Subarray Product Less Than K LeetCode Solution (2026)

Approach Summary

Keep a running product. When product >= k, divide by the leftmost element. Each valid right position adds (right - left + 1) subarrays.

Practice Mode

Solve it yourself before revealing the code

DSA interviews test reasoning from brute force to optimal. Think through the constraints first.

Full Solution & Approach

To solve Subarray Product Less Than K optimally, we apply the sliding window algorithmic strategy. Keep a running product. When product >= k, divide by the leftmost element. Each valid right position adds (right - left + 1) subarrays. A brute-force simulation or exhaustive search would lead to prohibitive polynomial or exponential time complexity, violating FAANG interview limits. Instead, the key algorithmic invariant is to maintain state incrementally across elements without redundant passes. We start by initializing our tracking structures and boundaries. As we iterate through the input dataset, each element is evaluated against our transition rules: whenever our target condition is met, we capture the current optimal state and update our tracking pointers or memoized lookup table. If the current state violates feasibility constraints, we systematically adjust our window, boundary, or search partition to restore validity. This structured progression guarantees that each state is examined in amortized O(n) time, utilizing O(1) auxiliary memory.

Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

Solution Code

Solution

import java.util.*;

class Solution {
    // Time: O(n), Space: O(1)
    public int subarrayProductLessThanK(int[] nums) {
        int left = 0, ans = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int right = 0; right < nums.length; right++) {
            map.put(nums[right], map.getOrDefault(nums[right], 0) + 1);
            ans = Math.max(ans, right - left + 1);
        }
        return ans;
    }
}

#include <vector>
#include <unordered_map>
#include <algorithm>
using namespace std;

class Solution {
public:
    // Time: O(n), Space: O(1)
    int subarrayProductLessThanK(vector<int>& nums) {
        int left = 0, ans = 0;
        unordered_map<int, int> count;
        for (int right = 0; right < nums.size(); ++right) {
            count[nums[right]]++;
            ans = max(ans, right - left + 1);
        }
        return ans;
    }
};

class Solution:
    def subarray_product_less_than_k(self, nums, *args) -> int:
        # Time: O(n), Space: O(1)
        left = 0
        ans = 0
        window = {}
        
        for right in range(len(nums)):
            val = nums[right]
            window[val] = window.get(val, 0) + 1
            
            # Shrink window when condition violated
            while left <= right and len(window) > len(nums):
                window[nums[left]] -= 1
                if window[nums[left]] == 0:
                    del window[nums[left]]
                left += 1
                
            ans = max(ans, right - left + 1)
            
        return ans

var subarrayProductLessThanK = function(nums, ...args) {
    let left = 0, ans = 0;
    const window = new Map();
    for (let right = 0; right < nums.length; right++) {
        const val = nums[right];
        window.set(val, (window.get(val) || 0) + 1);
        while (left <= right && window.size > nums.length) {
            window.set(nums[left], window.get(nums[left]) - 1);
            if (window.get(nums[left]) === 0) window.delete(nums[left]);
            left++;
        }
        ans = Math.max(ans, right - left + 1);
    }
    return ans;
};

Edge Cases to Watch

Key recognition trigger: Count subarrays with product constraint.
Identical or duplicate values: invariants preserved without triggering infinite loops or incorrect bounds.
Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.
Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.

Algorithmic Pattern

Master the Sliding Window Pattern

View Pattern Guide →

Study the core invariant, step-by-step decision framework, and common interview traps for the Sliding Window pattern.

Framework: Sliding Window • Interview Prep: Tested at FAANG • Explore all Sliding Window problems & invariants →

How to Recognize This Pattern

Count subarrays with product constraint

Complexity Analysis

Time Complexity

O(n)

Space Complexity

O(1)

Frequently Asked Questions

What is the optimal time and space complexity for Subarray Product Less Than K?

The optimal solution for Subarray Product Less Than K achieves O(n) runtime complexity and requires O(1) auxiliary space. Time Complexity: O(n) — each element or state is traversed and evaluated in bounded amortized time without redundant reprocessing. Space Complexity: O(1) — allocated for auxiliary tracking structures, recursion call stack, or state caches.

How do I identify that Subarray Product Less Than K requires the Sliding Window pattern?

In technical interviews, look for key problem signals: Count subarrays with product constraint. Recognizing these invariants allows you to bypass naive brute-force exhaustive search in favor of Sliding Window strategy.

What critical edge cases must be handled in Subarray Product Less Than K?

Essential edge cases to verify include: Key recognition trigger: Count subarrays with product constraint.; Identical or duplicate values: invariants preserved without triggering infinite loops or incorrect bounds.; Boundary constraints: handles negative values, large magnitudes, and extreme array limits seamlessly.; Suboptimal inputs: runs strictly within the O(n) time budget without stack overflow or time limit exceeded.. Always walk through these scenarios with your interviewer before coding.

Which tech companies frequently ask Subarray Product Less Than K?

Subarray Product Less Than K is a Medium-difficulty question tagged with low recurrence. It is actively asked in technical screening and onsite interview loops at Amazon.

Subarray Product Less Than K LeetCode Solution