Leetcode #2892: Minimizing Array After Replacing Pairs With Their Product
In this guide, we solve Leetcode #2892 Minimizing Array After Replacing Pairs With Their Product in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given an integer array nums and an integer k, you can perform the following operation on the array any number of times: Select two adjacent elements of the array like x and y, such that x * y <= k, and replace both of them with a single element with value x * y (e.g. in one operation the array [1, 2, 2, 3] with k = 5 can become [1, 4, 3] or [2, 2, 3], but can't become [1, 2, 6]).
Quick Facts
- Difficulty: Medium
- Premium: Yes
- Tags: Greedy, Array, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: nums = [2,3,3,7,3,5], k = 20
Output: 3
Explanation: We perform these operations:
1. [2,3,3,7,3,5] -> [6,3,7,3,5]
2. [6,3,7,3,5] -> [18,7,3,5]
3. [18,7,3,5] -> [18,7,15]
It can be shown that 3 is the minimum length possible to achieve with the given operation.
Python Solution
class Solution:
def minArrayLength(self, nums: List[int], k: int) -> int:
ans, y = 1, nums[0]
for x in nums[1:]:
if x == 0:
return 1
if x * y <= k:
y *= x
else:
y = x
ans += 1
return ans
Complexity
The time complexity is , where n is the length of the array. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.