LeetCode 2892 - Minimizing Array After Replacing Pairs With Their Product

The problem gives us an integer array nums and a value k. We are allowed to repeatedly merge adjacent elements under one condition: - If two adjacent values x and y satisfy x y <= k, then we may replace them with a single value equal to x y.

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LeetCode Problem 2892

Difficulty: 🟡 Medium
Topics: Array, Dynamic Programming, Greedy

Solution

Problem Understanding

The problem gives us an integer array nums and a value k. We are allowed to repeatedly merge adjacent elements under one condition:

  • If two adjacent values x and y satisfy x * y <= k, then we may replace them with a single value equal to x * y.

Each operation reduces the array length by exactly one because two elements become one.

Our goal is to minimize the final length of the array after performing any number of valid operations.

The important detail is that after merging two numbers, the resulting product becomes a new array element and may participate in future merges. This means operations can chain together. For example, if we merge 2 and 3 into 6, then 6 may later merge with another adjacent value if the new product still stays within the limit k.

The input size is large:

  • nums.length can be up to 10^5
  • Values can be as large as 10^9

These constraints immediately rule out expensive recursive search or simulation approaches that explore many possible merge sequences.

Another important observation is that all numbers are non-negative because:

0 <= nums[i] <= 10^9

This matters because products never become negative, and multiplication behaves monotonically.

Several edge cases are important:

  • Arrays containing 0, because 0 * anything = 0, which is always <= k
  • Arrays where no adjacent pair can merge
  • Arrays where the entire array can collapse into one element
  • Large values near 10^9, where overflow concerns matter in some languages
  • Long chains of merges, where repeatedly combining elements is optimal

The problem guarantees that k >= 1, so comparisons against the product are always meaningful.

Approaches

Brute Force Approach

A brute force solution would try every possible valid merge sequence.

At each step:

  1. Scan the array for adjacent pairs whose product is <= k
  2. Try merging each possible pair
  3. Recursively continue from the new array state
  4. Return the minimum achievable length across all possibilities

This works because it explores every valid sequence of operations, so it cannot miss the optimal answer.

However, the number of possible states grows exponentially. Even a modest array can generate many different merge combinations. Since every merge changes adjacency relationships, the branching factor becomes enormous.

For an array of length n, the total number of possible merge sequences can become exponential in n, making this approach completely impractical for n = 10^5.

Key Insight for the Optimal Solution

The crucial observation is that multiplication is associative:

(a * b) * c = a * (b * c)

Because of this, if several consecutive numbers can eventually be merged into one value whose product is <= k, then the order of merging inside that segment does not matter.

This transforms the problem into:

Partition the array into the fewest possible contiguous groups such that the product of each group is <= k.

Each valid group can ultimately collapse into one number.

So instead of simulating operations, we greedily build the largest possible segment whose product remains within k.

We scan from left to right while maintaining the current segment product.

  • If multiplying the next number keeps the product <= k, we extend the segment.
  • Otherwise, we must start a new segment.

The number of segments becomes the minimum possible final array length.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force Exponential Exponential Explores all possible merge sequences
Optimal Greedy O(n) O(1) Greedily forms the largest valid mergeable segments

Algorithm Walkthrough

Step 1: Initialize the answer

We begin with:

  • current_product = 1
  • groups = 0

The variable current_product stores the product of the current mergeable segment.

The variable groups counts how many final segments we need.

Step 2: Iterate through the array

We process each number from left to right.

For every value num:

  • Check whether current_product * num <= k

If true, we can safely include num in the current segment.

Step 3: Extend the current segment when possible

If the new product remains valid:

current_product *= num

This means all numbers in the current segment can eventually collapse into one value.

Step 4: Start a new segment when necessary

If adding num would make the product exceed k, then the current segment cannot absorb this value.

So:

  1. Finalize the current segment
  2. Increment groups
  3. Start a new segment with num

Conceptually:

groups += 1
current_product = num

Step 5: Count the final segment

After the loop finishes, there is still one unfinished segment remaining.

So we add one more group to the answer.

Why it works

The greedy strategy works because once a segment product exceeds k, no future operations can reduce it. All numbers are non-negative, so multiplying additional elements can only maintain or increase the product.

Therefore, when the current segment cannot include the next element, any valid solution must start a new segment at that point.

By always extending the current segment as much as possible, we minimize the number of total segments, which directly minimizes the final array length.

Python Solution

from typing import List

class Solution:
    def minArrayLength(self, nums: List[int], k: int) -> int:
        groups = 0
        current_product = 1

        for num in nums:
            if current_product * num <= k:
                current_product *= num
            else:
                groups += 1
                current_product = num

        return groups + 1

The implementation follows the greedy strategy directly.

We maintain current_product, which represents the product of the current mergeable segment.

For each number:

  • If multiplying it keeps the product within the limit, we extend the current segment.
  • Otherwise, we finalize the current segment and begin a new one.

The variable groups counts completed segments. Since the final segment is still active after the loop ends, we return groups + 1.

