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];

• Step 1: Pre-calculate Subarray Costs (Efficiently)
  1. Create a costs array of size n - x + 1.
  2. Use a sliding window of size x over nums.
  3. Maintain two heaps (a max-heap for the smaller half, a min-heap for the larger half) to track elements in the window.
  4. For the first window, populate the heaps and calculate costs[0].
  5. For subsequent windows, slide the window by removing the old element and adding the new one. Update the heaps in O(log x) time.
  6. Calculate the cost for each window using the heaps' state and store it in the costs array.
• Step 2: Dynamic Programming
  1. This step is identical to the previous approach. Use the efficiently calculated costs array to fill the O(n*k) DP table.
  2. The recurrence dp[i][j] = min(dp[i-1][j], costs[i-1] + dp[i-x][j-1]) is used.
  3. The final answer is dp[n-x+1][k].

Detailed Steps:

1. Efficient Cost Calculation with Sliding Window Median: The main improvement is in calculating the costs array. We slide a window of size x across the nums array. To find the median of the elements in the window at each step efficiently, we use two heaps:

   • A max-heap (small) to store the smaller half of the window's elements.
   • A min-heap (large) to store the larger half.

   We keep the heaps balanced so that their sizes differ by at most 1. The median is always large.peek(). We also maintain the sum of elements in each heap to calculate the cost in O(1). When the window slides, we remove one element and add another. These heap operations take O(log x) time. This reduces the total time for cost pre-calculation to O(n log x).

   1// Note: In Java, PriorityQueue.remove(Object) is O(N). A true O(log N) removal2// requires a more complex implementation (e.g., with a HashMap for lazy removal)3// or using a balanced binary search tree (TreeMap).4// The following snippet illustrates the idea but has O(x) removal time.5private long[] calculateAllCosts(int[] nums, int x) {6    int n = nums.length;7    int m = n - x + 1;8    long[] costs = new long[m];9    10    // Two heaps for sliding window median11    PriorityQueue<Integer> small = new PriorityQueue<>(Collections.reverseOrder());12    PriorityQueue<Integer> large = new PriorityQueue<>();13    long smallSum = 0, largeSum = 0;14 15    // Helper to balance heaps16    Runnable balance = () -> {17        while (small.size() > large.size()) {18            int val = small.poll(); smallSum -= val;19            large.add(val); largeSum += val;20        }21        while (large.size() > small.size() + 1) {22            int val = large.poll(); largeSum -= val;23            small.add(val); smallSum += val;24        }25    };26 27    // Initial window28    for (int i = 0; i < x; i++) {29        small.add(nums[i]);30        smallSum += nums[i];31    }32    balance.run();33 34    // Sliding the window35    for (int i = 0; i < m; i++) {36        // Calculate cost for current window37        long median = large.peek();38        costs[i] = (median * small.size() - smallSum) + (largeSum - median * large.size());39 40        // Slide to next window41        if (i + x < n) {42            int toRemove = nums[i];43            int toAdd = nums[i + x];44 45            if (small.remove(toRemove)) smallSum -= toRemove;46            else if (large.remove(toRemove)) largeSum -= toRemove;47 48            if (!large.isEmpty() && toAdd >= large.peek()) {49                large.add(toAdd); largeSum += toAdd;50            } else {51                small.add(toAdd); smallSum += toAdd;52            }53            balance.run();54        }55    }56    return costs;57}

2. Dynamic Programming: This part is identical to Approach 1. We use the costs array generated above and fill the dp[m+1][k+1] table to find the minimum total cost. The time complexity for this part remains O(n*k).

• Step 1: Pre-calculate Subarray Costs (Efficiently)
  1. This step is identical to Approach 2. Use the sliding window with two heaps to compute the costs array in O(n log x) time.
• Step 2: Space-Optimized Dynamic Programming
  1. Observe that dp[i][j] only depends on dp values from column j and j-1.
  2. Instead of a 2D DP table, use two 1D arrays: dp_prev (for j-1 subarrays) and dp_curr (for j subarrays), each of size n-x+2.
  3. Iterate j from 1 to k.
  4. In each iteration, compute dp_curr using values from dp_prev and the already computed values of dp_curr. dp_curr[i] = min(dp_curr[i-1], costs[i-1] + dp_prev[i-x]).
  5. After the inner loop over i completes, dp_curr becomes the new dp_prev for the next j.
  6. The final answer is the last element of the DP array after k iterations.

Detailed Steps:

1. Efficient Cost Calculation: This step is identical to Approach 2. We generate the costs array in O(n log x) time using the sliding window median technique.

2. Space-Optimized Dynamic Programming: The key insight is that the DP state dp[i][j] only depends on dp[i-1][j] and dp[i-x][j-1]. This means we only need to store the DP results for the current number of subarrays (j) and the previous one (j-1).

   We use two 1D arrays, dp_prev and dp_curr, of size m+1 (where m = n-x+1).

   • dp_prev stores the minimum costs for selecting j-1 subarrays.
   • dp_curr is used to compute the minimum costs for selecting j subarrays.

   The outer loop runs from j = 1 to k. In each iteration, we compute dp_curr using dp_prev. At the end of the iteration, the content of dp_curr is copied to dp_prev to be used in the next iteration.

   1public long minOperations(int[] nums, int x, int k) {2    int n = nums.length;3    int m = n - x + 1;4    // Assume costs array is calculated efficiently as in Approach 25    long[] costs = calculateAllCosts(nums, x); 6 7    long[] dpPrev = new long[m + 1]; // for j-1 subarrays8    long[] dpCurr = new long[m + 1]; // for j subarrays9    long INF = Long.MAX_VALUE / 2;10 11    // dpPrev is implicitly all 0s, for the base case j=0.12 13    for (int j = 1; j <= k; j++) {14        Arrays.fill(dpCurr, INF);15        for (int i = 1; i <= m; i++) {16            // Option 1: Don't take subarray starting at i-117            long option1 = dpCurr[i - 1];18            19            // Option 2: Take subarray starting at i-120            long option2 = INF;21            if (i - x >= 0) {22                if (dpPrev[i - x] != INF) {23                    option2 = costs[i - 1] + dpPrev[i - x];24                }25            } else { // This can only be the first subarray chosen26                if (j == 1) {27                    option2 = costs[i - 1];28                }29            }30            dpCurr[i] = Math.min(option1, option2);31        }32        // Prepare for next iteration: current becomes previous33        System.arraycopy(dpCurr, 0, dpPrev, 0, m + 1);34    }35 36    return dpPrev[m];37}38// calculateAllCosts function from Approach 2 would be needed here.