LeetCode #2585 — HARD

Number of Ways to Earn Points

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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] = [count_i, marks_i] indicates that there are count_i questions of the i^th type, and each one of them is worth marks_i points.

Return the number of ways you can earn exactly target points in the exam. Since the answer may be too large, return it modulo 10⁹ + 7.

Note that questions of the same type are indistinguishable.

For example, if there are 3 questions of the same type, then solving the 1^st and 2^nd questions is the same as solving the 1^st and 3^rd questions, or the 2^nd and 3^rd questions.

Example 1:

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 0^th type: 1 + 1 + 1 + 1 + 1 + 1 = 6
- Solve 4 questions of the 0^th type and 1 question of the 1^st type: 1 + 1 + 1 + 1 + 2 = 6
- Solve 2 questions of the 0^th type and 2 questions of the 1^st type: 1 + 1 + 2 + 2 = 6
- Solve 3 questions of the 0^th type and 1 question of the 2^nd type: 1 + 1 + 1 + 3 = 6
- Solve 1 question of the 0^th type, 1 question of the 1^st type and 1 question of the 2^nd type: 1 + 2 + 3 = 6
- Solve 3 questions of the 1^st type: 2 + 2 + 2 = 6
- Solve 2 questions of the 2^nd type: 3 + 3 = 6

Example 2:

Input: target = 5, types = [[50,1],[50,2],[50,5]]
Output: 4
Explanation: You can earn 5 points in one of the four ways:
- Solve 5 questions of the 0^th type: 1 + 1 + 1 + 1 + 1 = 5
- Solve 3 questions of the 0^th type and 1 question of the 1^st type: 1 + 1 + 1 + 2 = 5
- Solve 1 questions of the 0^th type and 2 questions of the 1^st type: 1 + 2 + 2 = 5
- Solve 1 question of the 2^nd type: 5

Example 3:

Input: target = 18, types = [[6,1],[3,2],[2,3]]
Output: 1
Explanation: You can only earn 18 points by answering all questions.

Constraints:

1 <= target <= 1000
n == types.length
1 <= n <= 50
types[i].length == 2
1 <= count_i, marks_i <= 50

Step 01

Brute Force Baseline

Problem summary: 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. Return the number of ways you can earn exactly target points in the exam. Since the answer may be too large, return it modulo 109 + 7. Note that questions of the same type are indistinguishable. For example, if there are 3 questions of the same type, then solving the 1st and 2nd questions is the same as solving the 1st and 3rd questions, or the 2nd and 3rd questions.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Dynamic Programming

Example 1

6
[[6,1],[3,2],[2,3]]

Example 2

5
[[50,1],[50,2],[50,5]]

Example 3

18
[[6,1],[3,2],[2,3]]

Core Insight

What unlocks the optimal approach

Use Dynamic Programming
Let ways[i][points] be the number of ways to score a given number of points after solving some questions of the first i types.
ways[i][points] is equal to the sum of ways[i-1][points - solved * marks[i] over 0 <= solved <= count_i

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2585: Number of Ways to Earn Points
class Solution {
    public int waysToReachTarget(int target, int[][] types) {
        int n = types.length;
        final int mod = (int) 1e9 + 7;
        int[][] f = new int[n + 1][target + 1];
        f[0][0] = 1;
        for (int i = 1; i <= n; ++i) {
            int count = types[i - 1][0], marks = types[i - 1][1];
            for (int j = 0; j <= target; ++j) {
                for (int k = 0; k <= count; ++k) {
                    if (j >= k * marks) {
                        f[i][j] = (f[i][j] + f[i - 1][j - k * marks]) % mod;
                    }
                }
            }
        }
        return f[n][target];
    }
}

// Accepted solution for LeetCode #2585: Number of Ways to Earn Points
func waysToReachTarget(target int, types [][]int) int {
	n := len(types)
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, target+1)
	}
	f[0][0] = 1
	const mod = 1e9 + 7
	for i := 1; i <= n; i++ {
		count, marks := types[i-1][0], types[i-1][1]
		for j := 0; j <= target; j++ {
			for k := 0; k <= count; k++ {
				if j >= k*marks {
					f[i][j] = (f[i][j] + f[i-1][j-k*marks]) % mod
				}
			}
		}
	}
	return f[n][target]
}

# Accepted solution for LeetCode #2585: Number of Ways to Earn Points
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]

// Accepted solution for LeetCode #2585: Number of Ways to Earn Points
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2585: Number of Ways to Earn Points
// class Solution {
//     public int waysToReachTarget(int target, int[][] types) {
//         int n = types.length;
//         final int mod = (int) 1e9 + 7;
//         int[][] f = new int[n + 1][target + 1];
//         f[0][0] = 1;
//         for (int i = 1; i <= n; ++i) {
//             int count = types[i - 1][0], marks = types[i - 1][1];
//             for (int j = 0; j <= target; ++j) {
//                 for (int k = 0; k <= count; ++k) {
//                     if (j >= k * marks) {
//                         f[i][j] = (f[i][j] + f[i - 1][j - k * marks]) % mod;
//                     }
//                 }
//             }
//         }
//         return f[n][target];
//     }
// }

// Accepted solution for LeetCode #2585: Number of Ways to Earn Points
function waysToReachTarget(target: number, types: number[][]): number {
    const n = types.length;
    const mod = 10 ** 9 + 7;
    const f: number[][] = Array.from({ length: n + 1 }, () => Array(target + 1).fill(0));
    f[0][0] = 1;
    for (let i = 1; i <= n; ++i) {
        const [count, marks] = types[i - 1];
        for (let j = 0; j <= target; ++j) {
            for (let k = 0; k <= count; ++k) {
                if (j >= k * marks) {
                    f[i][j] = (f[i][j] + f[i - 1][j - k * marks]) % mod;
                }
            }
        }
    }
    return f[n][target];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × target × count)

Space

O(n × target)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.