713. Subarray Product Less Than K

Description

Given an array of integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly 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.

Example 2:

Input: nums = [1,2,3], k = 0
Output: 0

Constraints:

1 <= nums.length <= 3 * 10⁴
1 <= nums[i] <= 1000
0 <= k <= 10⁶

Solutions

Solution 1: Two Pointers

Thinking

Count contiguous subarrays whose product is strictly less than \(k\). \(n\) can be \(3\times 10^4\), so enumerating both ends is \(O(n^2)\).

All values are positive, so the window product is monotone as the right end grows. Once it reaches \(k\), only advancing the left end restores a legal window—each pointer moves at most once.

Keep product \(p\) and left \(l\). After multiplying by \(x\), divide out \(\textit{nums}[l]\) while \(p\ge k\). Then \(r-l+1\) subarrays end at \(r\). One pass suffices.

We can use two pointers to maintain a sliding window, where the product of all elements in the window is less than \(k\).

Define two pointers \(l\) and \(r\) pointing to the left and right boundaries of the sliding window, initially \(l = r = 0\). We use a variable \(p\) to record the product of all elements in the window, initially \(p = 1\).

Each time, we move \(r\) one step to the right, adding the element \(x\) pointed to by \(r\) to the window, and update \(p = p \times x\). Then, if \(p \geq k\), we move \(l\) one step to the right in a loop and update \(p = p \div \text{nums}[l]\) until \(p < k\) or \(l \gt r\). Thus, the number of contiguous subarrays ending at \(r\) with a product less than \(k\) is \(r - l + 1\). We then add this number to the answer and continue moving \(r\) until \(r\) reaches the end of the array.

The time complexity is \(O(n)\), where \(n\) is the length of the array. The space complexity is \(O(1)\).

Python3JavaC++GoTypeScriptRustJavaScriptKotlinC#

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        ans = l = 0
        p = 1
        for r, x in enumerate(nums):
            p *= x
            while l <= r and p >= k:
                p //= nums[l]
                l += 1
            ans += r - l + 1
        return ans

class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int ans = 0, l = 0;
        int p = 1;
        for (int r = 0; r < nums.length; ++r) {
            p *= nums[r];
            while (l <= r && p >= k) {
                p /= nums[l++];
            }
            ans += r - l + 1;
        }
        return ans;
    }
}

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int ans = 0, l = 0;
        int p = 1;
        for (int r = 0; r < nums.size(); ++r) {
            p *= nums[r];
            while (l <= r && p >= k) {
                p /= nums[l++];
            }
            ans += r - l + 1;
        }
        return ans;
    }
};

func numSubarrayProductLessThanK(nums []int, k int) (ans int) {
    l, p := 0, 1
    for r, x := range nums {
        p *= x
        for l <= r && p >= k {
            p /= nums[l]
            l++
        }
        ans += r - l + 1
    }
    return
}

function numSubarrayProductLessThanK(nums: number[], k: number): number {
    const n = nums.length;
    let [ans, l, p] = [0, 0, 1];
    for (let r = 0; r < n; ++r) {
        p *= nums[r];
        while (l <= r && p >= k) {
            p /= nums[l++];
        }
        ans += r - l + 1;
    }
    return ans;
}

impl Solution {
    pub fn num_subarray_product_less_than_k(nums: Vec<i32>, k: i32) -> i32 {
        let mut ans = 0;
        let mut l = 0;
        let mut p = 1;

        for (r, &x) in nums.iter().enumerate() {
            p *= x;
            while l <= r && p >= k {
                p /= nums[l];
                l += 1;
            }
            ans += (r - l + 1) as i32;
        }

        ans
    }
}

/**
 * @param {number[]} nums
 * @param {number} k
 * @return {number}
 */
var numSubarrayProductLessThanK = function (nums, k) {
    const n = nums.length;
    let [ans, l, p] = [0, 0, 1];
    for (let r = 0; r < n; ++r) {
        p *= nums[r];
        while (l <= r && p >= k) {
            p /= nums[l++];
        }
        ans += r - l + 1;
    }
    return ans;
};

class Solution {
    fun numSubarrayProductLessThanK(nums: IntArray, k: Int): Int {
        var ans = 0
        var l = 0
        var p = 1

        for (r in nums.indices) {
            p *= nums[r]
            while (l <= r && p >= k) {
                p /= nums[l]
                l++
            }
            ans += r - l + 1
        }

        return ans
    }
}

public class Solution {
    public int NumSubarrayProductLessThanK(int[] nums, int k) {
        int ans = 0, l = 0;
        int p = 1;
        for (int r = 0; r < nums.Length; ++r) {
            p *= nums[r];
            while (l <= r && p >= k) {
                p /= nums[l++];
            }
            ans += r - l + 1;
        }
        return ans;
    }
}

713. Subarray Product Less Than K

Description

Solutions

Solution 1: Two Pointers

Comments