Leetcode #2934: Minimum Operations to Maximize Last Elements in Arrays
In this guide, we solve Leetcode #2934 Minimum Operations to Maximize Last Elements in Arrays 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
You are given two 0-indexed integer arrays, nums1 and nums2, both having length n. You are allowed to perform a series of operations (possibly none).
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Enumeration
Intuition
The constraints allow a direct scan that keeps only the essential state.
By translating the requirements into a clean loop, the logic stays easy to reason about.
Approach
Iterate through the data once, updating the state needed to compute the answer.
Return the final state after the traversal is complete.
Steps:
- Parse the input.
- Iterate and update state.
- Return the computed answer.
Example
Input: nums1 = [1,2,7], nums2 = [4,5,3]
Output: 1
Explanation: In this example, an operation can be performed using index i = 2.
When nums1[2] and nums2[2] are swapped, nums1 becomes [1,2,3] and nums2 becomes [4,5,7].
Both conditions are now satisfied.
It can be shown that the minimum number of operations needed to be performed is 1.
So, the answer is 1.
Python Solution
class Solution:
def minOperations(self, nums1: List[int], nums2: List[int]) -> int:
def f(x: int, y: int) -> int:
cnt = 0
for a, b in zip(nums1[:-1], nums2[:-1]):
if a <= x and b <= y:
continue
if not (a <= y and b <= x):
return -1
cnt += 1
return cnt
a, b = f(nums1[-1], nums2[-1]), f(nums2[-1], nums1[-1])
return -1 if a + b == -2 else min(a, b + 1)
Complexity
The time complexity is , where 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.