2438. Range Product Queries of Powers
Explanation
To solve this problem, we first need to generate the array powers that represents the minimum number of powers of 2 to sum up to n. We then iterate through each query and calculate the product of elements in the range specified by the query in the powers array. Finally, we return the answers modulo 10^9 + 7.
- Generate the
powersarray representing the minimum powers of 2 to sum up ton. - Iterate through each query and calculate the product of elements in the specified range.
- Return the answers modulo 10^9 + 7.
Time Complexity:
Generating the powers array takes O(log n) time. Handling each query takes O(1) time. Thus, the overall time complexity is O(log n + q), where q is the number of queries.
Space Complexity:
The space complexity is O(log n) to store the powers array.
class Solution {
public int[] rangeProductQueriesOfPowers(int n, int[][] queries) {
int[] powers = new int[(int)(Math.log(n) / Math.log(2)) + 1];
powers[0] = 1;
for (int i = 1; i < powers.length; i++) {
powers[i] = 2 * powers[i - 1];
}
int[] answers = new int[queries.length];
int mod = 1000000007;
for (int i = 0; i < queries.length; i++) {
int left = queries[i][0];
int right = queries[i][1];
long product = 1;
for (int j = left; j <= right; j++) {
product = (product * powers[j]) % mod;
}
answers[i] = (int)product;
}
return answers;
}
}Code Editor (Testing phase)
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