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Formatted question description: https://leetcode.ca/all/1735.html

1735. Count Ways to Make Array With Product

Level

Hard

Description

You are given a 2D integer array, queries. For each queries[i], where queries[i] = [n_i, k_i], find the number of different ways you can place positive integers into an array of size ni such that the product of the integers is k_i. As the number of ways may be too large, the answer to the i-th query is the number of ways modulo 10^9 + 7.

Return an integer array answer where answer.length == queries.length, and answer[i] is the answer to the i-th query.

Example 1:

Input: queries = [2,6,5,1,73,660]

Output: [4,1,50734910]

Explanation: Each query is independent.

Example 2:

Input: queries = [[1,1],[2,2],[3,3],[4,4],[5,5]]

Output: [1,2,3,10,5]

Constraints:

  • 1 <= queries.length <= 10^4
  • 1 <= n_i, k_i <= 10^4

Solution

First, get all combination numbers for all possible values of n and k. Next, for each query (n, k), do prime factorization for k and calculate the number of ways using the combination numbers and the prime factorization results of k.

  • class Solution {
    int[] primes = new int[14];
    int[] exponents = new int[14];
    int index = 0;
    long[][] combinations = new long[11000][20];
    static final int MODULO = 1000000007;

    public int[] waysToFillArray(int[][] queries) {
        for (int i = 0; i < 10500; i++) {
            combinations[i][0] = 1;
            if (i < 15)
                combinations[i][i] = 1;
            int end = Math.min(i, 15);
            for (int j = 1; j < end; j++)
                combinations[i][j] = (combinations[i - 1][j - 1] + combinations[i - 1][j]) % MODULO;
        }
        int queriesCount = queries.length;
        int[] ways = new int[queriesCount];
        for (int i = 0; i < queriesCount; i++) {
            int n = queries[i][0], k = queries[i][1];
            long way = 1;
            fact(k);
            for (int j = 1; j <= index; j++)
                way = way * combinations[n + exponents[j] - 1][exponents[j]] % MODULO;
            ways[i] = (int) way;
        }
        return ways;
    }

    public void fact(int n) {
        index = 0;
        for (int i = 2; i * i <= n; i++) {
            if (n % i == 0) {
                index++;
                primes[index] = i;
                exponents[index] = 0;
                while (n % i == 0) {
                    exponents[index]++;
                    n /= i;
                }
            }
        }
        if (n != 1) {
            index++;
            primes[index] = n;
            exponents[index] = 1;
        }
    }
}

  • int N = 10020;
int MOD = 1e9 + 7;
long f[10020];
long g[10020];
vector<int> p[10020];

long qmi(long a, long k, long p) {
    long res = 1;
    while (k != 0) {
        if ((k & 1) == 1) {
            res = res * a % p;
        }
        k >>= 1;
        a = a * a % p;
    }
    return res;
}

int init = []() {
    f[0] = 1;
    g[0] = 1;
    for (int i = 1; i < N; ++i) {
        f[i] = f[i - 1] * i % MOD;
        g[i] = qmi(f[i], MOD - 2, MOD);
        int x = i;
        for (int j = 2; j <= x / j; ++j) {
            if (x % j == 0) {
                int cnt = 0;
                while (x % j == 0) {
                    ++cnt;
                    x /= j;
                }
                p[i].push_back(cnt);
            }
        }
        if (x > 1) {
            p[i].push_back(1);
        }
    }
    return 0;
}();

int comb(int n, int k) {
    return (f[n] * g[k] % MOD) * g[n - k] % MOD;
}

class Solution {
public:
    vector<int> waysToFillArray(vector<vector<int>>& queries) {
        vector<int> ans;
        for (auto& q : queries) {
            int n = q[0], k = q[1];
            long long t = 1;
            for (int x : p[k]) {
                t = t * comb(x + n - 1, n - 1) % MOD;
            }
            ans.push_back(t);
        }
        return ans;
    }
};

  • N = 10020
MOD = 10**9 + 7
f = [1] * N
g = [1] * N
p = defaultdict(list)
for i in range(1, N):
    f[i] = f[i - 1] * i % MOD
    g[i] = pow(f[i], MOD - 2, MOD)
    x = i
    j = 2
    while j <= x // j:
        if x % j == 0:
            cnt = 0
            while x % j == 0:
                cnt += 1
                x //= j
            p[i].append(cnt)
        j += 1
    if x > 1:
        p[i].append(1)


def comb(n, k):
    return f[n] * g[k] * g[n - k] % MOD


class Solution:
    def waysToFillArray(self, queries: List[List[int]]) -> List[int]:
        ans = []
        for n, k in queries:
            t = 1
            for x in p[k]:
                t = t * comb(x + n - 1, n - 1) % MOD
            ans.append(t)
        return ans


  • const n = 1e4 + 20
const mod = 1e9 + 7

var f = make([]int, n)
var g = make([]int, n)
var p = make([][]int, n)

func qmi(a, k, p int) int {
	res := 1
	for k != 0 {
		if k&1 == 1 {
			res = res * a % p
		}
		k >>= 1
		a = a * a % p
	}
	return res
}

func init() {
	f[0], g[0] = 1, 1
	for i := 1; i < n; i++ {
		f[i] = f[i-1] * i % mod
		g[i] = qmi(f[i], mod-2, mod)
		x := i
		for j := 2; j <= x/j; j++ {
			if x%j == 0 {
				cnt := 0
				for x%j == 0 {
					cnt++
					x /= j
				}
				p[i] = append(p[i], cnt)
			}
		}
		if x > 1 {
			p[i] = append(p[i], 1)
		}
	}
}

func comb(n, k int) int {
	return (f[n] * g[k] % mod) * g[n-k] % mod
}

func waysToFillArray(queries [][]int) (ans []int) {
	for _, q := range queries {
		n, k := q[0], q[1]
		t := 1
		for _, x := range p[k] {
			t = t * comb(x+n-1, n-1) % mod
		}
		ans = append(ans, t)
	}
	return
}

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