What is the time complexity of Number of Dice Rolls With Target Sum?

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

What is the space complexity of Number of Dice Rolls With Target Sum?

The space complexity is O(target).

What data structures are used to solve Number of Dice Rolls With Target Sum?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Number of Dice Rolls With Target Sum?

Insight 1: Dynamic programming is well-suited for this problem because the number of ways to reach a certain sum with a certain number of dice can be built upon the number of ways to reach smaller sums with fewer dice. Insight 2: We can optimize space by using a rolling array technique to store only the results from the previous iteration, rather than a full 2D table.

Which companies ask LeetCode 1155. Number of Dice Rolls With Target Sum in interviews?

This problem is frequently asked at StackAdapt.

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Number of Dice Rolls With Target Sum - LeetCode 1155 Solution

Number of Dice Rolls With Target Sum - Complete Solution Guide

Number of Dice Rolls With Target Sum is LeetCode problem 1155, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You have n dice, and each dice has k faces numbered from 1 to k . Given three integers n , k , and target , return the number of possible ways (out of the k n total ways) to roll the dice, so the sum of the face-up numbers equals target . Since the answer may be too large, return it modulo 10 9 + 7 . Example 1: Input: n = 1, k = 6, target = 3 Output: 1 Explanation: You throw one die with 6 faces. There is only one way to get a sum of 3. Example 2: Input: n = 2, k = 6, target = 7 Output: 6 Explan

Detailed Explanation

The problem asks us to find the number of ways to roll 'n' dice, each with 'k' faces (numbered 1 to k), such that the sum of the face-up numbers equals 'target'. Since the number of ways can be very large, we need to return the result modulo 10^9 + 7. For example, if we have 2 dice with 6 faces each, and the target sum is 7, then the possible combinations are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1), so the answer is 6.

Solution Approach

The provided solution uses dynamic programming with space optimization. It builds a table `dp` where `dp[i]` represents the number of ways to achieve a sum of `i` using a certain number of dice. The algorithm iterates through each die. For each die, it updates the `dp` array based on the possible outcomes of rolling that die. The space optimization is achieved by using a `new_dp` array that is updated in each iteration and then assigned to `dp`, effectively reusing the same space. The number of ways is accumulated by summing the number of ways to reach a sum 'j-1' from the previous dice and adding them to the current total. The 'rolling' feature ensures that results from the *previous* dice roll only are used, and sums greater than *k* are removed.

Step-by-Step Algorithm

Step 1: Initialize a `dp` array of size `target + 1` with all elements set to 0. `dp[0]` is set to 1, representing the base case where a sum of 0 can be achieved with 0 dice (or before any dice are rolled).
Step 2: Iterate `n` times, where `n` is the number of dice.
Step 3: In each iteration, create a `new_dp` array of size `target + 1` and initialize all elements to 0.
Step 4: Iterate from `j = 1` to `target`. For each `j`, calculate the number of ways to achieve the sum `j` by considering the possible values of the current die. This is done by summing the values of `dp[j-1]`. Accumulating `dp[j-1]` in `new_dp[j]`.
Step 5: If `j > k` (where k is the number of faces on the die), it means we have considered sums that are impossible with the current die (since the minimum value is 1, sums can be greater than k). We remove the contribution from `dp[j - k - 1]` because `j - (k+1) + k >= j` implies that value can be reached. We do the modulus and addition to ensure no underflow errors.
Step 6: After the inner loop completes, update `dp` to `new_dp` to prepare for the next iteration.
Step 7: After `n` iterations, the value of `dp[target]` will contain the number of ways to achieve the target sum using `n` dice. Return `dp[target]`.

Key Insights

Insight 1: Dynamic programming is well-suited for this problem because the number of ways to reach a certain sum with a certain number of dice can be built upon the number of ways to reach smaller sums with fewer dice.
Insight 2: We can optimize space by using a rolling array technique to store only the results from the previous iteration, rather than a full 2D table.
Insight 3: The modulo operation is crucial to prevent integer overflow as the number of combinations can grow very quickly.

Complexity Analysis

Time Complexity: O(n*target)

Space Complexity: O(target)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: StackAdapt.

