# 3183. The Number of Ways to Make the Sum ðŸ”’

## Description

You have an infinite number of coins with values 1, 2, and 6, and only 2 coins with value 4.

Given an integer n, return the number of ways to make the sum of n with the coins you have.

Since the answer may be very large, return it modulo 109 + 7.

Note that the order of the coins doesn't matter and [2, 2, 3] is the same as [2, 3, 2].

Example 1:

Input: n = 4

Output: 4

Explanation:

Here are the four combinations: [1, 1, 1, 1], [1, 1, 2], [2, 2], [4].

Example 2:

Input: n = 12

Output: 22

Explanation:

Note that [4, 4, 4] is not a valid combination since we cannot use 4 three times.

Example 3:

Input: n = 5

Output: 4

Explanation:

Here are the four combinations: [1, 1, 1, 1, 1], [1, 1, 1, 2], [1, 2, 2], [1, 4].

Constraints:

• 1 <= n <= 105

## Solutions

### Solution 1: Dynamic Programming (Complete Knapsack)

We can start by ignoring coin $4$, defining the coin array coins = [1, 2, 6], and then using the idea of the complete knapsack problem. We define $f[j]$ as the number of ways to make up amount $j$ using the first $i$ types of coins, initially $f[0] = 1$. Then, we iterate through the coin array coins, and for each coin $x$, we iterate through amounts from $x$ to $n$, updating $f[j] = f[j] + f[j - x]$.

Finally, $f[n]$ is the number of ways to make up amount $n$ using coins $1, 2, 6$. Then, if $n \geq 4$, we consider choosing one coin $4$, so the number of ways becomes $f[n] + f[n - 4]$, and if $n \geq 8$, we consider choosing two coins $4$, so the number of ways becomes $f[n] + f[n - 4] + f[n - 8]$.

Note the modulus operation for the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the amount.

• class Solution {
public int numberOfWays(int n) {
final int mod = (int) 1e9 + 7;
int[] coins = {1, 2, 6};
int[] f = new int[n + 1];
f[0] = 1;
for (int x : coins) {
for (int j = x; j <= n; ++j) {
f[j] = (f[j] + f[j - x]) % mod;
}
}
int ans = f[n];
if (n >= 4) {
ans = (ans + f[n - 4]) % mod;
}
if (n >= 8) {
ans = (ans + f[n - 8]) % mod;
}
return ans;
}
}

• class Solution {
public:
int numberOfWays(int n) {
const int mod = 1e9 + 7;
int coins[3] = {1, 2, 6};
int f[n + 1];
memset(f, 0, sizeof(f));
f[0] = 1;
for (int x : coins) {
for (int j = x; j <= n; ++j) {
f[j] = (f[j] + f[j - x]) % mod;
}
}
int ans = f[n];
if (n >= 4) {
ans = (ans + f[n - 4]) % mod;
}
if (n >= 8) {
ans = (ans + f[n - 8]) % mod;
}
return ans;
}
};

• class Solution:
def numberOfWays(self, n: int) -> int:
mod = 10**9 + 7
coins = [1, 2, 6]
f = [0] * (n + 1)
f[0] = 1
for x in coins:
for j in range(x, n + 1):
f[j] = (f[j] + f[j - x]) % mod
ans = f[n]
if n >= 4:
ans = (ans + f[n - 4]) % mod
if n >= 8:
ans = (ans + f[n - 8]) % mod
return ans


• func numberOfWays(n int) int {
const mod int = 1e9 + 7
coins := []int{1, 2, 6}
f := make([]int, n+1)
f[0] = 1
for _, x := range coins {
for j := x; j <= n; j++ {
f[j] = (f[j] + f[j-x]) % mod
}
}
ans := f[n]
if n >= 4 {
ans = (ans + f[n-4]) % mod
}
if n >= 8 {
ans = (ans + f[n-8]) % mod
}
return ans
}

• function numberOfWays(n: number): number {
const mod = 10 ** 9 + 7;
const f: number[] = Array(n + 1).fill(0);
f[0] = 1;
for (const x of [1, 2, 6]) {
for (let j = x; j <= n; ++j) {
f[j] = (f[j] + f[j - x]) % mod;
}
}
let ans = f[n];
if (n >= 4) {
ans = (ans + f[n - 4]) % mod;
}
if (n >= 8) {
ans = (ans + f[n - 8]) % mod;
}
return ans;
}