Number of Dice Rolls With Target Sum - LeetCode javascript solutions

Problem

You have n dice, and each die 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

Solution

/**
 * @param {number} n
 * @param {number} k
 * @param {number} target
 * @return {number}
 */
var numRollsToTarget = function(n, k, target) {
    var dp = Array(n + 1).fill(0).map(() => ({}));
    return helper(n, k, target, dp);
};

var helper = function(n, k, target, dp) {
    if (dp[n][target] !== undefined) return dp[n][target];
    if (n === 0 && target === 0) return 1;
    if (n <= 0 || target <= 0) return 0;
    var res = 0;
    var mod = Math.pow(10, 9) + 7;
    for (var i = 1; i <= k; i++) {
        if (target < i) break;
        res += helper(n - 1, k, target - i, dp);
        res %= mod;
    }
    dp[n][target] = res;
    return res;
};

Explain:

nope.

Complexity:

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