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Problem solution · Java

Minimum Operations to Make Elements Within K Subarrays Equal

Minimum Operations to Make Elements Within K Subarrays Equal: a Java solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
100 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Direct simulation

For Minimum Operations to Make Elements Within K Subarrays Equal, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 100 lines of Java from the credited upstream file 3505.java.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 1 loop block detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMinimum Operations to Make Elements Within K Subarrays Equal · JavaJava
Use this to learn the idea, then write your own version.
class MyMap {  public TreeMap<Integer, Integer> map = new TreeMap<>();  public int size = 0;  public long sum = 0;} class Solution {  public long minOperations(int[] nums, int x, int k) {    // minOps[i] := the minimum number of operations needed to make    // nums[i..i + x - 1] equal to the median    List<Long> minOps = getMinOps(nums, x);    Long[][] mem = new Long[nums.length + 1][k + 1];    return minOperations(nums, x, 0, k, minOps, mem);  }   private static final long INF = Long.MAX_VALUE / 2;   // Returns the minimum operations needed to have at least k non-overlapping  // subarrays of size x in nums[i..n - 1].  private long minOperations(int[] nums, int x, int i, int k, List<Long> minOps, Long[][] mem) {    if (k == 0)      return 0;    if (i == nums.length)      return INF;    if (mem[i][k] != null)      return mem[i][k];    final long skip = minOperations(nums, x, i + 1, k, minOps, mem);    final long pick = i + x <= nums.length                          ? minOps.get(i) + minOperations(nums, x, i + x, k - 1, minOps, mem)                          : INF;    return mem[i][k] = Math.min(skip, pick);  }   // Returns the minimum operations needed to make all elements in the window of  // size x equal to the median.  private List<Long> getMinOps(int[] nums, int x) {    List<Long> minOps = new ArrayList<>();    MyMap lower = new MyMap();    MyMap upper = new MyMap();    for (int i = 0; i < nums.length; ++i) {      if (lower.map.isEmpty() || nums[i] <= lower.map.lastKey()) {        lower.map.merge(nums[i], 1, Integer::sum);        lower.sum += nums[i];        ++lower.size;      } else {        upper.map.merge(nums[i], 1, Integer::sum);        upper.sum += nums[i];        ++upper.size;      }      if (i >= x) {        final int outNum = nums[i - x];        if (lower.map.containsKey(outNum)) {          lower.map.merge(outNum, -1, Integer::sum);          if (lower.map.get(outNum) == 0)            lower.map.remove(outNum);          lower.sum -= outNum;          --lower.size;        } else {          upper.map.merge(outNum, -1, Integer::sum);          if (upper.map.get(outNum) == 0)            upper.map.remove(outNum);          upper.sum -= outNum;          --upper.size;        }      }      // Balance the two maps s.t.      // |lower| >= |upper| and |lower| - |upper| <= 1.      if (lower.size < upper.size) {        final int val = upper.map.firstKey();        upper.map.merge(val, -1, Integer::sum);        if (upper.map.get(val) == 0)          upper.map.remove(val);        lower.map.merge(val, 1, Integer::sum);        upper.sum -= val;        lower.sum += val;        --upper.size;        ++lower.size;      } else if (lower.size - upper.size > 1) {        final int val = lower.map.lastKey();        lower.map.merge(val, -1, Integer::sum);        if (lower.map.get(val) == 0)          lower.map.remove(val);        upper.map.merge(val, 1, Integer::sum);        lower.sum -= val;        upper.sum += val;        --lower.size;        ++upper.size;      }      // Calculate operations needed to make all elements in the window equal      // to the median.      if (i >= x - 1) {        final int median = lower.map.lastKey();        final long ops = (median * lower.size - lower.sum) + (upper.sum - median * upper.size);        minOps.add(ops);      }    }    return minOps;  }}

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