713. Subarray Product Less Than K - Explanation

Description

You are 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

Example 3:

Input: nums = [1,1,1], k = 1

Output: 0

Constraints:

1 <= nums.length <= 30,000
1 <= nums[i] <= 1000
0 <= k <= 1,000,000

Topics

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Prerequisites

Before attempting this problem, you should be comfortable with:

Sliding Window - The optimal solution uses a variable-size window to track valid subarrays with products below the threshold
Prefix Sums - The binary search approach converts products to sums using logarithms, requiring prefix sum knowledge
Binary Search - One solution uses binary search on prefix sums to find valid subarray endpoints

1. Brute Force

Intuition

The most straightforward approach is to check every possible subarray. For each starting index, extend the subarray one element at a time while tracking the running product. If the product stays below k, count it. Once the product reaches or exceeds k, no further extensions from this starting point will work (since all elements are positive, the product only grows).

Algorithm

Initialize a counter res = 0.
For each starting index i from 0 to n - 1:
- Set curProd = 1.
- For each ending index j from i to n - 1:
  - Multiply curProd by nums[j].
  - If curProd >= k, break out of the inner loop.
  - Otherwise, increment res.
Return res.

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        n, res = len(nums), 0

        for i in range(n):
            curProd = 1
            for j in range(i, n):
                curProd *= nums[j]
                if curProd >= k:
                    break
                res += 1

        return res

public class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        int n = nums.length, res = 0;

        for (int i = 0; i < n; i++) {
            int curProd = 1;
            for (int j = i; j < n; j++) {
                curProd *= nums[j];
                if (curProd >= k) break;
                res++;
            }
        }

        return res;
    }
}

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        int n = nums.size(), res = 0;

        for (int i = 0; i < n; i++) {
            int curProd = 1;
            for (int j = i; j < n; j++) {
                curProd *= nums[j];
                if (curProd >= k) break;
                res++;
            }
        }

        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @param {number} k
     * @return {number}
     */
    numSubarrayProductLessThanK(nums, k) {
        let n = nums.length,
            res = 0;

        for (let i = 0; i < n; i++) {
            let curProd = 1;
            for (let j = i; j < n; j++) {
                curProd *= nums[j];
                if (curProd >= k) break;
                res++;
            }
        }

        return res;
    }
}

public class Solution {
    public int NumSubarrayProductLessThanK(int[] nums, int k) {
        int n = nums.Length, res = 0;

        for (int i = 0; i < n; i++) {
            long curProd = 1;
            for (int j = i; j < n; j++) {
                curProd *= nums[j];
                if (curProd >= k) break;
                res++;
            }
        }

        return res;
    }
}

func numSubarrayProductLessThanK(nums []int, k int) int {
    n := len(nums)
    res := 0

    for i := 0; i < n; i++ {
        curProd := 1
        for j := i; j < n; j++ {
            curProd *= nums[j]
            if curProd >= k {
                break
            }
            res++
        }
    }

    return res
}

class Solution {
    fun numSubarrayProductLessThanK(nums: IntArray, k: Int): Int {
        val n = nums.size
        var res = 0

        for (i in 0 until n) {
            var curProd = 1
            for (j in i until n) {
                curProd *= nums[j]
                if (curProd >= k) break
                res++
            }
        }

        return res
    }
}

class Solution {
    func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
        let n = nums.count
        var res = 0

        for i in 0..<n {
            var curProd = 1
            for j in i..<n {
                curProd *= nums[j]
                if curProd >= k { break }
                res += 1
            }
        }

        return res
    }
}

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$

2. Binary Search

Intuition

Products grow multiplicatively, which makes binary search tricky with raw values. However, taking logarithms converts products into sums: log(a * b) = log(a) + log(b). This means we can build a prefix sum of logarithms and binary search for the rightmost position where the prefix sum difference is less than log(k). For each starting index, we find how many valid ending positions exist.

