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3224. Minimum Array Changes to Make Differences Equal


Problem Description

You are given an integer array nums of size n where n is even, and an integer k. You can perform some changes on the array, where in one change you can replace any element in the array with any integer in the range from 0 to k.

You need to perform some changes (possibly none) such that the final array satisfies the following condition:

There exists an integer X such that abs(a[i] - a[n - i - 1]) = X for all (0 <= i < n).

Return the minimum number of changes required to satisfy the above condition.

Intuition

The core idea of the solution is based on ensuring that for each pair (nums[i], nums[n-i-1]), the difference is a consistent value, X. To achieve this, we explore reformulating the problem in terms of required changes to make the absolute differences consistent across all elements.

Understanding the Problem

Given the array nums and range 0 to k, we must ensure that the difference abs(nums[i] - nums[n-i-1]) is the same for all index pairs (i, n-i-1). Achieving this might require changing some elements within the allowed range 0 to k.

Solution Approach

  1. Pairwise Differences: For each pair (nums[i], nums[n-i-1]), determine the current difference and what changes are necessary. Let's say x is the smaller of these two numbers and y is the larger one.

  2. Difference Array Construction: We'll use a difference array to record the number of changes needed at each possible difference value:

    • If no change is needed, y - x = s.
    • If one change is made, s should be <= max(y, k-x) either by setting x to 0 or y to k.
    • If two changes are made, s > max(y, k-x).
  3. Scenarios for Changes:

    • For difference range [0, y-x-1], one change is needed.
    • For difference y-x, no changes are needed.
    • For range [y-x+1, max(y, k-x)], one change is required.
    • For range [max(y, k-x)+1, k], two changes are necessary.
  4. Accumulate Changes: After populating the difference array with these scenarios for every pair, compute the prefix sums to determine the minimum number of changes required across all possible consistent difference values.

  5. Determine Minimum Changes: The minimum value in the prefix sum array gives the least number of changes needed to achieve a uniform difference across all pairs.

This structured approach ensures an optimal number of modifications, leveraging difference arrays to efficiently handle overlapping scenarios of required changes.

Learn more about Prefix Sum patterns.

Solution Approach

The given problem can be efficiently solved by utilizing a difference array technique. This approach allows for calculating and minimizing the required changes in an optimal manner. Here’s a detailed breakdown of the solution implementation:

  1. Initialize a Difference Array:

    • Create an array d of size k + 2 initialized with zeros to keep track of the incremental changes needed for each possible difference value.
  2. Iterate Over Pair Indexes:

    • For each i in the range from 0 to n/2 (since n is even), consider the pair of elements at positions nums[i] and nums[n-i-1].
    • Denote x as the smaller and y as the larger of the two values.
  3. Update Difference Array for Each Pair:

    • If x and y need no changes, set y - x = s; otherwise:
      • Increment d[0] by 1 as a baseline for the number of changes.
      • For the range where exactly one change is beneficial, decrement d[y - x] and increment d[y - x + 1] to mark intervals where 0 changes transition to 1 change.
      • Adjust d values for ranges beyond max(y, k-x), indicating where the second change becomes necessary, with adjustments ensuring transitions between different change requirements are recorded.
  4. Calculate Minimum Changes:

    • Use the accumulate function to compute prefix sums over the d array, which gives cumulative changes needed across different range partitions.
    • The smallest value in the resulting prefix sum array indicates the minimum number of changes needed to achieve the uniform difference condition.

The implementation leverages the efficient space and time characteristics of difference arrays, allowing for a comprehensive yet optimal determination of change requirements:

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))

This method ensures that the final number of changes computed is optimal and meets the criteria specified by the problem efficiently, using O(n) time and space complexity.

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Example Walkthrough

Let's walk through an example to illustrate the solution approach described above.

Consider the array nums = [1, 5, 3, 7] and k = 9.

  1. Identify Pairs: Since n = 4, the pairs of indices are (0, 3) and (1, 2). We need to ensure that the absolute difference between these pairs is consistent.

  2. Process Each Pair:

    • For pair (nums[0], nums[3]) which are (1, 7), let x = 1 and y = 7.

      • The current difference is y - x = 6.
      • We initialize the difference array d, marking that one change might be needed by setting d[0] += 1.
      • Since y - x = 6, d[6] -= 1 indicates no change is required, d[7] += 1 for transitions from no changes to one change.
      • For the range [8, 9], we adjust further for possible two changes with d[8] -= 1 and d[10] += 2.
    • For pair (nums[1], nums[2]) which are (5, 3), let x = 3 and y = 5.

      • The current difference is y - x = 2.
      • Again, we increment d[0] += 1 and adjust for ranges: d[2] -= 1, d[3] += 1, and considering max(y, k-x) + 1 = 8, adjust: d[6] -= 1, d[9] += 2.
  3. Accumulate Changes:

    • The difference array d is initialized and updated for all pairs. After iterating over all pairs, d = [2, 0, -1, 0, 0, 0, -1, 1, -1, 2].
  4. Compute Minimum Changes:

    • Using prefix sums over d, compute cumulative changes: [2, 2, 1, 1, 1, 1, 0, 1, 0, 2].
  5. Result:

    • The minimum value from the prefix sum array is 0, indicating that no changes are required to make the array meet the condition for a uniform X difference.

This breakdown exemplifies the difference array technique and illustrates the minimum number of modifications required to fulfill the problem requirements.

