Minimum Deletions to Make Array Divisible

Minimum Deletions to Make Array Divisible - Problem

Array Math Sorting Heap (Priority Queue) Number Theory Hard

You are given two positive integer arrays nums and numsDivide. You can delete any number of elements from nums.

Return the minimum number of deletions such that the smallest element in nums divides all the elements of numsDivide. If this is not possible, return -1.

Note: An integer x divides y if y % x == 0.

Input & Output

Example 1 — Basic Case

$ Input: nums = [2,3,2,4,6,7,4], numsDivide = [9,3,12,6]

› Output: 2

💡 Note: We need to find the smallest element in nums that divides all elements in numsDivide. After sorting nums = [2,2,3,4,4,6,7], we test: 2 doesn't divide 9, but 3 divides all elements (9÷3=3, 3÷3=1, 12÷3=4, 6÷3=2). Since 3 is at index 2 in sorted array, we need 2 deletions.

Example 2 — No Solution

$ Input: nums = [7,21,14], numsDivide = [5,10,15]

› Output: -1

💡 Note: No element in nums can divide all elements in numsDivide. For example, 7 doesn't divide 5 (5÷7 has remainder), 14 doesn't divide 5, and 21 doesn't divide 5. Since no valid divisor exists, return -1.

Example 3 — First Element Works

$ Input: nums = [1,2,3], numsDivide = [2,4,6]

› Output: 0

💡 Note: The smallest element 1 divides all elements in numsDivide (2÷1=2, 4÷1=4, 6÷1=6). Since 1 is already the smallest, no deletions needed.

Constraints

1 ≤ nums.length, numsDivide.length ≤ 10⁵
1 ≤ nums[i], numsDivide[i] ≤ 10⁹

Visualization

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Asked in

G Google 15 f Meta 8

The key insight is that any divisor of all elements in numsDivide must also divide their GCD. Best approach computes GCD of numsDivide, sorts nums, then finds the first element that divides the GCD. Time: O(n log n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Elements

⏱️ Time: O(n log n + n × m) Space: O(1)

Sort the nums array to process elements from smallest to largest. For each element, check if it divides all elements in numsDivide. Return the count of elements to remove when we find a valid divisor.

Mathematical Optimization using GCD

⏱️ Time: O(n log n + m log(max(numsDivide))) Space: O(1)

Any number that divides all elements in numsDivide must also divide their GCD. So we first compute GCD of numsDivide, then find the smallest element in nums that divides this GCD.

Brute Force - Try All Elements — Algorithm Steps

Sort nums array in ascending order
For each element in sorted nums, check if it divides all elements in numsDivide
Return the index (number of deletions) when we find a valid divisor

Visualization

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Step-by-Step Walkthrough

Sort nums

Arrange elements from smallest to largest

Test each element

Check if current element divides all numsDivide elements

Return count

Number of elements to delete when valid divisor found

Code -

solution.c — C

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

int compare(const void *a, const void *b) {
    return (*(int*)a - *(int*)b);
}

int solution(int* nums, int numsSize, int* numsDivide, int numsDivideSize) {
    qsort(nums, numsSize, sizeof(int), compare);
    
    for (int i = 0; i < numsSize; i++) {
        int current = nums[i];
        int canDivideAll = 1;
        
        for (int j = 0; j < numsDivideSize; j++) {
            if (numsDivide[j] % current != 0) {
                canDivideAll = 0;
                break;
            }
        }
        
        if (canDivideAll) {
            return i;
        }
    }
    
    return -1;
}

int main() {
    char line[10000];
    fgets(line, sizeof(line), stdin);
    
    // Parse first 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");
    }
    
    fgets(line, sizeof(line), stdin);
    
    // Parse second array
    int numsDivide[1000];
    int numsDivideSize = 0;
    token = strtok(line, "[],\n");
    while (token != NULL) {
        if (strlen(token) > 0) {
            numsDivide[numsDivideSize++] = atoi(token);
        }
        token = strtok(NULL, "[],\n");
    }
    
    printf("%d\n", solution(nums, numsSize, numsDivide, numsDivideSize));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n log n + n × m)

O(n log n) for sorting nums, then O(n × m) for checking each element against all numsDivide elements

⚡ Linearithmic

Space Complexity

O(1)

Only using constant extra space for variables

✓ Linear Space

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Common Approaches

Brute Force - Try All Elements — Algorithm Steps

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Code -

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