Leetcode #2585: Number of Ways to Earn Points
In this guide, we solve Leetcode #2585 Number of Ways to Earn Points in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
There is a test that has n types of questions. You are given an integer target and a 0-indexed 2D integer array types where types[i] = [counti, marksi] indicates that there are counti questions of the ith type, and each one of them is worth marksi points.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Array, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: target = 6, types = [[6,1],[3,2],[2,3]]
Output: 7
Explanation: You can earn 6 points in one of the seven ways:
- Solve 6 questions of the 0th type: 1 + 1 + 1 + 1 + 1 + 1 = 6
- Solve 4 questions of the 0th type and 1 question of the 1st type: 1 + 1 + 1 + 1 + 2 = 6
- Solve 2 questions of the 0th type and 2 questions of the 1st type: 1 + 1 + 2 + 2 = 6
- Solve 3 questions of the 0th type and 1 question of the 2nd type: 1 + 1 + 1 + 3 = 6
- Solve 1 question of the 0th type, 1 question of the 1st type and 1 question of the 2nd type: 1 + 2 + 3 = 6
- Solve 3 questions of the 1st type: 2 + 2 + 2 = 6
- Solve 2 questions of the 2nd type: 3 + 3 = 6
Python Solution
class Solution:
def waysToReachTarget(self, target: int, types: List[List[int]]) -> int:
n = len(types)
mod = 10**9 + 7
f = [[0] * (target + 1) for _ in range(n + 1)]
f[0][0] = 1
for i in range(1, n + 1):
count, marks = types[i - 1]
for j in range(target + 1):
for k in range(count + 1):
if j >= k * marks:
f[i][j] = (f[i][j] + f[i - 1][j - k * marks]) % mod
return f[n][target]
Complexity
The time complexity is and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.