Uniform K-Subarray Sums

The first line of input contains two space-separated integers, 'N' and 'K'.

The second line contains 'N' space-separated integers, representing the elements of the array 'arr'.

Print a single integer representing the minimum number of replacements required.

The condition that all K-length subarray sums are equal implies a periodic property. If the sum of {a_i, ..., a_{i+K-1}} must equal the sum of {a_{i+1}, ..., a_{i+K}}, it forces a_i to be equal to a_{i+K}. This means all elements at indices j, j+K, j+2K, ... must be the same for the sums to be equal with minimal changes.

Sample Input 1:

5 2
3 4 3 5 6

Sample Output 1:

2

Explanation for Sample 1:

To make all subarrays of length 2 have the same sum, we need `arr[i] = arr[i+2]` for all valid `i`. This creates two groups of elements that must be made equal:
  Group 0 (indices 0, 2, 4): `{3, 3, 6}`. To make them equal with minimum changes, we change them to the most frequent element, which is 3. This requires 1 change (6 -> 3).
  Group 1 (indices 1, 3): `{4, 5}`. To make them equal, we can change one to match the other. This requires 1 change (e.g., 5 -> 4).
Total minimum changes = 1 + 1 = 2.

Sample Input 2:

5 3
1 2 3 1 2

Sample Output 2:

0

Explanation for Sample 2:

The subarrays of length 3 are `{1, 2, 3}`, `{2, 3, 1}`, and `{3, 1, 2}`. All of these already have a sum of 6. No operations are needed.
Following the periodic property:
   Group 0 (indices 0, 3): `{1, 1}`. Already equal. 0 changes.
   Group 1 (indices 1, 4): `{2, 2}`. Already equal. 0 changes.
   Group 2 (index 2): `{3}`. Already equal. 0 changes.
Total changes = 0.

Expected Time Complexity:

The expected time complexity is O(N).

Constraints:

1 <= K <= N <= 10^5
1 <= arr[i] <= 10^9

Time limit: 1sec