What is the time complexity of Number of Ways to Earn Points?

The optimal solution has a time complexity of O(t*n).

What data structures are used to solve Number of Ways to Earn Points?

This problem uses Array, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Ways to Earn Points?

Insight 1: The problem can be modeled as a variation of the classic knapsack problem, where we aim to find the number of ways to reach a target sum using a given set of items (question types). Insight 2: Dynamic programming is the most efficient approach to solve this problem, as it avoids redundant calculations by storing and reusing intermediate results.

Which companies ask LeetCode 2585. Number of Ways to Earn Points in interviews?

This problem is frequently asked at TuSimple.

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Number of Ways to Earn Points - LeetCode 2585 Solution

Number of Ways to Earn Points - Complete Solution Guide

Number of Ways to Earn Points is LeetCode problem 2585, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 10 9 + 7 . Note that questions of the same type are indistinguishable. For example, if there are 3 questions o

Detailed Explanation

The problem asks us to find the number of ways to achieve a specific target score in a test, given a list of question types. Each question type has a certain number of available questions and a specific point value per question. We need to find the number of combinations of questions (from different types) that sum up to the target score, returning the result modulo 10^9 + 7. Questions of the same type are considered indistinguishable.

Solution Approach

The solution uses dynamic programming to determine the number of ways to achieve the target score. A 1D DP array `dp` is initialized, where `dp[i]` represents the number of ways to achieve a score of `i`. The `dp` array is updated iteratively for each question type. For each question type, we iterate through the possible scores (from `0` to `target`). At each score `j`, we calculate the number of ways to achieve it by either not using the current question type (inheriting the value from the previous DP state) or using one or more questions of the current type. The recurrence relation `new_dp[j] = new_dp[j-marks] + dp[j] - dp[j - (count+1)*marks]` efficiently handles this calculation. The `new_dp` stores results after processing the current type, and is used to update `dp` for next type.

Step-by-Step Algorithm

Step 1: Initialize a DP array `dp` of size `target + 1` with all elements set to 0. Set `dp[0] = 1`, indicating that there is one way to achieve a score of 0 (by not answering any questions).
Step 2: Iterate through each question type in the `types` array. For each question type, extract the number of available questions (`count`) and the point value per question (`marks`).
Step 3: Create a new DP array `new_dp` of size `target + 1` initialized with zeros. This array will hold the updated number of ways to reach each target after considering the current question type.
Step 4: Iterate through all possible scores `j` from 0 to `target`. For each score `j`, calculate the number of ways to reach it considering current question type. This calculation has three terms:
Step 5: `term1`: If `j >= marks`, then new_dp[j-marks] is the number of ways to reach `j` by answering at least one question of the current type. If `j < marks` then this term is 0, because there is no way to answer at least one question.
Step 6: `term2`: `dp[j]` is the number of ways to reach `j` without answering any questions of the current type (inheriting the result from previous types).
Step 7: `term3`: `dp[j - (count+1)*marks]`. Since we are only permitted to answer `count` questions of the current type, we need to eliminate cases where we use more than `count` questions of the current type. If `j >= (count + 1) * marks`, we need to subtract out the number of ways to reach j, which require more than `count` questions of the current type. Otherwise, this term is 0.
Step 8: Update `new_dp[j]` with the sum of `term1`, `term2` and the subtraction of `term3`, all taken modulo `10^9 + 7`. Formula: `new_dp[j] = (term1 + term2 - term3 + MOD) % MOD`
Step 9: After processing all possible scores `j` for the current question type, update the `dp` array with the values from `new_dp` and free the memory allocated for new_dp.
Step 10: After processing all question types, the value `dp[target]` will contain the number of ways to achieve the target score. Return this value.

Key Insights

Insight 1: The problem can be modeled as a variation of the classic knapsack problem, where we aim to find the number of ways to reach a target sum using a given set of items (question types).
Insight 2: Dynamic programming is the most efficient approach to solve this problem, as it avoids redundant calculations by storing and reusing intermediate results.
Insight 3: We need to consider the constraint on the number of questions available for each type. We can't use more questions than available for each type.

Complexity Analysis

Time Complexity: O(t*n)

Space Complexity: O(t)

Topics

This problem involves: Array, Dynamic Programming.

Companies

Asked at: TuSimple.

Number of Ways to Earn Points | LeetCode 2585

Hard

Array Dynamic Programming

Time: O(t*n)Space: O(t)

Solution Code

Available in:

class Solution {
    public int waysToReachTarget(int target, int int[][] types) {
        int MOD = (int)1e9 + 7;
        
        int[] dp = new int[target + 1];
        dp[0] = 1;
        
        for (int[] type : types) {
            int count = type[0];
            int marks = type[1];
            int[] new_dp = new int[target + 1];
            for (int j = 0; j <= target; j++) {
                // The recurrence for the number of ways to make sum j using
                // the current type and previous types (new_dp) based on the ways
                // from just previous types (dp) is:
                // new_dp[j] = new_dp[j-marks] + dp[j] - dp[j - (count+1)*marks]
                
