Problem Description

You are given an integer array nums and an integer k. You can perform operations on the array to reduce its length.

In each operation, you can:

• Select two adjacent elements in the array (let's call them x and y)
• Check if their product x * y is less than or equal to k
• If yes, replace both elements with a single element whose value is x * y

For example, with array [1, 2, 2, 3] and k = 5:

• You could merge the first two elements 1 and 2 (since 1 * 2 = 2 ≤ 5) to get [2, 2, 3]
• Or you could merge the middle two elements 2 and 2 (since 2 * 2 = 4 ≤ 5) to get [1, 4, 3]
• But you cannot merge 2 and 3 (since 2 * 3 = 6 > 5)

The goal is to find the minimum possible length of the array after performing any number of such operations. You can perform as many operations as you want, in any order, as long as the merge condition (x * y ≤ k) is satisfied.

The key insight is to greedily merge adjacent elements whenever possible. Starting from the left, we try to accumulate the product of consecutive elements as long as the product stays within k. When adding another element would exceed k, we start a new segment. The number of such segments gives us the minimum array length.

Special case: If any element in the array is 0, the entire array can be reduced to length 1 since 0 multiplied by any number is 0, which is always ≤ k.

Intuition

The key observation is that we want to merge as many adjacent elements as possible to minimize the final array length. Think of it like trying to pack items into boxes - we want to fit as many items as possible into each box before starting a new one.

Since we can only merge adjacent elements, and merging creates a new element with their product, we should ask: what's the best strategy for deciding when to merge?

Consider walking through the array from left to right. At each position, we have a choice: either merge the current element with the previous group (if possible), or start a new group. The constraint is that the product of all elements in a group cannot exceed k.

This naturally leads to a greedy approach: keep accumulating the product of consecutive elements as long as the product stays within k. When we encounter an element that would cause the product to exceed k, we have no choice but to start a new group with that element.

Why does this greedy strategy work? Because if we can merge elements now, there's no benefit to keeping them separate. Merging reduces the array length immediately, and since we're processing left to right, it doesn't affect our ability to make future merges. Each group we create represents a maximal sequence of elements that can be merged together.

The special case of 0 is interesting: if any element is 0, then 0 * anything = 0 ≤ k, meaning we can merge the 0 with its neighbors repeatedly until the entire array becomes a single element with value 0. This is why the presence of any 0 immediately tells us the answer is 1.

The algorithm essentially counts how many "breaking points" we have - positions where we must start a new group because the accumulated product would exceed k. The number of groups equals the minimum array length.

Learn more about Greedy and Dynamic Programming patterns.

Solution Approach

The implementation uses a greedy algorithm with two key variables:

• ans: tracks the minimum array length (initially 1 since we start with at least one group)
• y: maintains the running product of the current group (initially nums[0])

We traverse the array starting from the second element. For each element x:

Case 1: Element is Zero (x == 0)

• If we encounter a 0, we immediately return 1
• This is because 0 * anything = 0 ≤ k, allowing us to merge the entire array into a single element

Case 2: Can Merge (x * y ≤ k)

• We can safely merge x with the current group
• Update the running product: y = y * x
• The array length (ans) remains unchanged since we're extending the current group

Case 3: Cannot Merge (x * y > k)

• The product would exceed k, so we cannot merge
• Start a new group with x: set y = x
• Increment the array length: ans = ans + 1

The algorithm processes each element exactly once, making decisions based on whether merging would violate the constraint product ≤ k.

Example walkthrough with nums = [2, 3, 3, 7, 3, 5] and k = 20:

1. Initialize: ans = 1, y = 2
2. Process 3: 2 * 3 = 6 ≤ 20 → merge, y = 6
3. Process 3: 6 * 3 = 18 ≤ 20 → merge, y = 18
4. Process 7: 18 * 7 = 126 > 20 → cannot merge, start new group, ans = 2, y = 7
5. Process 3: 7 * 3 = 21 > 20 → cannot merge, start new group, ans = 3, y = 3
6. Process 5: 3 * 5 = 15 ≤ 20 → merge, y = 15
7. Return ans = 3

The time complexity is O(n) where n is the length of the array, and space complexity is O(1) as we only use a constant amount of extra space.

Example Walkthrough

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

Example: nums = [3, 2, 4, 1] and k = 10

We'll use the greedy approach, maintaining a running product and counting how many groups we need.