Number of Dice Rolls With Target Sum | LeetCode 1155

Medium

Dynamic Programming

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

Solution Code

Available in:

class Solution {
    public int numRollsToTarget(int n, int k, int target) {
        int MOD = (int)1e9 + 7;

        int[] dp = new int[target + 1];
        dp[0] = 1;

        for (int i = 0; i < n; i++) {
            int[] new_dp = new int[target + 1];
            for (int j = 1; j <= target; j++) {
                new_dp[j] = (new_dp[j - 1] + dp[j - 1]) % MOD;
                if (j > k) {
                    new_dp[j] = (new_dp[j] - dp[j - k - 1] + MOD) % MOD;
                }
            }
            dp = new_dp;
        }

        return dp[target];
    }
}

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Problem Info

Description

Problem 1155: Number of Dice Rolls With Target Sum - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

StackAdapt

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:1155

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

7 steps

1
Step 1: Initialize a `dp` array of size `target + 1` with all elements set to 0. `dp[0]` is set to 1, representing the base case where a sum of 0 can be achieved with 0 dice (or before any dice are rolled).
2
Step 2: Iterate `n` times, where `n` is the number of dice.
3
Step 3: In each iteration, create a `new_dp` array of size `target + 1` and initialize all elements to 0.
4
Step 4: Iterate from `j = 1` to `target`. For each `j`, calculate the number of ways to achieve the sum `j` by considering the possible values of the current die. This is done by summing the values of `dp[j-1]`. Accumulating `dp[j-1]` in `new_dp[j]`.
5
Step 5: If `j > k` (where k is the number of faces on the die), it means we have considered sums that are impossible with the current die (since the minimum value is 1, sums can be greater than k). We remove the contribution from `dp[j - k - 1]` because `j - (k+1) + k >= j` implies that value can be reached. We do the modulus and addition to ensure no underflow errors.
6
Step 6: After the inner loop completes, update `dp` to `new_dp` to prepare for the next iteration.
7
Step 7: After `n` iterations, the value of `dp[target]` will contain the number of ways to achieve the target sum using `n` dice. Return `dp[target]`.

💡

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(n*target)

💾

Space Complexity

O(target)

📝

Why?

The nested loops determine the time complexity. The outer loop iterates 'n' times, and the inner loop iterates 'target' times in each iteration of the outer loop. The `dp` arrays being of size 'target + 1' determines space complexity.

📖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

2D Dynamic Programming

A more intuitive approach is to use a 2D DP table where `dp[i][j]` represents the number of ways to achieve a sum of `j` using `i` dice. The time complexity remains the same, but the space complexity increases.

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

Trade-offs: This approach uses more space (O(n*target)) but can be easier to understand than the rolling array optimization. It is helpful for visualizing the dynamic programming transitions.

✨

Key Insights

1Insight 1: Dynamic programming is well-suited for this problem because the number of ways to reach a certain sum with a certain number of dice can be built upon the number of ways to reach smaller sums with fewer dice.
2Insight 2: We can optimize space by using a rolling array technique to store only the results from the previous iteration, rather than a full 2D table.
3Insight 3: The modulo operation is crucial to prevent integer overflow as the number of combinations can grow very quickly.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: `n = 0` or `target = 0`. The code correctly handles this since dp[0] is initialized to 1, and the loops will execute appropriately.
Edge case 2: `target` is less than `n`. In such cases, it's impossible to reach the target, and the code will return 0, due to the initial setup of dp array as all 0 except dp[0]=1.
Edge case 3: `target` is greater than `n * k`. Similarly, it is impossible to reach the target.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to apply the modulo operation after each addition. This can lead to integer overflow and incorrect results.

✗Common mistake 2: Incorrectly initializing the base case `dp[0] = 1`. This is crucial for the dynamic programming recurrence to work correctly.

✗Common mistake 3: Not handling the case where `j > k` correctly. Failing to subtract the invalid contribution would result in an incorrect answer.

✅

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: This dynamic programming approach can be applied to other counting problems where the solution can be built recursively from smaller subproblems, such as coin change or knapsack problems.

🏦Real-world scenario 2: It can also be used in situations involving probabilities, such as calculating the probability of a certain outcome in a series of independent events with a finite number of possibilities.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

👨‍💻 From Experience

This problem pattern appears frequently in real codebases. I've used similar logic when building production systems.

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 15, 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

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.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 Dice Rolls With Target Sum"?

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

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