                // Term from using one or more of the current type question
                // This corresponds to new_dp[j-marks]
                int term1 = 0;
                if (j >= marks) {
                    term1 = new_dp[j - marks];
                }
                
                // Term from not using the current type, relying on previous types
                // This corresponds to dp[j]
                int term2 = dp[j];
                
                // Term to subtract for combinations that use more than `count`
                // questions of the current type. This corresponds to dp[j-(count+1)*marks]
                int term3 = 0;
                int limit = (count + 1) * marks;
                if (j >= limit) {
                    term3 = dp[j - limit];
                }
                
                new_dp[j] = (int)(((long)term1 + term2 - term3 + MOD) % MOD);
            }
            
            dp = new_dp;
            
        }
        
        return dp[target];
    }
}

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Problem Info

Description

Problem 2585: Number of Ways to Earn Points - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming

Companies

TuSimple

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2585

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

10 steps

1
Step 1: Initialize a DP array `dp` of size `target + 1` with all elements set to 0. Set `dp[0] = 1`, indicating that there is one way to achieve a score of 0 (by not answering any questions).
2
Step 2: Iterate through each question type in the `types` array. For each question type, extract the number of available questions (`count`) and the point value per question (`marks`).
3
Step 3: Create a new DP array `new_dp` of size `target + 1` initialized with zeros. This array will hold the updated number of ways to reach each target after considering the current question type.
4
Step 4: Iterate through all possible scores `j` from 0 to `target`. For each score `j`, calculate the number of ways to reach it considering current question type. This calculation has three terms:
5
Step 5: `term1`: If `j >= marks`, then new_dp[j-marks] is the number of ways to reach `j` by answering at least one question of the current type. If `j < marks` then this term is 0, because there is no way to answer at least one question.
6
Step 6: `term2`: `dp[j]` is the number of ways to reach `j` without answering any questions of the current type (inheriting the result from previous types).
7
Step 7: `term3`: `dp[j - (count+1)*marks]`. Since we are only permitted to answer `count` questions of the current type, we need to eliminate cases where we use more than `count` questions of the current type. If `j >= (count + 1) * marks`, we need to subtract out the number of ways to reach j, which require more than `count` questions of the current type. Otherwise, this term is 0.
8
Step 8: Update `new_dp[j]` with the sum of `term1`, `term2` and the subtraction of `term3`, all taken modulo `10^9 + 7`. Formula: `new_dp[j] = (term1 + term2 - term3 + MOD) % MOD`
9
Step 9: After processing all possible scores `j` for the current question type, update the `dp` array with the values from `new_dp` and free the memory allocated for new_dp.
10
Step 10: After processing all question types, the value `dp[target]` will contain the number of ways to achieve the target score. Return this value.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(t*n)

💾

Space Complexity

O(t)

📝

Why?

The time complexity is determined by the nested loops, which iterate through the question types and the possible scores. The space complexity is determined by the size of the DP array, which depends on the target score.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach with Memoization

A recursive approach with memoization can also be used to solve this problem. This involves defining a recursive function that takes the current target score and the index of the current question type as input. The function explores all possible ways to achieve the target score by either including or excluding the current question type. Memoization is used to store the results of the recursive calls to avoid redundant calculations.

Time: O(n*target)Space: O(n*target)

Trade-offs: While conceptually simpler, recursive solutions often have higher overhead due to function call stack. The explicit DP is often slightly faster. The space complexity can be higher due to recursion depth.

✨

Key Insights

1Insight 1: The problem can be modeled as a variation of the classic knapsack problem, where we aim to find the number of ways to reach a target sum using a given set of items (question types).
2Insight 2: Dynamic programming is the most efficient approach to solve this problem, as it avoids redundant calculations by storing and reusing intermediate results.
3Insight 3: We need to consider the constraint on the number of questions available for each type. We can't use more questions than available for each type.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `target = 0`. In this case, there's only one way to achieve the target (by answering no questions), which is handled by initializing `dp[0] = 1`.
Edge case 2: `types` array is empty. The loop doesn't execute, and the function returns `dp[target]`, which will be 0 if target > 0 and 1 if target == 0.
Edge case 3: The sum of all possible points from available questions is less than the target. The solution will correctly compute the number of possible ways (which will likely be zero).
Edge case 4: Large counts and marks potentially overflow int, so we need to perform operations using `long` datatype, and convert it to `int` after the modulo operation.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to take the modulo `10^9 + 7` at each step, which can lead to integer overflow and incorrect results.

✗Common mistake 2: Not correctly handling the constraint on the number of available questions for each type. Failing to prevent using more questions than available can lead to overcounting.

✗Common mistake 3: Incorrectly initializing the DP array, particularly `dp[0]`. Initializing it to 0 instead of 1 will result in an incorrect count.

✗Common mistake 4: Not accounting for the cases where the current marks is larger than the current j

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Inventory management: Determining the number of ways to fulfill a customer order given a limited stock of different products.

🏦Real-world scenario 2: Financial planning: Calculating the number of investment strategies to reach a specific financial goal, given different investment options with varying returns and limitations.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 5, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Ways to Earn Points"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?