Initial State:

• ans = 1 (we start with at least one group)
• y = 3 (the running product starts with the first element)

Step 1: Process element 2 (index 1)

• Check if we can merge: y * 2 = 3 * 2 = 6
• Since 6 ≤ 10, we can merge
• Update running product: y = 6
• Array conceptually becomes: [6, 4, 1]
• ans remains 1 (still one group)

Step 2: Process element 4 (index 2)

• Check if we can merge: y * 4 = 6 * 4 = 24
• Since 24 > 10, we cannot merge with the current group
• Start a new group with this element: y = 4
• Increment group count: ans = 2
• Array conceptually becomes: [6] [4, 1] (two separate groups)

Step 3: Process element 1 (index 3)

• Check if we can merge: y * 1 = 4 * 1 = 4
• Since 4 ≤ 10, we can merge
• Update running product: y = 4
• Array conceptually becomes: [6] [4] (two groups)
• ans remains 2

Final Result: The minimum array length is 2.

The algorithm efficiently determines that we can merge elements [3, 2] into one group (product = 6) and elements [4, 1] into another group (product = 4), but we cannot merge across these groups because 6 * 4 = 24 > 10.

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the input array nums. This is because the algorithm iterates through the array exactly once using a single for loop that processes each element from nums[1:], performing constant-time operations for each element (multiplication, comparison, and assignment operations).

The space complexity is O(1). The algorithm only uses a fixed amount of extra space regardless of the input size, storing just three variables: ans to track the result count, y to maintain the current product value, and x as the loop variable. No additional data structures that scale with input size are created.

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

Common Pitfalls

Pitfall 1: Integer Overflow in Product Calculation

The Problem: When checking if current_num * current_product <= k, the multiplication current_num * current_product can cause integer overflow for large values. In Python, while integers can grow arbitrarily large, in other languages or even in Python when dealing with very large numbers, this can lead to incorrect comparisons or performance issues.

Example Scenario:

nums = [10^9, 10^9, 2]
k = 10^9

The product 10^9 * 10^9 = 10^18 might overflow in languages with fixed integer sizes.

Solution: Instead of computing the product directly, rearrange the comparison to avoid overflow:

# Instead of: current_num * current_product <= k
# Use: current_num <= k // current_product (when current_product > 0)
if current_product > 0 and current_num <= k // current_product:
    current_product *= current_num
else:
    current_product = current_num
    result_length += 1

Pitfall 2: Not Handling Zero at the Beginning

The Problem: The current code checks for zero during iteration but initializes current_product with nums[0]. If nums[0] is zero, the algorithm doesn't immediately return 1, potentially leading to incorrect behavior when multiplying subsequent elements.

Example Scenario:

nums = [0, 5, 3]
k = 10

Starting with current_product = 0, all subsequent multiplications will be 0.

Solution: Check for zero elements at the very beginning or during initialization:

def minArrayLength(self, nums: List[int], k: int) -> int:
    # Check for any zeros upfront
    if 0 in nums:
        return 1
  
    result_length = 1
    current_product = nums[0]
  
    for current_num in nums[1:]:
        if current_num * current_product <= k:
            current_product *= current_num
        else:
            current_product = current_num
            result_length += 1
  
    return result_length

Pitfall 3: Edge Case with Single Element Array

The Problem: While the current solution handles single-element arrays correctly (returns 1), the logic isn't immediately obvious since we're iterating from nums[1:], which would be empty for a single-element array.

Solution: Add explicit handling or documentation for clarity:

def minArrayLength(self, nums: List[int], k: int) -> int:
    # Edge case: single element
    if len(nums) == 1:
        return 1
  
    # ... rest of the code

Pitfall 4: Misunderstanding the Greedy Strategy

The Problem: Developers might think they need to find the optimal merging strategy by considering different orderings or using dynamic programming. However, the greedy left-to-right approach is actually optimal because once you decide not to merge at a position, you cannot "go back" and merge across that boundary later.

Solution: Document why the greedy approach works:

def minArrayLength(self, nums: List[int], k: int) -> int:
    """
    Greedy approach: Always merge when possible from left to right.
    This is optimal because:
    1. Merging reduces array length
    2. Once we decide not to merge at position i, we cannot merge 
       elements across position i later
    3. Therefore, merging eagerly from left to right gives minimum segments
    """
    # ... implementation