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3687. Library Late Fee Calculator 🔒

Description

You are given an integer array daysLate where daysLate[i] indicates how many days late the ith book was returned.

The penalty is calculated as follows:

• If daysLate[i] == 1, penalty is 1.
• If 2 <= daysLate[i] <= 5, penalty is 2 * daysLate[i].
• If daysLate[i] > 5, penalty is 3 * daysLate[i].

Return the total penalty for all books.

 

Example 1:

Input: daysLate = [5,1,7]

Output: 32

Explanation:

• daysLate[0] = 5: Penalty is 2 * daysLate[0] = 2 * 5 = 10.
• daysLate[1] = 1: Penalty is 1.
• daysLate[2] = 7: Penalty is 3 * daysLate[2] = 3 * 7 = 21.
• Thus, the total penalty is 10 + 1 + 21 = 32.

Example 2:

Input: daysLate = [1,1]

Output: 2

Explanation:

• daysLate[0] = 1: Penalty is 1.
• daysLate[1] = 1: Penalty is 1.
• Thus, the total penalty is 1 + 1 = 2.

 

Constraints:

• 1 <= daysLate.length <= 100
• 1 <= daysLate[i] <= 100

Solutions

Solution 1: Simulation

We define a function $\text{f}(x)$ to calculate the late fee for each book:

\[\text{f}(x) = \begin{cases} 1 & x = 1 \\ 2x & 2 \leq x \leq 5 \\ 3x & x > 5 \end{cases}\]

Then, for each element $x$ in the array $\textit{daysLate}$, we compute $\text{f}(x)$ and sum them up to get the total late fee.

The time complexity is $O(n)$, where $n$ is the length of the array $\textit{daysLate}$. The space complexity is $O(1)$.

• Java
• C++
• Python
• Go
• TypeScript

• class Solution {
      public int lateFee(int[] daysLate) {
          IntUnaryOperator f = x -> {
              if (x == 1) {
                  return 1;
              } else if (x > 5) {
                  return 3 * x;
              } else {
                  return 2 * x;
              }
          };
  
          int ans = 0;
          for (int x : daysLate) {
              ans += f.applyAsInt(x);
          }
          return ans;
      }
  }
  
  

• class Solution {
  public:
      int lateFee(vector<int>& daysLate) {
          auto f = [](int x) {
              if (x == 1) {
                  return 1;
              } else if (x > 5) {
                  return 3 * x;
              } else {
                  return 2 * x;
              }
          };
  
          int ans = 0;
          for (int x : daysLate) {
              ans += f(x);
          }
          return ans;
      }
  };
  
  

• class Solution:
      def lateFee(self, daysLate: List[int]) -> int:
          def f(x: int) -> int:
              if x == 1:
                  return 1
              if x > 5:
                  return 3 * x
              return 2 * x
  
          return sum(f(x) for x in daysLate)
  
  

• func lateFee(daysLate []int) (ans int) {
  	f := func(x int) int {
  		if x == 1 {
  			return 1
  		} else if x > 5 {
  			return 3 * x
  		}
  		return 2 * x
  	}
  	for _, x := range daysLate {
  		ans += f(x)
  	}
  	return
  }
  
  

• function lateFee(daysLate: number[]): number {
      const f = (x: number): number => {
          if (x === 1) {
              return 1;
          } else if (x > 5) {
              return 3 * x;
          }
          return 2 * x;
      };
      return daysLate.reduce((acc, days) => acc + f(days), 0);
  }
  
  

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