Question

1672. Richest Customer Wealth

中文文档

Description

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 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.length
• n == accounts[i].length
• 1 <= m, n <= 50
• 1 <= accounts[i][j] <= 100

Solutions

Solution 1: Summation

We traverse accounts and find the maximum sum of each row.

The time complexity is $O(m \times n)$, where $m$ and $n$ are the number of rows and columns in the grid, respectively. The space complexity is $O(1)$.

class Solution:
  def maximumWealth(self, accounts: List[List[int]]) -> int:
    return max(sum(v) for v in accounts)

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;
  }
}

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;
  }
};

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
}

function maximumWealth(accounts: number[][]): number {
  return accounts.reduce(
    (r, v) =>
      Math.max(
        r,
        v.reduce((r, v) => r + v),
      ),
    0,
  );
}

impl Solution {
  pub fn maximum_wealth(accounts: Vec<Vec<i32>>) -> i32 {
    accounts
      .iter()
      .map(|v| v.iter().sum())
      .max()
      .unwrap()
  }
}

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;
  }
}

#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;
}

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
  }
}