Minimum Flips in Binary Tree to Get Result

Hard

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Topics:

TreesDynamic ProgrammingRecursion

You are given the root of a binary tree that consists of nodes with values 0 or 1. You are also given an integer result, which is the desired bitwise OR of all values in the binary tree.

You can flip the value of any node in the binary tree. A flip changes 0 to 1 or 1 to 0.

Return the minimum number of flips required to make the bitwise OR of all node values in the tree equal to result. If it is impossible to make the bitwise OR equal to result, return -1.

Example 1:

Input: root = [1,1,0], result = 1
Output: 0
Explanation: The bitwise OR of all node values is 1. The result we want equals the bitwise OR, so we don't have to flip any nodes.

Example 2:

Input: root = [1,1,0], result = 0
Output: 2
Explanation: We have to flip the values of the root and one of its children to make the bitwise OR equal to 0.

Example 3:

Input: root = [0,0,0], result = 1
Output: 3
Explanation: We have to flip the value of each node to make the bitwise OR equal to 1.

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
The value of each node is 0 or 1.
result is either 0 or 1.

Try coding it in LeetCode

Minimum Flips in Binary Tree to Get Result

Hard

Asked by:

5 views

Topics:

TreesDynamic ProgrammingRecursion

You are given the root of a binary tree that consists of nodes with values 0 or 1. You are also given an integer result, which is the desired bitwise OR of all values in the binary tree.

You can flip the value of any node in the binary tree. A flip changes 0 to 1 or 1 to 0.

Return the minimum number of flips required to make the bitwise OR of all node values in the tree equal to result. If it is impossible to make the bitwise OR equal to result, return -1.

Example 1:

Input: root = [1,1,0], result = 1
Output: 0
Explanation: The bitwise OR of all node values is 1. The result we want equals the bitwise OR, so we don't have to flip any nodes.

Example 2:

Input: root = [1,1,0], result = 0
Output: 2
Explanation: We have to flip the values of the root and one of its children to make the bitwise OR equal to 0.

Example 3:

Input: root = [0,0,0], result = 1
Output: 3
Explanation: We have to flip the value of each node to make the bitwise OR equal to 1.

Constraints:

The number of nodes in the tree is in the range [1, 10⁵].
The value of each node is 0 or 1.
result is either 0 or 1.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What are the possible values for the nodes in the binary tree, and what data type are they?
If the target result is unachievable through flips, what should I return?
Can the binary tree be empty or contain only a single node?
Is the operator associated with each internal node guaranteed to be either 'AND' or 'OR', or could there be other operators?
What is the structure of the binary tree input; is it a standard binary tree represented as a nested data structure, and what is the maximum depth it could have?

Brute Force Solution

Approach

The brute force strategy involves exploring every possible combination of flips at each node in the binary tree. For each combination, we evaluate the entire expression tree to see if it produces the desired result. We then choose the combination with the fewest number of flips that satisfies the desired output.

Here's how the algorithm would work step-by-step:

Start by considering the root node of the tree.
Imagine you can either flip the value of the root node or not flip it.
For each of these two possibilities (flip or no flip), move to the left child.
Again, for the left child, consider the two possibilities: flip or no flip.
Repeat this process for every node in the tree, always considering the two possibilities: flip or no flip.
Each time you reach the end of a branch (a leaf node), you've created one possible 'world' where some nodes are flipped and others aren't.
Evaluate the entire expression tree for that specific 'world' to see what final result you get (0 or 1).
Keep track of how many flips it took to create that particular 'world'.
Do this for *every* single possible combination of flips in the entire tree.
After checking every possibility, find the one that gives you the desired result (0 or 1) with the fewest number of flips.

Code Implementation

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

def minimum_flips_brute_force(root, desired_result):
    minimum_number_of_flips = float('inf')

    def evaluate_tree(node):
        if node is None:
            return None
        if node.left is None and node.right is None:
            return node.value

        left_value = evaluate_tree(node.left)
        right_value = evaluate_tree(node.right)

        if node.value == 2: # AND operation
            return left_value & right_value
        else: # OR operation
            return left_value | right_value

    def explore_flips(node, current_flips):
        nonlocal minimum_number_of_flips

        if node is None:
            result = evaluate_tree(root)
            if result == desired_result:
                minimum_number_of_flips = min(minimum_number_of_flips, current_flips)
            return

