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SubarrayProductLessThanK.java
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SubarrayProductLessThanK.java
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package Leetcode;
/**
* @author kalpak
*
* Your are given an array of positive integers nums.
*
* Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less than k.
*
* 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:
*
* 0 < nums.length <= 50000.
* 0 < nums[i] < 1000.
* 0 <= k < 10^6.
*
*/
public class SubarrayProductLessThanK {
public static int numSubarrayProductLessThanK(int[] nums, int k) {
/**
* The idea is always keep an max-product-window less than K;
* Every time add a new number on the right(j),
* reduce numbers on the left(i), until the subarray product fit less than k again, (subarray could be empty);
* Each step introduces x new subarrays, where x is the size of the current window (j - i + 1);
* Say now we have {1,2,3} and add {4} into it. Apparently, the new subarray introduced here are:
* {1,2,3,4}, {2,3,4}, {3,4}, {4}, which is the number of elements in the new list.
* If we also remove some at the left, say we we remove 1, then subarrays are:
* {2,3,4}, {3,4}, {4}. It is easy to get the result is j - i + 1.
*/
if (nums == null || nums.length == 0 || k == 0)
return 0;
int result = 0;
int product = 1;
for(int i = 0, j = 0; j < nums.length; j++)
{
product *= nums[j];
while(i <= j && product >= k)
{
product /= nums[i];
i++;
}
result += j - i + 1;
}
return result;
}
public static void main(String[] args) {
int[] arr = new int[]{10,5,2,6};
System.out.println(numSubarrayProductLessThanK(arr, 100));
}
}