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Leetcode 1672. Richest Customer Wealth

Problem Statement

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 account. Return the wealth that the richest customer has.

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.

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 = 1 + 5 = 6
- 2nd customer has wealth = 7 + 3 = 10
- 3rd customer has wealth = 3 + 5 = 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

Clarifying Questions

1. What is the maximum size of the accounts array?
   • The size constraints are not specified here, but typically for LeetCode problems, let’s assume a manageable size for computations within a typical online judge.
2. Can there be negative values in the accounts array?
   • The problem does not specify the presence of negative values, so we will assume all values are non-negative.
3. Should the solution handle empty subarrays?
   • We assume that the input will always contain proper subarrays with at least one integer.

Strategy

1. Initialize a variable max_wealth to keep track of the maximum wealth found.
2. Iterate over each customer in the accounts array.
3. For each customer, calculate their total wealth by summing the values in their subarray.
4. Update max_wealth if the current customer’s wealth is greater than the previously found maximum wealth.
5. Return the max_wealth after iterating through all customers.

Code

We will implement the strategy in C++ as follows:

#include <vector>
#include <algorithm>

class Solution {
public:
    int maximumWealth(std::vector<std::vector<int>>& accounts) {
        int max_wealth = 0;
        
        for (const auto& customer : accounts) {
            int current_wealth = 0;
            for (int money : customer) {
                current_wealth += money;
            }
            max_wealth = std::max(max_wealth, current_wealth);
        }
        
        return max_wealth;
    }
};

Time Complexity

• Time Complexity: O(m * n), where m is the number of customers and n is the number of bank accounts per customer. This is because we need to iterate through each element in the accounts grid exactly once.
• Space Complexity: O(1), as no additional space proportional to the input size is used. Only a constant amount of extra space is required for variables like max_wealth and current_wealth.

This solution ensures that we efficiently compute the maximum wealth with optimal space and time complexities.

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