Algorithm

Handle the edge case: if k <= 1, return 0 (no positive product can be less than 1 or less).
Build a prefix sum array of logarithms: logs[i+1] = logs[i] + log(nums[i]).
For each starting index i:
- Binary search for the smallest index j where logs[j] >= logs[i] + log(k).
- The count of valid subarrays starting at i is j - (i + 1).
Sum all counts and return.

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        if k <= 1:
            return 0

        n = len(nums)
        res = 0
        logs = [0] * (n + 1)
        logK = log(k)
        for i in range(n):
            logs[i + 1] = logs[i] + log(nums[i])

        for i in range(n):
            l, r = i + 1, n + 1
            while l < r:
                mid = (l + r) >> 1
                if logs[mid] < logs[i] + logK - 1e-9:
                    l = mid + 1
                else:
                    r = mid
            res += l - (i + 1)

        return res

public class Solution {
    public int numSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) return 0;
        int n = nums.length;
        double[] logs = new double[n + 1];
        double logK = Math.log(k);
        for (int i = 0; i < n; i++) {
            logs[i + 1] = logs[i] + Math.log(nums[i]);
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            int l = i + 1, r = n + 1;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (logs[mid] < logs[i] + logK - 1e-12) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            }
            res += l - (i + 1);
        }
        return res;
    }
}

class Solution {
public:
    int numSubarrayProductLessThanK(vector<int>& nums, int k) {
        if (k <= 1) return 0;
        int n = nums.size();
        vector<double> logs(n + 1, 0.0);
        double logK = log(k);
        for (int i = 0; i < n; i++) {
            logs[i + 1] = logs[i] + log(nums[i]);
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            int l = i + 1, r = n + 1;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (logs[mid] < logs[i] + logK - 1e-12) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            }
            res += l - (i + 1);
        }
        return res;
    }
};

class Solution {
    /**
     * @param {number[]} nums
     * @param {number} k
     * @return {number}
     */
    numSubarrayProductLessThanK(nums, k) {
        if (k <= 1) return 0;
        const n = nums.length;
        const logs = new Array(n + 1).fill(0);
        const logK = Math.log(k);
        for (let i = 0; i < n; i++) {
            logs[i + 1] = logs[i] + Math.log(nums[i]);
        }
        let res = 0;
        for (let i = 0; i < n; i++) {
            let l = i + 1, r = n + 1;
            while (l < r) {
                let mid = (l + r) >> 1;
                if (logs[mid] < logs[i] + logK - 1e-12) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            }
            res += l - (i + 1);
        }
        return res;
    }
}

public class Solution {
    public int NumSubarrayProductLessThanK(int[] nums, int k) {
        if (k <= 1) return 0;
        int n = nums.Length;
        double[] logs = new double[n + 1];
        double logK = Math.Log(k);
        for (int i = 0; i < n; i++) {
            logs[i + 1] = logs[i] + Math.Log(nums[i]);
        }
        int res = 0;
        for (int i = 0; i < n; i++) {
            int l = i + 1, r = n + 1;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (logs[mid] < logs[i] + logK - 1e-12) {
                    l = mid + 1;
                } else {
                    r = mid;
                }
            }
            res += l - (i + 1);
        }
        return res;
    }
}

func numSubarrayProductLessThanK(nums []int, k int) int {
    if k <= 1 {
        return 0
    }
    n := len(nums)
    logs := make([]float64, n+1)
    logK := math.Log(float64(k))
    for i := 0; i < n; i++ {
        logs[i+1] = logs[i] + math.Log(float64(nums[i]))
    }
    res := 0
    for i := 0; i < n; i++ {
        l, r := i+1, n+1
        for l < r {
            mid := (l + r) >> 1
            if logs[mid] < logs[i]+logK-1e-12 {
                l = mid + 1
            } else {
                r = mid
            }
        }
        res += l - (i + 1)
    }
    return res
}

