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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. 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 richest customer is the 2nd customer with a wealth of 10.

Example 3:

Input: accounts = [[2,8,7],[7,1,3],[1,9,5]]
Output: 17

Clarifying Questions

1. What constraints should be considered?
   • The length of accounts and the length of each subarray can vary, typically within common input sizes for coding challenges.
2. Can the values in accounts be negative?
   • Based on problem context, all values are non-negative integers as they represent amounts of money.
3. Are there any performance constraints?
   • Standard efficient algorithms for finding maximum values and summing arrays should be sufficient.

Strategy

1. Iterate through each customer in the accounts list.
2. For each customer, calculate the sum of all amounts in their accounts.
3. Keep track of the maximum sum encountered.
4. Return the maximum sum found.

Code

Here is the Python implementation of the solution:

def maximumWealth(accounts):
    max_wealth = 0
    
    for customer in accounts:
        customer_wealth = sum(customer)
        if customer_wealth > max_wealth:
            max_wealth = customer_wealth
    
    return max_wealth

Time Complexity

• The time complexity of this algorithm is O(m * n), where m is the number of customers and n is the number of accounts per customer.
  • We need to iterate through each customer (m iterations).
  • For each customer, summing their wealth involves iterating through n accounts.
• The space complexity is O(1), as we are using a constant amount of extra space, only storing the maximum wealth found.

This implementation is efficient and straightforward, suitable for the problem constraints typically encountered in coding interviews.

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