Minimum Product of K Consecutive
Find the minimum product of any k consecutive elements.
Use a sliding window of size k. Initialize the window state with the first k elements. Slide the window one position at a time, updating the state efficiently. Track the minimum across all windows.
The sliding window technique processes each element exactly twice (once when it enters, once when it leaves), giving O(n) time complexity with O(1) extra space.
Edge cases include empty arrays (return appropriate default like 0), k larger than array length (return default), and single-element arrays where k=1.
Maintain the running product of each k-length window. Track the minimum product across all windows.
Example Input & Output
Algorithm Flow

Solution Approach
Initialize the sliding window with the first k elements. Compute the initial minimum. For each position i from k to n-1, update the window by removing element i-k and adding element i. Update the result if the new window value is better.
Best Answers
class Solution {
public int solution(int[] nums, int k) {
if (nums.length==0) return 0;
int w=0;for(int i=0;i<k;i++)w+=nums[i];
int b=w;
for(int i=k;i<nums.length;i++){w+=nums[i]-nums[i-k];b=Math.min(b,w);}
return b;
}
}Comments (0)
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