3224. Minimum Array Changes to Make Differences Equal

Problem Statement

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.

Example 1:

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

Output: 2

Explanation: We can perform the following changes:

Replace nums[1] by 2. The resulting array is nums = [1,2,1,2,4,3].
Replace nums[3] by 3. The resulting array is nums = [1,2,1,3,4,3].

The integer X will be 2.

Example 2:

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

Output: 2

Explanation: We can perform the following operations:

Replace nums[3] by 0. The resulting array is nums = [0,1,2,0,3,6,5,4].
Replace nums[4] by 4. The resulting array is nums = [0,1,2,0,4,6,5,4].

The integer X will be 4.

Constraints:

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

Intuition

Approach: 
Keep track of actual difference -> Zero changes
Also keep track of Maximum achievable difference, that can be formed by changing 
element

Now, If we get Max achievable diff for certain pair as 4, then we can acheieve 
a diff of 3 2 1 0 All 

So 
Next step 

For finding answer 
Take Diff = 18
One Diff(18) to get - Zero_changes to get 18 = Only those extra pairs we need 
To change by updating one value in a pair

And remaining goes to 2 Pair

Video Links

Approach 1:

C++

class Solution {
public:
    int minChanges(vector<int>& nums, int k) {
        int n = nums.size();
        unordered_map<int, int> curDiff;
        vector<int> one_diff(k+1, 0);

        for(int i=0; i<n/2; i++){
            int diff = abs(nums[i] - nums[n-i-1]);
            curDiff[diff] ++;
            
            int mini = min(nums[i], nums[n-i-1]);
            int maxi = max(nums[i], nums[n-i-1]);
            int max_diff = max(k-mini, maxi-0);
            one_diff[max_diff] ++;
        }

        // Propogate values
        for(int i=k-1; i>=0; i--){
            one_diff[i] += one_diff[i+1];
        }

        //find_ans
        int ans = INT_MAX;
        for(auto &[diff, count]: curDiff){
            int one_extra = one_diff[diff] - count;
            int two = n/2 - one_extra - count;
            ans = min(ans, one_extra + two*2);
        }

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++