Description
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
Constraints:
m == accounts.lengthn == accounts[i].length1 <= m, n <= 501 <= accounts[i][j] <= 100
Solutions
Solution 1: Summation
We traverse accounts and find the maximum sum of each row.
The time complexity is O(m × n), where m and n are the number of rows and columns in the grid, respectively. The space complexity is O(1).
PythonJavaC++GoTypeScriptRustPHPCKotlin
class Solution:
def maximumWealth(self, accounts: List[List[int]]) -> int:
return max(sum(v) for v in accounts)(code-box)
class Solution {
public int maximumWealth(int[][] accounts) {
int ans = 0;
for (var e : accounts) {
// int s = Arrays.stream(e).sum();
int s = 0;
for (int v : e) {
s += v;
}
ans = Math.max(ans, s);
}
return ans;
}
}(code-box)
class Solution {
public:
int maximumWealth(vector<vector<int>>& accounts) {
int ans = 0;
for (auto& v : accounts) {
ans = max(ans, accumulate(v.begin(), v.end(), 0));
}
return ans;
}
};(code-box)
func maximumWealth(accounts [][]int) int {
ans := 0
for _, e := range accounts {
s := 0
for _, v := range e {
s += v
}
if ans < s {
ans = s
}
}
return ans
}(code-box)
function maximumWealth(accounts: number[][]): number {
return accounts.reduce(
(r, v) =>
Math.max(
r,
v.reduce((r, v) => r + v),
),
0,
);
}(code-box)
impl Solution {
pub fn maximum_wealth(accounts: Vec<Vec<i32>>) -> i32 {
accounts.iter().map(|v| v.iter().sum()).max().unwrap()
}
}(code-box)
class Solution {
/**
* @param Integer[][] $accounts
* @return Integer
*/
function maximumWealth($accounts) {
$rs = 0;
for ($i = 0; $i < count($accounts); $i++) {
$sum = 0;
for ($j = 0; $j < count($accounts[$i]); $j++) {
$sum += $accounts[$i][$j];
}
if ($sum > $rs) {
$rs = $sum;
}
}
return $rs;
}
}(code-box)
#define max(a, b) (((a) > (b)) ? (a) : (b))
int maximumWealth(int** accounts, int accountsSize, int* accountsColSize) {
int ans = INT_MIN;
for (int i = 0; i < accountsSize; i++) {
int sum = 0;
for (int j = 0; j < accountsColSize[i]; j++) {
sum += accounts[i][j];
}
ans = max(ans, sum);
}
return ans;
}(code-box)
class Solution {
fun maximumWealth(accounts: Array<IntArray>): Int {
var max = 0
for (account in accounts) {
val sum = account.sum()
if (sum > max) {
max = sum
}
}
return max
}
}(code-box)
