Formatted question description: https://leetcode.ca/all/713.html
Level
Medium
Description
You 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
.
Solution
Initialize product = 1
, count = 0
, start = 0
and end = 0
. Loop over nums
using end
as the index. Each time let product *= nums[end]
. If product >= k
, then do product /= nums[start++]
until product < k
. Then calculate the length of the current subarray, which is end - start + 1
. The length of the current subarray is also the number of subarrays with product less than k
and ends at index end
. Add the length of the current subarray to count
. Finally, return count
.
Java Code
public class Subarray_Product_Less_Than_K {
public static void main(String[] args) {
Subarray_Product_Less_Than_K out = new Subarray_Product_Less_Than_K();
Solution s = out.new Solution();
System.out.println(s.numSubarrayProductLessThanK(new int[]{10, 5, 2, 6}, 100));
}
class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k == 0) {
return 0;
}
// log transformation: log(a*b) = log(a) + log(b)
double logk = Math.log(k);
double[] prefix = new double[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
prefix[i+1] = prefix[i] + Math.log(nums[i]);
}
int count = 0;
for (int i = 0; i < prefix.length; i++) {
int lo = i + 1, hi = prefix.length;
while (lo < hi) {
int mi = lo + (hi - lo) / 2;
if (prefix[mi] < prefix[i] + logk - 1e-9) {
lo = mi + 1;
} else {
hi = mi;
}
}
count += lo - i - 1;
}
return count;
}
}
// this is just a flip, instead of compare for <k, to compare >=k then count.
// I don't like this one, seems hack. Will still fail according to input
// what if test case is [9,9,9, ... ,9] (many) and k=1
class Solution_ACed {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k < 2) {
return 0;
}
int result = 0;
int product = 1;
for (int i = 0, right = 0; right < nums.length; right++) {
product *= nums[right];
while (i < nums.length && product >= k) {
product /= nums[i++];
}
result += right - i + 1;
}
return result;
}
}
/*
failed by:
nums=[1,1,1,1,1, ... ,1] (billion 1s)
k=9
*/
class Solution_wrong {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (nums == null || nums.length == 0) {
return 0;
}
int count = 0;
int left = 0;
// basic idea, moving right to right direction, if current multiple is >= k , then no need to go right further
// assumption: all positive
// variation of this question, what if there is 0, or negative numbers
while (left < nums.length) {
int current = 1;
int right = left;
// @note: possible overflow from * operation
while (right < nums.length && current * nums[right] < k) { // will stop if current >= k
count++;
current *= nums[right];
right++;
}
left++;
}
return count;
}
}
}
Java
class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (nums == null || nums.length == 0 || k <= 1)
return 0;
int product = 1;
int count = 0;
int start = 0, end = 0;
int length = nums.length;
while (end < length) {
product *= nums[end];
while (product >= k) {
product /= nums[start];
start++;
}
count += end - start + 1;
end++;
}
return count;
}
}