Difficulty: 🟢 Easy
You are given an m x n
integer grid accounts where accounts[i][j]
is the
amount of money the ith customer has in the jth bank.
Return the wealth that the richest customer has.
A customer's wealth is the amount 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 = 6
2nd customer has wealth = 10
3rd customer has wealth = 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
m == accounts.length
n == accounts[i].length
1 <= m, n <= 50
1 <= accounts[i][j] <= 100
class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(sum(account) for account in accounts)
The given solution uses a list comprehension to calculate the wealth of each customer by summing the amounts in their bank accounts. It then returns the maximum wealth from the calculated values.
The algorithm works as follows:
- Iterate through each
account
in theaccounts
matrix. - Calculate the wealth of each customer by summing the amounts in their bank accounts using the
sum
function. - Return the maximum wealth among all the customers.
The time complexity of this algorithm is O(m * n), where m is the number of rows and n is the number of columns in the accounts
matrix. The algorithm iterates through each element in the matrix to calculate the wealth of each customer.
The space complexity of the algorithm is O(1) as it only uses a constant amount of space to store intermediate variables.
The given solution calculates the wealth of each customer in the accounts
matrix by summing the amounts in their bank accounts. It then returns the maximum wealth among all the customers. The algorithm has a time complexity of O(m * n) and a space complexity of O(1), making it an efficient solution for the problem at hand.
NB: If you want to get community points please suggest solutions in other languages as merge requests.