LeetCode #1155 — MEDIUM

Number of Dice Rolls With Target Sum

Move from brute-force thinking to an efficient approach using dynamic programming strategy.

The Problem

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 kⁿ 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 10⁹ + 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
Explanation: You throw two dice, each with 6 faces.
There are 6 ways to get a sum of 7: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1.

Example 3:

Input: n = 30, k = 30, target = 500
Output: 222616187
Explanation: The answer must be returned modulo 10⁹ + 7.

Constraints:

1 <= n, k <= 30
1 <= target <= 1000

Step 01

Brute Force Baseline

Problem summary: 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 kn 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 109 + 7.

Baseline thinking

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

Pattern signal: Dynamic Programming

Example 1

1
6
3

Example 2

2
6
7

Example 3

30
30
500

Core Insight

What unlocks the optimal approach

Use dynamic programming. The states are how many dice are remaining, and what sum total you have rolled so far.

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

Upper-end input sizes

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 #1155: Number of Dice Rolls With Target Sum
class Solution {
    public int numRollsToTarget(int n, int k, int target) {
        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) {
            for (int j = 1; j <= Math.min(target, i * k); ++j) {
                for (int h = 1; h <= Math.min(j, k); ++h) {
                    f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod;
                }
            }
        }
        return f[n][target];
    }
}

// Accepted solution for LeetCode #1155: Number of Dice Rolls With Target Sum
func numRollsToTarget(n int, k int, target int) int {
	const mod int = 1e9 + 7
	f := make([][]int, n+1)
	for i := range f {
		f[i] = make([]int, target+1)
	}
	f[0][0] = 1
	for i := 1; i <= n; i++ {
		for j := 1; j <= min(target, i*k); j++ {
			for h := 1; h <= min(j, k); h++ {
				f[i][j] = (f[i][j] + f[i-1][j-h]) % mod
			}
		}
	}
	return f[n][target]
}

# Accepted solution for LeetCode #1155: Number of Dice Rolls With Target Sum
class Solution:
    def numRollsToTarget(self, n: int, k: int, target: int) -> int:
        f = [[0] * (target + 1) for _ in range(n + 1)]
        f[0][0] = 1
        mod = 10**9 + 7
        for i in range(1, n + 1):
            for j in range(1, min(i * k, target) + 1):
                for h in range(1, min(j, k) + 1):
                    f[i][j] = (f[i][j] + f[i - 1][j - h]) % mod
        return f[n][target]

// Accepted solution for LeetCode #1155: Number of Dice Rolls With Target Sum
impl Solution {
    pub fn num_rolls_to_target(n: i32, k: i32, target: i32) -> i32 {
        let _mod = 1_000_000_007;
        let n = n as usize;
        let k = k as usize;
        let target = target as usize;
        let mut f = vec![vec![0; target + 1]; n + 1];
        f[0][0] = 1;

        for i in 1..=n {
            for j in 1..=target.min(i * k) {
                for h in 1..=j.min(k) {
                    f[i][j] = (f[i][j] + f[i - 1][j - h]) % _mod;
                }
            }
        }

        f[n][target]
    }
}

// Accepted solution for LeetCode #1155: Number of Dice Rolls With Target Sum
function numRollsToTarget(n: number, k: number, target: number): number {
    const f = Array.from({ length: n + 1 }, () => Array(target + 1).fill(0));
    f[0][0] = 1;
    const mod = 1e9 + 7;
    for (let i = 1; i <= n; ++i) {
        for (let j = 1; j <= Math.min(i * k, target); ++j) {
            for (let h = 1; h <= Math.min(j, k); ++h) {
                f[i][j] = (f[i][j] + f[i - 1][j - h]) % 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 × k × target)

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.

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.