LeetCode-in-Java
1735. Count Ways to Make Array With Product
Hard
You are given a 2D integer array, queries. For each queries[i], where queries[i] = [ni, ki], 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 ki. As the number of ways may be too large, the answer to the ith query is the number of ways modulo 109 + 7.
Return an integer array answer where answer.length == queries.length, and answer[i] is the answer to the ith 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 <= 1041 <= ni, ki <= 104
Solution
import java.util.ArrayList;
import java.util.List;
public class Solution {
private static final int MOD = 1_000_000_007;
private long[][] tri;
private List<Integer> primes;
public int[] waysToFillArray(int[][] queries) {
int len = queries.length;
int[] res = new int[len];
primes = getPrimes(100);
tri = getTri(10015, 15);
for (int i = 0; i < len; i++) {
res[i] = calculate(queries[i][0], queries[i][1]);
}
return res;
}
private List<Integer> getPrimes(int limit) {
boolean[] notPrime = new boolean[limit + 1];
List<Integer> res = new ArrayList<>();
for (int i = 2; i <= limit; i++) {
if (!notPrime[i]) {
res.add(i);
for (int j = i * i; j <= limit; j += i) {
notPrime[j] = true;
}
}
}
return res;
}
private long[][] getTri(int m, int n) {
long[][] res = new long[m + 1][n + 1];
for (int i = 0; i <= m; i++) {
res[i][0] = 1;
for (int j = 1; j <= Math.min(n, i); j++) {
res[i][j] = (res[i - 1][j - 1] + res[i - 1][j]) % MOD;
}
}
return res;
}
private int calculate(int n, int target) {
long res = 1;
for (int prime : primes) {
if (prime > target) {
break;
}
int cnt = 0;
while (target % prime == 0) {
cnt++;
target /= prime;
}
res = (res * tri[cnt + n - 1][cnt]) % MOD;
}
return target > 1 ? (int) (res * n % MOD) : (int) res;
}
}