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.
Idea 1. Brute force, any subarray nums[i..j] can be represented by a given pair of integers 0 <= i <= j < nums.length.
Time complexity: O(n2)
Space complexity: O(1)
1 class Solution { 2 public int numSubarrayProductLessThanK(int[] nums, int k) { 3 int count = 0; 4 for(int i = 0; i < nums.length; ++i) { 5 int product = 1; 6 for(int j = i; j < nums.length; ++j) { 7 product *= nums[j]; 8 if(product < k) { 9 ++count; 10 } 11 else break; 12 } 13 } 14 return count; 15 } 16 }
Idea 2. Binary search. Taking advantage:
1. log(x*y) = log(x) + log(y)
2. positive numbers mean monoto increasing for subarray product(or log subarray sum), the prefix product/sum is sorted
reducing subarry product problem to subarray sum problem. For each index i, find the rightmost index j such that prefix[j] - nums[i-1] < Math.log(k). Note the comparison for double, reduce log(k) by 1e-9, to avoid searching the wrong half.
Take nums= {10,3,3,7,2,9,7,4,7,2,8,6,5,1,5}, k = 30 for example,
when searching the subarray {6, 5, 1, 5}, 6* 5 * 1 == 30, if not adding 1e-9, the search will move to {5} because of doble comparison, hence make logK slightly smaller, if the product is equal to k, move to the lower half.
Time complexity: O(nlogn)
Space complexity: O(n)
1 class Solution { 2 int findMaxIndex(double[] prefix, int left, int right, int k) { 3 double logK = Math.log(k); 4 int i = left, j = right; 5 while(i < j) { 6 int mid = i + (j - i) / 2; 7 if(prefix[mid] - prefix[left-1] < logK - 1e-9) i = mid + 1; 8 else j = mid; 9 } 10 return i; 11 } 12 13 public int numSubarrayProductLessThanK(int[] nums, int k) { 14 int count = 0; 15 16 double[] prefix = new double[nums.length + 1]; 17 for(int i = 1; i < prefix.length; ++i) { 18 prefix[i] = prefix[i-1] + Math.log(nums[i-1]); 19 } 20 21 for(int i = 1; i < prefix.length; ++i) { 22 int maxIndex = findMaxIndex(prefix, i, prefix.length, k); 23 count += maxIndex - i; 24 } 25 return count; 26 } 27 }
Idea 3. Slicing widnow. Keeping a max-product-windown less than k. Since the elements in the array is positive, the subarray product is a monotone increasing fuction, so we use a sliding window.
Time complexity: O(n)
Space complexity: O(1)
3.a Fixed the left point of a subarray, for each left, find the largest right so that the product of the subarray nums[left] + nums[left+1] + ... + nums[right-1] is less than k. Hence the product of each subarray starting at left and ending at rightEnd (left <= rightEnd < right) is less than k, there are right - left such subarrays. For every left, we update right and prod to maintain the invariant.
Starting with index left ( left = 0), find the first largest right so that the product * nums[right] is larger than k, add right - left to the count.
Shifting the window by moving left to the right by element, update product (product = product/nums[left]), the new subarry will start with index left = left + 1,
if the product*nums[right] is still larger than k, there are right - left subarrys with product less than k; Note: if right == left, it's empty array, no need to update product, product should remain as 1, it means nums[right] > k, we need to move right forward to maintain the invariant left <= right.
otherwise, expand right until product * nums[right] > k, continue the above process until left reaches the end of the array.
Take nums = [10, 5, 2, 100, 6], k = 100 for example,
left = 0, right = 2, product = 100, [10, 5, 2] , there are 2 (right - left) subarrays starting at 0 with product less than 100: [10], [10, 5]
left moving forward, left = 1, product = 100/nums[0] = 10, continue expand right until product = 1000 >= k, right = 3, [5, 2, 100], 2 (right - left) subarrays starting with 5: [5], [5, 2]
left moving forward, left = 2, product = 1000/nums[1] = 200, since product is already larger than k, no need to expand right, keep right= 3, 1 (right - left) subarray starting with 2: [2]
left moving forward, left = 3, right = 3, since left < right, as we define right as the excluded boundary, there is no subarry in the range (right - left) = 0, right is not in the product, we need to move right, right = 4
left moving forward, left = 4, product = 1, 1 * 6 < 100, expanding right = 5, 1 subarray starting with 6, [6]
(10)
(10, 5, 2)
(5, 2, 100)
(2, 100)
(100)
(6)
1 class Solution { 2 public int numSubarrayProductLessThanK(int[] nums, int k) { 3 int count = 0; 4 int product = 1; 5 for(int left = 0, right = 0; left < nums.length; ++left) { 6 for(;right < nums.length && product * nums[right] < k; ++right) { 7 product *= nums[right]; 8 } 9 10 count += right - left; 11 if(left < right) { 12 product = product/nums[left]; 13 } 14 else { 15 ++right; 16 } 17 } 18 return count; 19 } 20 }
3.b Fix the right point, for each right point, find the smallest left such that the product is less than k.
Shifting the window by adding a new element on the right,
if the product is less than k, then all the subarray nums[left...right] satisfy the requrement, there are right - left + 1 such subarrays ending at index right;
otherwise, shrinking the subarray by moving left index to the right until the subarray product less than k; Note if nums[right] >= k, left will move one step before right, the arry ending at right is empty with right - left + 1 = 0.
(10)
(10, 5)
(5, 2)
(100) right = left + 1, count = 0
(6)
1 class Solution { 2 public int numSubarrayProductLessThanK(int[] nums, int k) { 3 int count = 0; 4 int product = 1; 5 for(int left = 0, right = 0; right < nums.length; ++right) { 6 product *= nums[right]; 7 while(left <= right && product >= k) { 8 product /= nums[left]; 9 ++left; 10 } 11 count += right - left + 1; 12 } 13 return count; 14 } 15 }