Number of Subarrays With LCM Equal to K - Practice Coding Problems

Number of Subarrays With LCM Equal to K - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,6,4], k = 6

› Output: 2

💡 Note: The subarrays [6] and [2,6] have LCM equal to 6. LCM of [2,6,4] = 12, LCM of [6,4] = 12, LCM of [4] = 4.

Example 2 — Single Element

$ Input: nums = [4,4], k = 1

› Output: 0

💡 Note: No subarray has LCM equal to 1. Both [4] and [4,4] have LCM = 4.

Example 3 — Multiple Matches

$ Input: nums = [3,6,2,3], k = 6

› Output: 4

💡 Note: Subarrays with LCM = 6: [6], [3,6], [6,2], [3,6,2] all have LCM equal to 6.

Constraints

1 ≤ nums.length ≤ 1000
1 ≤ nums[i], k ≤ 1000

Visualization

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Understanding the Visualization

Input Array

Array of integers and target LCM value k

Check Subarrays

Find all contiguous subarrays and calculate their LCM

Count Matches

Count subarrays where LCM equals k

Key Takeaway

🎯 Key Insight: Only elements that divide k can contribute to subarrays with LCM = k

Asked in

G Google 15 M Microsoft 12 a Amazon 8

The key insight is that for a subarray to have LCM = k, every element must be a divisor of k. The optimal approach uses incremental LCM calculation with early termination. Time: O(n²), Space: O(1)

Common Approaches

Approach	Time	Space	Notes
✓ Optimized - Incremental LCM Calculation	O(n²)	O(1)	Calculate LCM incrementally as we extend subarrays to avoid redundant computation
Brute Force - Check All Subarrays	O(n³)	O(1)	Generate all possible subarrays and calculate LCM for each

Optimized - Incremental LCM Calculation — Algorithm Steps

For each starting position i
Initialize LCM with nums[i]
Extend subarray and update LCM incrementally
Early termination when LCM > k

Visualization

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Step-by-Step Walkthrough

Start Position

Begin with each element as starting point

Extend & Calculate

Extend subarray and update LCM incrementally

Early Termination

Stop when LCM > k or element doesn't divide k

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

long long gcd(long long a, long long b) {
    while (b != 0) {
        long long temp = b;
        b = a % b;
        a = temp;
    }
    return a;
}

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

int solution(int* nums, int numsSize, int k) {
    int count = 0;
    
    for (int i = 0; i < numsSize; i++) {
        if (k % nums[i] != 0) continue;
        
        long long currentLcm = nums[i];
        if (currentLcm == k) {
            count++;
        }
        
        for (int j = i + 1; j < numsSize; j++) {
            if (k % nums[j] != 0) break;
            
            currentLcm = lcm(currentLcm, nums[j]);
            if (currentLcm > k) {
                break;
            } else if (currentLcm == k) {
                count++;
            }
        }
    }
    
    return count;
}

int main() {
    char line[1000];
    fgets(line, sizeof(line), stdin);
    
    // Parse array
    int nums[1000];
    int numsSize = 0;
    char* token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0) {
            nums[numsSize++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    int k;
    scanf("%d", &k);
    
    int result = solution(nums, numsSize, k);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops to generate subarrays, but LCM calculated incrementally

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Optimized - Incremental LCM Calculation — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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