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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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

✓ Brute Force - Check All Subarrays

⏱️ Time: O(n³) Space: O(1)

For each starting position, extend the subarray one element at a time and calculate the LCM. Count how many subarrays have LCM equal to k.

Optimized - Incremental LCM Calculation

⏱️ Time: O(n²) Space: O(1)

For each starting position, maintain the running LCM as we extend the subarray. Use the property that if current LCM > k, extending further will only increase LCM.

Brute Force - Check All Subarrays — Algorithm Steps

For each starting index i
For each ending index j >= i
Calculate LCM of subarray nums[i:j+1]
If LCM equals k, increment counter

Visualization

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Step-by-Step Walkthrough

Generate Subarrays

Create all possible contiguous subarrays

Calculate LCM

Find LCM for each subarray

Count Matches

Count subarrays where LCM equals 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++) {
        for (int j = i; j < numsSize; j++) {
            long long currentLcm = nums[i];
            for (int x = i + 1; x <= j; x++) {
                currentLcm = lcm(currentLcm, nums[x]);
                if (currentLcm > k) break;
            }
            if (currentLcm == k) {
                count++;
            }
        }
    }
    
    return count;
}

void parseArray(const char* str, int* arr, int* size) {
    *size = 0;
    const char* p = str;
    while (*p && *p != '[') p++;
    if (*p == '[') p++;
    while (*p && *p != ']') {
        while (*p == ' ' || *p == ',') p++;
        if (*p == ']' || *p == '\0') break;
        arr[(*size)++] = (int)strtol(p, (char**)&p, 10);
    }
}

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(n³)

Nested loops O(n²) to generate subarrays, O(n) to calculate LCM for each

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Number of Subarrays With LCM Equal to K - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Check All Subarrays — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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