Buttercola: Buddy System

Buttercola: Buddy System
Given a complete binary tree with nodes of values of either 1 or 0, the following rules always hold:
1. A node's value is 1 iff and only if all its subtree nodes' value are 1
2. A leaf node can have value either 1 or 0.

Implement the following 2 APIs:
set_bit(offset, length), set the bits at range from offset to offset + length - 1
clear_bit(offset, length), clear the bits at range from offset to offset + length - 1

这是一个二维数组，不是一个heap(heap就是说这个树存在一个一维数组里，满足A的child是A[2i+1]和A[2i+2])，问题背景起源于内存分配问题
比如有N个level，第一个level有一个bit，第二个level有2个bit，第三个level有四个bit，调用第x个level的第y个bit直接用A[x][y]. 鍥磋鎴戜滑@1point 3 acres
那么A[x][y]的孩子包括A[x+1][2y] 和 A[x+1][2y+1]
题目要求完成的是：
例如ClearBits(A, 4,9) => 把第N个level的第4位到第9位清0. 当child清0之后， parent也要跟着清0，一直到root
e.g.                 0

                    /       \
                  0         0
                 /   \      /   \
                1   0    1    0
               / \   / \   / \   / \
              1 1 1 0 1 1 0 0  

After calling the clear_bit(1, 3), the binary tree becomes

                        0

                    /       \
                  0         0
                 /   \      /   \
                0   0    1    0
               / \   / \   / \   / \
              1 0 0 0 1 1 0 0  


For this tree, after call set_bit(1, 5), the binary tree becomes:


Understand the problem:

What's a buddy system? Here the link gives good explanation:
https://www.cs.fsu.edu/~engelen/courses/COP402003/p827.pdf

  private int[][] bits;

  private int n;

  private int numNodesLastLevel;

   

  public Solution (int n) {

    this.n = n;

    this.numNodesLastLevel = (int) Math.pow(2, n - 1);

    bits = new int[n][numNodesLastLevel];

     

     

  }

   

  public void clearBits(int offset, int len) {

    if (offset < 0 || offset + len > numNodesLastLevel) {

      throw new IndexOutOfBoundsException();

    }

    clearBitsHelper(n - 1, offset, offset + len - 1);

  }

   

  private void clearBitsHelper(int level, int start, int end) {

    if (level < 0 || start > end) {

      return;

    }

     

    boolean containsOne = setToZero(level, start ,end);

    if (containsOne) {

      clearBitsHelper(level - 1, start / 2, end / 2);

    }

  }

   

  public void setBits(int offset, int len) {

    if (offset < 0 || offset + len > numNodesLastLevel) {

      throw new IndexOutOfBoundsException();

    }

    setBitsHelper(n - 1, offset, offset + len - 1);

  }

   

  private void setBitsHelper(int level, int start, int end) {

    if (level < 0 || start > end) {

      return;

    }

     

    setToOne(level, start, end);

     

    // if start index is odd

    if ((start & 1) == 1) {

      start--;

    }

     

    // if end index is even

    if ((end & 1) == 0) {

      end++;

    }

     

    // determine the start position of the next level

    int nextStart = Integer.MAX_VALUE;

    for (int i = start; i < end; i += 2) {

      if (bits[level][i] == 1 && bits[level][i + 1] == 1) {

        nextStart = i / 2;

        break;

      }

    }

     

    // determien the end position of the next level

    int nextEnd = Integer.MIN_VALUE;

    for (int i = end; i > start; i -= 2) {

      if (bits[level][i] == 1 && bits[level][i - 1] == 1) {

        nextEnd = i / 2;

        break;

      }

    }

     

    setBitsHelper(level - 1, nextStart, nextEnd);

  }

   

  private void setToOne(int level, int start, int end) {

    for (int i = start; i <= end; i++) {

      bits[level][i] = 1;

    }

  }

   

  private boolean setToZero(int level, int start, int end) {

    boolean containsOne = false;

    for (int i = start; i <= end; i++) {

      if (bits[level][i] == 1) {

        containsOne = true;

      }

       

      bits[level][i] = 0;

    }

     

    return containsOne;

  }

   

   

   

  private void printTree() {

    int nodes = 1;

    for (int[] level : bits) {

      for (int i = 0; i < nodes; i++) {

        System.out.print(level[i] + ", ");

      }

       

      System.out.println("");

      nodes *= 2;

    }

    System.out.println("");

  }

   

  public static void main(String[] args) {

    Solution sol = new Solution(4);

     

    sol.setBits(1, 5);

    sol.printTree();

     

    sol.clearBits(2, 6);

    sol.printTree();

  }

Several thoughts and optimizations:
1. Why we set and clear bits in level order instead of heap order? That is, for each bit, we set and update its parent and so on, then repeat for the next bit. T

hat is because data is stored continuous in one level of tree. So if we read one level and process then move to its parent level, it can achieve better data locality. 

2. For the clear_bits(), actually we don't have to repeat until we reach to the root. We can stop the recursion if all bits in the current level is zero. That's a prune of a recursion. 
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