#2934

Medium

LeetCode Problem Workspace

Minimum Operations to Maximize Last Elements in Arrays

Minimize the number of operations needed to maximize the last elements of two arrays with specific swaps.

Problem Statement

You are given two 0-indexed integer arrays, nums1 and nums2, both having length n. Your task is to find the minimum number of operations needed to maximize the last elements of both arrays. In each operation, you can swap the values at the same index of both arrays.

To complete the task, the goal is to ensure that both nums1[n-1] and nums2[n-1] are as large as possible, given that the rest of the elements remain unchanged. If it's impossible to achieve the goal, return -1.

Examples

Example 1

Input: nums1 = [1,2,7], nums2 = [4,5,3]

Output: 1

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.

Example 2

Input: nums1 = [2,3,4,5,9], nums2 = [8,8,4,4,4]

Output: 2

In this example, the following operations can be performed: First operation using index i = 4. When nums1[4] and nums2[4] are swapped, nums1 becomes [2,3,4,5,4], and nums2 becomes [8,8,4,4,9]. Another operation using index i = 3. When nums1[3] and nums2[3] are swapped, nums1 becomes [2,3,4,4,4], and nums2 becomes [8,8,4,5,9]. Both conditions are now satisfied. It can be shown that the minimum number of operations needed to be performed is 2. So, the answer is 2.

Example 3

Input: nums1 = [1,5,4], nums2 = [2,5,3]

Output: -1

In this example, it is not possible to satisfy both conditions. So, the answer is -1.

Constraints

1 <= n == nums1.length == nums2.length <= 1000
1 <= nums1[i] <= 109
1 <= nums2[i] <= 109

Solution Approach

Initial Check

Before starting any operations, verify if the last elements of nums1 and nums2 are already at their maximum possible values. If both elements are maximized, return 0, as no operation is needed.

Swapping Strategy

Consider performing swaps starting from the second-to-last elements towards the first, as this will allow you to incrementally adjust the last elements of both arrays while ensuring you are minimizing the number of operations. Track swaps carefully to ensure the condition is met.

Edge Case Handling

If no valid solution exists (e.g., when nums1[n-1] cannot be greater than or equal to nums2[n-1] after any number of swaps), return -1 to indicate that achieving the goal is impossible.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution depends on the number of possible swaps. In the worst case, it could be O(n), where n is the length of the arrays. Space complexity is O(1) as we only need a few variables to track swaps and indices.

What Interviewers Usually Probe

Look for understanding of array manipulation with minimal operations.
Assess if the candidate considers edge cases like impossible swaps.
Check for the candidate's ability to choose an efficient approach in terms of swap order.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle cases where no solution exists and returning -1 when necessary.
Failing to optimize the number of swaps by not selecting the best indices.
Overcomplicating the solution instead of focusing on the minimal operations needed.

Follow-up variants

Handling arrays with elements in descending or ascending order for a variety of swap conditions.
Dealing with larger arrays and considering performance optimizations.
Extending the problem to involve more than two arrays, requiring swaps across multiple arrays.

FAQ

What is the minimum number of operations to maximize the last elements in the arrays?

The minimum number of operations is the smallest number of swaps required to ensure both nums1[n-1] and nums2[n-1] are maximized.

Can the problem 'Minimum Operations to Maximize Last Elements in Arrays' be solved without swapping?

No, swapping is the only allowed operation in this problem to maximize the last elements.

What happens if no valid swap sequence can maximize the last elements?

If no valid sequence of swaps can maximize the last elements, the correct answer is -1.

What should be the approach if the arrays are already maximized?

If both arrays are already maximized, no operations are needed, and the answer is 0.

How does GhostInterview help in solving the 'Minimum Operations to Maximize Last Elements in Arrays' problem?

GhostInterview provides a structured approach to minimizing the number of operations and handling edge cases effectively.

terminal

Solution

Solution 1: Case Discussion + Greedy

We can discuss two cases:

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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)

Continue Practicing

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