2D list and finding the maximum wealth will work within the given constraints.

Steps:

• Initialize a variable `maxWealth` to store the maximum wealth found.

• Iterate through each customer in the `accounts` list.

• For each customer, calculate the sum of their wealth by summing the values in their row.

• Update `maxWealth` with the maximum wealth encountered.

• Return the `maxWealth` at the end.

Empty Inputs:

• There will always be at least one customer.

Large Inputs:

• The problem can handle up to 50 customers and 50 accounts for each.

Special Values:

• Balances will always be positive integers between 1 and 100.

Constraints:

int maximumWealth(vector<vector<int>>& accounts) {
    int richest = 0;
    for (auto &customer : accounts)
        richest = max (richest, accumulate(customer.begin(), customer.end(), 0));
    
    return richest;
}

1 : Function Declaration

int maximumWealth(vector<vector<int>>& accounts) {

This line defines the function `maximumWealth` that takes a 2D vector `accounts`, where each inner vector represents a customer's account balances, and returns an integer representing the maximum wealth.

2 : Variable Initialization

    int richest = 0;

Here, the variable `richest` is initialized to 0. This variable will hold the maximum wealth encountered as the program iterates over the `accounts` vector.

3 : Looping through Accounts

    for (auto &customer : accounts)

This for-loop iterates through each customer's account (represented as `customer`) in the `accounts` vector.

4 : Accumulation of Wealth

        richest = max (richest, accumulate(customer.begin(), customer.end(), 0));

This line calculates the sum of the balances in each customer's account using `accumulate()`. It then updates the `richest` variable if the current customer's wealth is greater than the previous maximum wealth.

5 : Return Statement

    return richest;

Once all customers have been processed, this line returns the maximum wealth found, stored in `richest`.

Best Case: O(m * n)

Average Case: O(m * n)

Worst Case: O(m * n)

Description: We iterate over each customer and their accounts, so the time complexity is proportional to the size of the input.

Best Case: O(1)

Worst Case: O(1)

Description: We only use a constant amount of extra space to track the maximum wealth.

Leetcode 1672: Richest Customer Wealth

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