Minimum Operations to Make Elements Within K Subarrays Equal

• Step 1: Pre-calculate Subarray Costs (Naively)
  1. Create a costs array of size n - x + 1.
  2. Iterate from i = 0 to n - x.
  3. For each i, extract the subarray nums[i...i+x-1].
  4. Sort the subarray to find its median (element at index (x-1)/2).
  5. Calculate the sum of absolute differences of each element from the median. This is costs[i].
• Step 2: Dynamic Programming
  1. Create a 2D DP table dp of size (n - x + 2) x (k + 1), initialized to a large value.
  2. Set the base case: dp[i][0] = 0 for all i (cost of 0 subarrays is 0).
  3. Iterate j from 1 to k (number of subarrays).
  4. Iterate i from 1 to n - x + 1 (number of possible starting positions).
  5. Calculate dp[i][j] using the recurrence: dp[i][j] = min(dp[i-1][j], costs[i-1] + dp[i-x][j-1]).
     • dp[i-1][j] represents not selecting the subarray starting at i-1.
     • costs[i-1] + dp[i-x][j-1] represents selecting the subarray at i-1. This is only possible if i-x >= 0.
  6. The final answer is dp[n-x+1][k].

Detailed Steps:

1. Pre-calculate Subarray Costs: We create a costs array to store the cost for each potential subarray. For every starting index i from 0 to n-x, we consider the subarray nums[i...i+x-1]. The minimum operations to make all elements in this subarray equal is to change them all to the subarray's median. We find the median by sorting a copy of the subarray. The cost is the sum of absolute differences between each element and the median. This process is repeated for all n-x+1 possible subarrays.

   1private long calculateCost(int[] nums, int start, int x) {2    int[] sub = new int[x];3    System.arraycopy(nums, start, sub, 0, x);4    Arrays.sort(sub);5    long median = sub[(x - 1) / 2];6    long cost = 0;7    for (int val : sub) {8        cost += Math.abs(val - median);9    }10    return cost;11}

2. Dynamic Programming: After computing all individual subarray costs, the problem reduces to selecting k non-overlapping entries from the costs array with minimum sum. Let dp[i][j] be the minimum cost to select j non-overlapping subarrays from the first i possible starting positions (costs[0...i-1]).

   The recurrence relation is: dp[i][j] = min(dp[i-1][j], costs[i-1] + dp[i-x][j-1])

   • dp[i-1][j]: This corresponds to the case where we do not select the subarray starting at i-1.
   • costs[i-1] + dp[i-x][j-1]: This corresponds to selecting the subarray starting at i-1. If we do, the previous j-1 subarrays must be chosen from starting positions up to i-1-x to ensure they are non-overlapping.

   The final answer is the value in dp[n-x+1][k].

   1// Part of the main function2int m = n - x + 1;3long[] costs = new long[m];4for (int i = 0; i < m; i++) {5    costs[i] = calculateCost(nums, i, x);6}7 8long[][] dp = new long[m + 1][k + 1];9long INF = Long.MAX_VALUE / 2;10for (long[] row : dp) {11    Arrays.fill(row, INF);12}13for (int i = 0; i <= m; i++) {14    dp[i][0] = 0;15}16 17for (int j = 1; j <= k; j++) {18    for (int i = 1; i <= m; i++) {19        // Option 1: Don't take subarray starting at i-120        long option1 = dp[i - 1][j];21        22        // Option 2: Take subarray starting at i-123        long option2 = INF;24        if (i - x >= 0) {25            if (dp[i - x][j - 1] != INF) {26                option2 = costs[i - 1] + dp[i - x][j - 1];27            }28        } else { // This can only be the first subarray chosen29            if (j == 1) {30                option2 = costs[i - 1];31            }32        }33        dp[i][j] = Math.min(option1, option2);34    }35}36return dp[m][k];