Solution Implementation

1from itertools import accumulate
2from typing import List
3
4class Solution:
5    def minChanges(self, nums: List[int], k: int) -> int:
6        # Initialize a difference array 'd' with zeros, size is (k + 2) to handle boundaries safely.
7        differences = [0] * (k + 2)
8        n = len(nums)
9      
10        # Iterate through the first half of the list. We're assuming 'nums' is mirrored around the center.
11        for i in range(n // 2):
12            # Assign x and y to be the i-th and corresponding element from the other half.
13            x, y = nums[i], nums[-i - 1]
14          
15            # Ensure x is the smaller or equal value for consistent logic
16            if x > y:
17                x, y = y, x
18          
19            # Increase the count for 0 in the differences array
20            differences[0] += 1
21          
22            # Adjust the boundaries for the differences
23            differences[y - x] -= 1
24            differences[y - x + 1] += 1
25          
26            # Calculate the maximum value that can be achieved without exceeding k
27            adjusted_max = max(y, k - x)
28          
29            # Adjust the difference for this range
30            differences[adjusted_max + 1] -= 1
31            differences[adjusted_max + 1] += 2
32      
33        # Use accumulate to transform the differences array into the result list and get the minimum value
34        return min(accumulate(differences))
35
1class Solution {
2    public int minChanges(int[] nums, int k) {
3        int[] deltaArray = new int[k + 2]; // Array to track increments and decrements at given indices.
4        int length = nums.length;
5      
6        // Iterate over the first half of the array
7        for (int i = 0; i < length / 2; ++i) {
8            // Determine the minimum and maximum of the current pair
9            int minVal = Math.min(nums[i], nums[length - i - 1]);
10            int maxVal = Math.max(nums[i], nums[length - i - 1]);
11          
12            // Update deltaArray to reflect the range of changes
13            deltaArray[0] += 1; // Increase the base count
14            deltaArray[maxVal - minVal] -= 1;
15            deltaArray[maxVal - minVal + 1] += 1;
16            int maxIndex = Math.max(maxVal, k - minVal) + 1;
17            deltaArray[maxIndex] -= 1;
18            deltaArray[maxIndex] += 2;
19        }
20      
21        int minChanges = length; // Assume initially that all elements need to be changed
22        int cumulativeSum = 0;
23      
24        // Calculate the minimum number of changes needed
25        for (int change : deltaArray) {
26            cumulativeSum += change; // Sum the increments and decrements
27            minChanges = Math.min(minChanges, cumulativeSum); // Keep track of the minimum number of changes
28        }
29      
30        return minChanges; // Return the minimum number of changes necessary
31    }
32}
33
1#include <vector>
2#include <cstring> // For memset
3#include <algorithm> // For min and max functions
4
5class Solution {
6public:
7    int minChanges(std::vector<int>& nums, int k) {
8        // Initialize an array d to perform operations
9        int d[k + 2];
10        memset(d, 0, sizeof(d)); // Set all elements of d to 0
11
12        int n = nums.size(); // Get the size of the input vector
13
14        // Iterate over half of the nums vector
15        for (int i = 0; i < n / 2; ++i) {
16            int x = std::min(nums[i], nums[n - i - 1]); // Smaller value between nums[i] and nums[n-i-1]
17            int y = std::max(nums[i], nums[n - i - 1]); // Larger value between nums[i] and nums[n-i-1]
18
19            // Update elements of d based on calculated x and y
20            d[0] += 1;
21            d[y - x] -= 1;
22            d[y - x + 1] += 1;
23            d[std::max(y, k - x) + 1] -= 1;
24            d[std::max(y, k - x) + 1] += 2;
25        }
26
27        int ans = n, s = 0; // Initialize answer with n and sum s with 0
28
29        // Accumulate values of d and determine the minimum changes
30        for (int value : d) {
31            s += value;
32            ans = std::min(ans, s);
33        }
34
35        return ans; // Return the minimum number of changes
36    }
37};
38
1function minChanges(nums: number[], k: number): number {
2    // Initialize a difference array `d` with a size of `k + 2` filled with zeros
3    const d: number[] = Array(k + 2).fill(0);
4    const n = nums.length;
5
6    // Loop through the first half of `nums`
7    for (let i = 0; i < n >> 1; ++i) {
8        // Determine the min and max elements between symmetric positions
9        const x = Math.min(nums[i], nums[n - 1 - i]);
10        const y = Math.max(nums[i], nums[n - 1 - i]);
11
12        // Increment total operation count by 1
13        d[0] += 1;
14
15        // Account for possible values with difference `y - x`
16        d[y - x] -= 1;
17        d[y - x + 1] += 1;
18
19        // Handle maximum value constraints and increment/decrement accordingly
20        const maxValue = Math.max(y, k - x) + 1;
21        d[maxValue] -= 1; // Reset point after the range
22        d[maxValue] += 2; // Further adjustment, specific purpose unclear without more context
23    }
24
25    // Initialize variables to calculate minimum changes needed
26    let ans = n;
27    let s = 0;
28
29    // Calculate prefix sums on the difference array `d` to get minimum changes
30    for (const x of d) {
31        s += x; // Accumulate the changes
32        ans = Math.min(ans, s); // Track the smallest change cost
33    }
34
35    return ans; // Return the minimum number of changes calculated
36}
37

Time and Space Complexity

The given code snippets can be analyzed as follows:

Time Complexity

The time complexity of the solution primarily revolves around the loop and the usage of the accumulate function from the itertools module:

  1. Loop through n // 2 elements: The loop iterates n // 2 times, where n is the length of nums. Thus, the loop contributes O(n) to the complexity.

  2. accumulate function: The accumulation process runs through the list d of size k + 2, leading to a time complexity of O(k).

Overall, the time complexity of the complete program can be considered as O(n + k).

Space Complexity

The space complexity is determined by the usage of the list d:

  • List d: A fixed-size list of size k + 2 is used. Therefore, the space complexity related to d is O(k).

Since no other significant space is used, the overall space complexity of the code is O(k).

Learn more about how to find time and space complexity quickly.


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