class Solution {
    fun numSubarrayProductLessThanK(nums: IntArray, k: Int): Int {
        if (k <= 1) return 0
        val n = nums.size
        val logs = DoubleArray(n + 1)
        val logK = kotlin.math.ln(k.toDouble())
        for (i in 0 until n) {
            logs[i + 1] = logs[i] + kotlin.math.ln(nums[i].toDouble())
        }
        var res = 0
        for (i in 0 until n) {
            var l = i + 1
            var r = n + 1
            while (l < r) {
                val mid = (l + r) shr 1
                if (logs[mid] < logs[i] + logK - 1e-12) {
                    l = mid + 1
                } else {
                    r = mid
                }
            }
            res += l - (i + 1)
        }
        return res
    }
}

class Solution {
    func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
        if k <= 1 { return 0 }
        let n = nums.count
        var logs = [Double](repeating: 0, count: n + 1)
        let logK = log(Double(k))
        for i in 0..<n {
            logs[i + 1] = logs[i] + log(Double(nums[i]))
        }
        var res = 0
        for i in 0..<n {
            var l = i + 1
            var r = n + 1
            while l < r {
                let mid = (l + r) >> 1
                if logs[mid] < logs[i] + logK - 1e-12 {
                    l = mid + 1
                } else {
                    r = mid
                }
            }
            res += l - (i + 1)
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n \log n)$
Space complexity: $O(n)$

3. Sliding Window

Intuition

Since all numbers are positive, the product of a subarray increases as we add elements and decreases as we remove them. This monotonic property makes sliding window ideal. We maintain a window [l, r] where the product is less than k. When adding nums[r] causes the product to exceed or equal k, we shrink from the left until the product drops below k again. Each valid window ending at r contributes r - l + 1 new subarrays.

Algorithm

Initialize l = 0, product = 1, and res = 0.
For each r from 0 to n - 1:
- Multiply product by nums[r].
- While product >= k and l <= r, divide product by nums[l] and increment l.
- Add r - l + 1 to res (this counts all subarrays ending at r with product < k).
Return res.

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        res = 0
        l = 0
        product = 1
        for r in range(len(nums)):
            product *= nums[r]
            while l <= r and product >= k:
                product //= nums[l]
                l += 1
            res += (r - l + 1)
        return res

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

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

class Solution {
    /**
     * @param {number[]} nums
     * @param {number} k
     * @return {number}
     */
    numSubarrayProductLessThanK(nums, k) {
        let res = 0,
            l = 0,
            product = 1;
        for (let r = 0; r < nums.length; r++) {
            product *= nums[r];
            while (l <= r && product >= k) {
                product = Math.floor(product / nums[l++]);
            }
            res += r - l + 1;
        }
        return res;
    }
}

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

func numSubarrayProductLessThanK(nums []int, k int) int {
    res := 0
    l := 0
    product := 1
    for r := 0; r < len(nums); r++ {
        product *= nums[r]
        for l <= r && product >= k {
            product /= nums[l]
            l++
        }
        res += r - l + 1
    }
    return res
}

class Solution {
    fun numSubarrayProductLessThanK(nums: IntArray, k: Int): Int {
        var res = 0
        var l = 0
        var product = 1L
        for (r in nums.indices) {
            product *= nums[r]
            while (l <= r && product >= k) {
                product /= nums[l]
                l++
            }
            res += r - l + 1
        }
        return res
    }
}

class Solution {
    func numSubarrayProductLessThanK(_ nums: [Int], _ k: Int) -> Int {
        var res = 0
        var l = 0
        var product = 1
        for r in 0..<nums.count {
            product *= nums[r]
            while l <= r && product >= k {
                product /= nums[l]
                l += 1
            }
            res += r - l + 1
        }
        return res
    }
}

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

Common Pitfalls

Forgetting the Edge Case When k <= 1

When k is 0 or 1, no positive product can be less than k. Failing to handle this edge case leads to incorrect results or infinite loops. Always check if k <= 1: return 0 at the start.

Using Strict Less-Than Comparison Incorrectly

The problem asks for products strictly less than k, not less than or equal. Using product <= k instead of product < k will overcount subarrays where the product equals exactly k.

Integer Overflow in Product Calculation

When multiplying elements together, the product can quickly exceed integer bounds. In languages like Java or C++, use long instead of int for the product variable to avoid overflow issues, especially with large arrays.