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SubarrayProductLessThanK.java
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SubarrayProductLessThanK.java
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/* (C) 2024 YourCompanyName */
package two_pointers;
/**
* Created by gouthamvidyapradhan on 17/02/2018. Your are given an array of positive integers nums.
*
* <p>Count and print the number of (contiguous) subarrays where the product of all the elements in
* the subarray is less than k.
*
* <p>Example 1: Input: nums = [10, 5, 2, 6], k = 100 Output: 8 Explanation: The 8 subarrays that
* have product less than 100 are: [10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6]. Note
* that [10, 5, 2] is not included as the product of 100 is not strictly less than k. Note:
*
* <p>0 < nums.length <= 50000. 0 < nums[i] < 1000. 0 <= k < 10^6.
*/
import java.util.ArrayDeque;
import java.util.Queue;
public class SubarrayProductLessThanK {
public static void main(String[] args) throws Exception {
int[] A = {10, 2, 2, 5, 4, 4, 4, 3, 7, 7};
System.out.println(new SubarrayProductLessThanK().numSubarrayProductLessThanK(A, 289));
}
public int numSubarrayProductLessThanK(int[] nums, int k) {
long prod = 1;
int count = 0;
Queue<Integer> queue = new ArrayDeque<>();
for (int i = 0; i < nums.length; i++) {
if (nums[i] < k) {
count++;
if ((prod * nums[i]) < k) {
prod *= nums[i];
if (!queue.isEmpty()) {
count += (i - queue.peek());
}
} else {
while (!queue.isEmpty()) {
int last = queue.poll();
prod /= nums[last];
if ((prod * nums[i]) < k) {
prod = prod * nums[i];
if (!queue.isEmpty()) {
count += (i - queue.peek());
}
break;
}
}
}
if (queue.isEmpty()) {
prod = nums[i];
}
queue.offer(i);
} else {
queue.clear();
}
}
return count;
}
}