#3224

Medium

auto_awesomeArray scanning plus hash lookup

LeetCode Problem Workspace

Minimum Array Changes to Make Differences Equal

Minimize changes to make array differences equal by replacing elements within a given range.

Problem Statement

You are given an integer array nums of size n (where n is even) and an integer k. You can perform changes on the array by replacing any element in nums with an integer between 0 and k, inclusive. Your goal is to make the differences between adjacent elements equal by performing the minimum number of such changes.

Formally, you need to ensure that for each i, nums[i] - nums[i - 1] is the same for all elements in the array. The task is to find the smallest number of changes to achieve this condition, where each change can replace an element in nums with any integer in the range from 0 to k.

Examples

Example 1

Input: nums = [1,0,1,2,4,3], k = 4

Output: 2

We can perform the following changes: The integer X will be 2.

Example 2

Input: nums = [0,1,2,3,3,6,5,4], k = 6

Output: 2

We can perform the following operations: The integer X will be 4.

Constraints

2 <= n == nums.length <= 105
n is even.
0 <= nums[i] <= k <= 105

Solution Approach

Array Scanning

First, perform a linear scan of the array to calculate the differences between adjacent elements. Use this information to determine how many different values of X can satisfy the condition of equal differences. This provides an efficient way to narrow down possible solutions.

Hash Table for Frequency Counting

Next, use a hash table to count the frequencies of the possible differences between adjacent elements. By tracking the number of occurrences of each difference, you can determine which difference occurs the most frequently, minimizing the number of changes required.

Optimal Replacement Strategy

With the frequency table in hand, identify the most frequent difference. Then, calculate how many elements need to be replaced to achieve this frequent difference across the entire array. This provides the minimal number of changes needed to meet the problem's conditions.

Complexity Analysis

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

The time complexity depends on the final approach, typically O(n) for scanning and hash table operations. The space complexity is also O(n) for storing differences and counts in the hash table.

What Interviewers Usually Probe

Candidate understands the importance of reducing the number of changes by focusing on the most frequent difference.
Candidate efficiently uses a hash table for counting occurrences and tracking differences between adjacent elements.
Candidate applies array scanning and hash lookup to solve the problem optimally, without over-complicating the solution.

Common Pitfalls or Variants

Common pitfalls

Not recognizing the optimal way to handle the differences, leading to unnecessary changes.
Failing to use a hash table effectively to count differences, which can slow down the solution.
Over-complicating the problem by trying to handle all potential differences instead of focusing on the most frequent one.

Follow-up variants

What if the array length was not even, and you still needed to perform minimal changes?
How would the approach change if negative numbers were allowed in the array?
What if the range of possible replacements for each element was not fixed but varied?

FAQ

What is the key insight for solving the "Minimum Array Changes to Make Differences Equal" problem?

The key insight is that there are at most k + 1 possible integer values for the difference between adjacent elements. This limits the number of different differences you need to consider.

How can hash tables help in solving this problem?

Hash tables are used to count the frequencies of the differences between adjacent elements, which allows you to determine which difference occurs the most and requires the fewest changes to achieve.

Why is array scanning necessary in this problem?

Array scanning helps identify the differences between adjacent elements, which is crucial for determining the minimal changes needed to make all differences equal.

How can I optimize the solution for large arrays?

By using an O(n) array scan and a hash table, you can efficiently compute the necessary changes without resorting to more expensive algorithms.

What is the space complexity of this problem?

The space complexity is O(n) because you need to store the differences and their frequencies, which scales linearly with the size of the input array.

terminal

Solution

Solution 1: Difference Array

Assume that in the final array, the difference between the pair $\textit{nums}[i]$ and $\textit{nums}[n-i-1]$ is $s$.

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class Solution:
    def minChanges(self, nums: List[int], k: int) -> int:
        d = [0] * (k + 2)
        n = len(nums)
        for i in range(n // 2):
            x, y = nums[i], nums[-i - 1]
            if x > y:
                x, y = y, x
            d[0] += 1
            d[y - x] -= 1
            d[y - x + 1] += 1
            d[max(y, k - x) + 1] -= 1
            d[max(y, k - x) + 1] += 2
        return min(accumulate(d))

Continue Practicing

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