713 Subarray Product Less Than K
1. Question
Your are given an array of positive integersnums
.
Count and print the number of (contiguous) subarrays where the product of all the elements in the subarray is less thank
.
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
.
2. Implementation
(1) Two Pointers
思路: 利用滑动窗口的思想,维护满足条件的两个pointer start和end。注意的是每当我们在右边增加一个数时,我们引入了end - start + 1个新的subarray。比如假设我们原来有[1, 2], 当我们再右边增加3时,我们比原来多了3个subarray,分别是:
[3], [1,3], [2,3]
class Solution {
public int numSubarrayProductLessThanK(int[] nums, int k) {
if (k <= 1) {
return 0;
}
int prod = 1;
int res = 0;
for (int start = 0, end = 0; end < nums.length; end++) {
prod *= nums[end];
while (prod >= k) {
prod /= nums[start];
++start;
}
res += end - start + 1;
}
return res;
}
}
3. Time & Space Complexity
Two Pointers: 时间复杂度O(n), 空间复杂度O(1)
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