Given a positive integer n, there exists a 0-indexed array called powers, composed of the minimum number of powers of 2 that sum to n. The array is sorted in non-decreasing order, and there is only one way to form the array.
You are also given a 0-indexed 2D integer array queries, where queries[i] = [lefti, righti]. Each queries[i] represents a query where you have to find the product of all powers[j] with lefti <= j <= righti.
Return an array answers, equal in length to queries, where answers[i] is the answer to the ith query. Since the answer to the ith query may be too large, each answers[i] should be returned modulo 109 + 7.
Example 1:
Input: n = 15, queries = [[0,1],[2,2],[0,3]] Output: [2,4,64] Explanation: For n = 15, powers = [1,2,4,8]. It can be shown that powers cannot be a smaller size. Answer to 1st query: powers[0] * powers[1] = 1 * 2 = 2. Answer to 2nd query: powers[2] = 4. Answer to 3rd query: powers[0] * powers[1] * powers[2] * powers[3] = 1 * 2 * 4 * 8 = 64. Each answer modulo 109 + 7 yields the same answer, so [2,4,64] is returned.
Example 2:
Input: n = 2, queries = [[0,0]] Output: [2] Explanation: For n = 2, powers = [2]. The answer to the only query is powers[0] = 2. The answer modulo 109 + 7 is the same, so [2] is returned.
Constraints:
1 <= n <= 1091 <= queries.length <= 1050 <= starti <= endi < powers.lengthclass Solution {
public int[] productQueries(int n, int[][] queries) {
int MOD = 1_000_000_007;
List<Integer> powers = new ArrayList<>();
for (int i = 0; n > 0; i++) {
if (n % 2 == 1)
powers.add(1 << i);
n = n >> 1;
}
int[] pdtQuery = new int[queries.length];
for(int i=0; i<queries.length; i++){
pdtQuery[i] = 1;
for(int j=queries[i][0]; j<=queries[i][1]; j++)
pdtQuery[i] = (int)((1l * pdtQuery[i] * powers.get(j)) % MOD);
}
return pdtQuery;
}
}class Solution {
public int[] productQueries(int n, int[][] queries) {
int MOD = 1_000_000_007;
List<Integer> powers = new ArrayList<>();
for (int i = 0; n > 0; i++) {
if (n % 2 == 1)
powers.add(1 << i);
n = n >> 1;
}
int m = powers.size();
int[][] prefixPdt = new int[m][m];
for (int i = 0; i < m; i++) {
long curr = 1;
for (int j = i; j < m; j++) {
curr *= powers.get(j);
prefixPdt[i][j] = (int) (curr %= MOD);
}
}
int[] pdtQuery = new int[queries.length];
int i = 0;
for (int[] q : queries)
pdtQuery[i++] = prefixPdt[q[0]][q[1]];
return pdtQuery;
}
}class Solution {
public int[] productQueries(int n, int[][] queries) {
int MOD = 1_000_000_007;
List<Integer> powers = new ArrayList<>();
for (int i = 0; n > 0; i++) {
if (n % 2 == 1)
powers.add(1 << i);
n = n >> 1;
}
int[] pdtQuery = new int[queries.length];
for(int i=0; i<queries.length; i++){
pdtQuery[i] = 1;
for(int j=queries[i][0]; j<=queries[i][1]; j++)
pdtQuery[i] = (int)((1l * pdtQuery[i] * powers.get(j)) % MOD);
}
return pdtQuery;
}
}