The solution only scans the array once and uses constant extra memory.

Go Solution

func minArrayLength(nums []int, k int) int {
    groups := 0
    currentProduct := 1

    for _, num := range nums {
        if currentProduct*num <= k {
            currentProduct *= num
        } else {
            groups++
            currentProduct = num
        }
    }

    return groups + 1
}

The Go implementation mirrors the Python solution closely.

One important consideration is integer overflow. The constraints allow values up to 10^9, so multiplication could become large. However, since we immediately compare against k <= 10^9, the product never needs to grow far beyond that threshold in practice.

Go's default int type is sufficient on LeetCode's 64 bit environment. If extra safety were desired, int64 could also be used.

Slices in Go are dynamic views over arrays, so no additional data structures are required.

Worked Examples

Example 1

nums = [2,3,3,7,3,5]
k = 20

We process the array greedily.

Step num Current Product Before Can Merge? Current Product After Groups
1 2 1 Yes 2 0
2 3 2 Yes 6 0
3 3 6 Yes 18 0
4 7 18 No 7 1
5 3 7 No 3 2
6 5 3 Yes 15 2

After finishing the loop:

groups + 1 = 3

Final answer:

3

The segments are effectively:

[2,3,3], [7], [3,5]

Each segment can collapse into one value.

Example 2

nums = [3,3,3,3]
k = 6
Step num Current Product Before Can Merge? Current Product After Groups
1 3 1 Yes 3 0
2 3 3 No 3 1
3 3 3 No 3 2
4 3 3 No 3 3

After the loop:

groups + 1 = 4

Final answer:

4

No adjacent pair can merge because:

3 * 3 = 9 > 6

Complexity Analysis

Measure Complexity Explanation
Time O(n) We scan the array exactly once
Space O(1) Only a few variables are used

The algorithm performs constant work for each array element. No nested loops, recursion, or auxiliary data structures are required.

Because the scan is linear and memory usage is constant, the solution easily handles the maximum constraint of 10^5 elements.

Test Cases

from typing import List

class Solution:
    def minArrayLength(self, nums: List[int], k: int) -> int:
        groups = 0
        current_product = 1

        for num in nums:
            if current_product * num <= k:
                current_product *= num
            else:
                groups += 1
                current_product = num

        return groups + 1

sol = Solution()

assert sol.minArrayLength([2,3,3,7,3,5], 20) == 3  # provided example
assert sol.minArrayLength([3,3,3,3], 6) == 4  # no merges possible

assert sol.minArrayLength([1], 10) == 1  # single element array
assert sol.minArrayLength([0,0,0], 1) == 1  # zeros can always merge
assert sol.minArrayLength([1,1,1,1], 1) == 1  # entire array merges
assert sol.minArrayLength([10,10,10], 50) == 3  # every pair too large
assert sol.minArrayLength([2,2,2,2], 16) == 1  # repeated successful merges
assert sol.minArrayLength([2,2,2,2], 8) == 2  # split into two groups
assert sol.minArrayLength([5,1,1,1], 5) == 1  # chain merging with ones
assert sol.minArrayLength([1000000000], 1000000000) == 1  # max value
assert sol.minArrayLength([0,5,0,5], 1) == 1  # zeros absorb everything
assert sol.minArrayLength([4,5,2], 20) == 2  # one valid merge segment

Test Case Summary

Test Why
[2,3,3,7,3,5], k=20 Validates standard merging behavior
[3,3,3,3], k=6 Validates no merges possible
[1], k=10 Smallest array size
[0,0,0], k=1 Ensures zero handling works
[1,1,1,1], k=1 Entire array collapses into one
[10,10,10], k=50 Every pair exceeds limit
[2,2,2,2], k=16 Full chain merging
[2,2,2,2], k=8 Requires partitioning into segments
[5,1,1,1], k=5 Ones allow repeated extensions
[1000000000], k=1000000000 Maximum numeric constraint
[0,5,0,5], k=1 Zero propagation through merges
[4,5,2], k=20 Mixed valid and invalid transitions

Edge Cases

Arrays Containing Zero

Zero is especially important because:

0 * anything = 0

Since 0 <= k always holds, zeros can merge with neighboring elements indefinitely.

For example:

[0, 100, 200]

can fully collapse into one segment even if k is very small.

The implementation handles this naturally because the running product becomes 0, and all future multiplications remain 0.

No Valid Merges

Consider:

nums = [10, 10, 10]
k = 50

Every adjacent product is:

10 * 10 = 100

which exceeds k.

A buggy implementation might incorrectly attempt merges after previous operations, but the greedy method correctly starts a new segment for every element.

The answer remains the original array length.

Entire Array Can Collapse

Consider:

nums = [1,1,1,1]
k = 1

The product never exceeds 1, so every element belongs to the same segment.

The algorithm continuously extends the current segment and ultimately returns 1.

This verifies that the implementation properly handles long merge chains without prematurely splitting segments.