LeetCode #2470 — MEDIUM

Number of Subarrays With LCM Equal to K

Move from brute-force thinking to an efficient approach using array strategy.

Solve on LeetCode
The Problem

Problem Statement

Given an integer array nums and an integer k, return the number of subarrays of nums where the least common multiple of the subarray's elements is k.

A subarray is a contiguous non-empty sequence of elements within an array.

The least common multiple of an array is the smallest positive integer that is divisible by all the array elements.

Example 1:

Input: nums = [3,6,2,7,1], k = 6
Output: 4
Explanation: The subarrays of nums where 6 is the least common multiple of all the subarray's elements are:
- [3,6,2,7,1]
- [3,6,2,7,1]
- [3,6,2,7,1]
- [3,6,2,7,1]

Example 2:

Input: nums = [3], k = 2
Output: 0
Explanation: There are no subarrays of nums where 2 is the least common multiple of all the subarray's elements.

Constraints:

  • 1 <= nums.length <= 1000
  • 1 <= nums[i], k <= 1000

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: Given an integer array nums and an integer k, return the number of subarrays of nums where the least common multiple of the subarray's elements is k. A subarray is a contiguous non-empty sequence of elements within an array. The least common multiple of an array is the smallest positive integer that is divisible by all the array elements.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math

Example 1

[3,6,2,7,1]
6

Example 2

[3]
2

Related Problems

  • Number of Subarrays With GCD Equal to K (number-of-subarrays-with-gcd-equal-to-k)
Step 02

Core Insight

What unlocks the optimal approach

  • The constraints on nums.length are small. It is possible to check every subarray.
  • To calculate LCM, you can use a built-in function or the formula lcm(a, b) = a * b / gcd(a, b).
  • As you calculate the LCM of more numbers, it can only become greater. Once it becomes greater than k, you know that any larger subarrays containing all the current elements will not work.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2470: Number of Subarrays With LCM Equal to K
class Solution {
    public int subarrayLCM(int[] nums, int k) {
        int n = nums.length;
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            int a = nums[i];
            for (int j = i; j < n; ++j) {
                int b = nums[j];
                int x = lcm(a, b);
                if (x == k) {
                    ++ans;
                }
                a = x;
            }
        }
        return ans;
    }

    private int lcm(int a, int b) {
        return a * b / gcd(a, b);
    }

    private int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a % b);
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(1)

Approach Breakdown

BRUTE FORCE
O(n²) time
O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED
O(n) time
O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

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