Problem Description

You have two arrays of positive integers: nums and numsDivide.

Your task is to delete elements from nums so that the smallest remaining element in nums can divide every element in numsDivide. You want to minimize the number of deletions.

Here's what you need to find:

• Delete some (possibly zero) elements from nums
• After deletion, the smallest element in nums must divide all elements in numsDivide
• Return the minimum number of deletions needed
• If it's impossible (no element in nums can divide all elements in numsDivide), return -1

Remember that an integer x divides y if y % x == 0 (remainder is zero).

For example:

• If nums = [2, 3, 2, 4, 3] and numsDivide = [9, 6, 9, 3, 15]
• The smallest element that can divide all elements in numsDivide is 3 (since 3 divides 9, 6, 9, 3, and 15)
• We need to delete the two 2s from nums to make 3 the smallest element
• So the answer is 2 deletions

Intuition

The key insight is that if a number needs to divide all elements in numsDivide, it must divide their Greatest Common Divisor (GCD). This is because any common divisor of a set of numbers must also divide their GCD.

Think about it this way: if a number x divides both a and b, then x must also divide gcd(a, b). Conversely, any divisor of gcd(a, b) will divide both a and b. This property extends to multiple numbers - any divisor of the GCD of all numbers will divide each individual number.

So instead of checking if each element in nums divides every single element in numsDivide (which would be inefficient), we can:

1. First find the GCD of all elements in numsDivide
2. Then find the smallest element in nums that divides this GCD

To minimize deletions, we want to keep the smallest valid element from nums. By sorting nums first, we can iterate through elements from smallest to largest. The first element that divides the GCD is our answer - we delete everything before it.

For example, if numsDivide = [6, 12, 18], their GCD is 6. Any element from nums that we keep must divide 6 (possible values: 1, 2, 3, or 6). If nums = [4, 3, 2, 6], after sorting we get [2, 3, 4, 6]. The first element that divides 6 is 2, so we delete 0 elements.

Learn more about Math, Sorting and Heap (Priority Queue) patterns.

Solution Approach

The implementation follows these steps:

1. Calculate the GCD of numsDivide: We start by finding the GCD of all elements in numsDivide. Initialize x with the first element, then iteratively compute the GCD with each subsequent element using gcd(x, v). This gives us the number that any valid element from nums must divide.

2. Sort the nums array: By sorting nums in ascending order, we ensure that we encounter potential candidates from smallest to largest. This is crucial because we want to minimize deletions, which means keeping the smallest valid element.

3. Find the first valid divisor: Iterate through the sorted nums array with both index i and value v. For each element, check if it divides the GCD using the condition x % v == 0.

   • If we find such an element, return its index i - this represents the number of smaller elements we need to delete
   • The index directly gives us the deletion count because all elements before index i need to be removed

4. Handle the impossible case: If we iterate through the entire nums array without finding any element that divides the GCD, return -1 to indicate it's impossible.

The algorithm efficiently reduces the problem from checking divisibility against all elements in numsDivide to checking against just their GCD. The time complexity is O(n log n + m) where n is the length of nums (for sorting) and m is the length of numsDivide (for GCD calculation). The space complexity is O(1) if we sort in-place.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Example: nums = [4, 3, 6, 2, 3] and numsDivide = [12, 9, 6]

Step 1: Calculate GCD of numsDivide

• Start with x = 12 (first element)
• x = gcd(12, 9) = 3
• x = gcd(3, 6) = 3
• The GCD of all elements in numsDivide is 3

This means any element we keep from nums must divide 3.

Step 2: Sort nums array

• Original: [4, 3, 6, 2, 3]
• Sorted: [2, 3, 3, 4, 6]

Step 3: Find first valid divisor Iterate through sorted nums with index and value:

• Index 0, value 2: Check if 3 % 2 == 0 → 1 ≠ 0 (doesn't divide, continue)
• Index 1, value 3: Check if 3 % 3 == 0 → 0 == 0 (divides! Found our answer)

Result: Return index 1

This means we need to delete 1 element (the 2 at index 0) from the sorted array. After deletion, the smallest element would be 3, which divides all elements in numsDivide since it divides their GCD.

Verification:

• 12 % 3 = 0 ✓
• 9 % 3 = 0 ✓
• 6 % 3 = 0 ✓

The minimum number of deletions needed is 1.

Solution Implementation

Time and Space Complexity

Time Complexity: O(m + n log n) where n is the length of nums and m is the length of numsDivide.

• Computing the GCD of all elements in numsDivide takes O(m * log(min(numsDivide))) time, where we iterate through m elements and each GCD operation takes O(log(min(a,b))) time. Since the logarithmic factor is typically much smaller than the array size, we can simplify this to O(m).
• Sorting the nums array takes O(n log n) time.
• Iterating through the sorted nums array takes O(n) time in the worst case.
• Overall: O(m) + O(n log n) + O(n) = O(m + n log n)

Space Complexity: O(1) or O(n) depending on the sorting algorithm implementation.

• The variable x uses O(1) space.
• The sorting operation may use O(1) space if an in-place sorting algorithm like heapsort is used, or O(n) space if the implementation uses a sorting algorithm like mergesort or if Python's Timsort requires auxiliary space.
• In Python's case, the sort() method sorts in-place but may use up to O(n) auxiliary space in the worst case.
• Overall: O(n) in the worst case for Python's implementation.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Not Understanding Why GCD Works

A common misconception is thinking you need to check if an element from nums divides every single element in numsDivide. The key insight is that if a number divides the GCD of all elements in numsDivide, it will divide every element in that array. This is because the GCD is the largest common divisor, and any divisor of the GCD is also a divisor of all the original numbers.

Solution: Remember that GCD(a, b, c) divides a, b, and c. So if x divides GCD(a, b, c), then x also divides a, b, and c.

2. Modifying the Original Array

The code sorts nums in-place with nums.sort(). If the problem requires preserving the original array or if you need to track original indices, this approach will fail.

Solution: Create a copy before sorting:

sorted_nums = sorted(nums)
for index, value in enumerate(sorted_nums):
    if target_gcd % value == 0:
        return index

3. Edge Case: Single Element in numsDivide

When numsDivide has only one element, the loop for value in numsDivide[1:] won't execute, leaving target_gcd as just numsDivide[0]. While this is actually correct behavior, it might not be immediately obvious.

Solution: The current implementation handles this correctly, but for clarity, you could use:

target_gcd = numsDivide[0]
for i in range(1, len(numsDivide)):
    target_gcd = gcd(target_gcd, numsDivide[i])

4. Duplicate Values Confusion

When nums contains duplicates of the smallest valid divisor, you might think you need special handling. For example, if nums = [3, 3, 3, 6] and the smallest valid divisor is 3, all three 3's are at indices 0, 1, and 2, but returning index 0 is correct since we don't need to delete any of them.

Solution: The current approach handles this correctly - it returns the index of the first occurrence, which means all duplicates of that value remain in the array.