        # Explore without flipping
        original_value = node.value
        explore_flips(node.left, current_flips)

        # Explore with flipping
        node.value = 1 - node.value if node.value in [0, 1] else node.value

        # This flips operation nodes
        if node.value == 2:
            node.value = 3
        elif node.value == 3:
            node.value = 2

        # We increment the flip count
        explore_flips(node.left, current_flips + 1)
        node.value = original_value

        # Restore original value for the parent node

        if original_value == 2:
            node.value = 2
        elif original_value == 3:
            node.value = 3

        node.value = original_value

        # Recursively traverse the right subtree for both flipped and non-flipped scenarios
        explore_flips(node.right, current_flips)
        # Flip and count increase here
        node.value = 1 - node.value if node.value in [0, 1] else node.value
        if node.value == 2:
            node.value = 3
        elif node.value == 3:
            node.value = 2

        explore_flips(node.right, current_flips + 1)
        node.value = original_value

        if original_value == 2:
            node.value = 2
        elif original_value == 3:
            node.value = 3

        node.value = original_value

    explore_flips(root, 0)

    if minimum_number_of_flips == float('inf'):
        return -1
    else:
        return minimum_number_of_flips

Big(O) Analysis

Time Complexity

O(2^n) – The algorithm explores every possible combination of flips for each node in the binary tree. Since there are n nodes, and each node has two choices (flip or no flip), there are 2^n possible combinations to consider. For each of these combinations, the algorithm evaluates the expression tree, which takes O(n) time in the worst case. Therefore, the overall time complexity is O(n * 2^n), but since the exponential term dominates, it simplifies to O(2^n).

Space Complexity

O(H) – The brute force approach described uses recursion to explore all possible combinations of flips. The maximum depth of the recursion is determined by the height (H) of the binary tree, which represents the maximum number of nested function calls on the call stack at any given time. Each recursive call adds a new frame to the call stack, storing local variables and the return address. Therefore, the auxiliary space complexity is proportional to the height of the tree, as the recursion stack consumes O(H) space. In the worst case (skewed tree), H can be equal to N (the number of nodes), but in a balanced tree, H is log(N).

Optimal Solution

Approach

The best way to solve this is to work from the bottom of the tree upwards. For each part of the tree, we figure out the fewest flips needed to make it result in either a 0 or a 1, based on the desired final result at the very top.

Here's how the algorithm would work step-by-step:

Start by looking at the very bottom pieces of the tree first, the leaves.
For each leaf, see if its value is already what we want it to be. If not, it costs one flip to change it.
Move up to the next level. For each operation (like AND or OR), figure out the cheapest way to make it result in a 0 and the cheapest way to make it result in a 1. Keep track of these costs.
When calculating the costs for an AND or OR, consider the cheapest costs of its children. For instance, if it's an AND, to get a 1, both children need to be 1. See the best costs saved in the step before.
Continue this process, working your way up the tree, always figuring out the cheapest cost to achieve a 0 or 1 at each operation.
Once you reach the root of the tree, look at the cheapest cost to achieve the target value (0 or 1) required by the question. That's your answer!

Code Implementation

def minimum_flips(root, result):    def solve(node):      if not node:        return (float('inf'), float('inf'))
      if node.val == 0 or node.val == 1:
        if node.val == 0:
          return (0, 1)
        else:
          return (1, 0)
      left_zero_cost, left_one_cost = solve(node.left)
      right_zero_cost, right_one_cost = solve(node.right)
      if node.val == 2: # AND operation
        zero_cost = min(left_zero_cost + right_zero_cost, \
                        left_zero_cost + right_one_cost, \
                        left_one_cost + right_zero_cost)
        one_cost = left_one_cost + right_one_cost        return (zero_cost, one_cost)
      elif node.val == 3: # OR operation
        zero_cost = left_zero_cost + right_zero_cost
        one_cost = min(left_zero_cost + right_one_cost, \
                       left_one_cost + right_zero_cost, \
                       left_one_cost + right_one_cost)        return (zero_cost, one_cost)
      elif node.val == 4: # XOR operation
        zero_cost = min(left_zero_cost + right_zero_cost, \
                       left_one_cost + right_one_cost)
        one_cost = min(left_zero_cost + right_one_cost, \
                      left_one_cost + right_zero_cost)
        return (zero_cost, one_cost)
    # Calculate min flips for 0 and 1 at root.
    zero_flips, one_flips = solve(root)
    # Return the minimum flips needed for the final result.
    if result == 1:
      return one_flips
    else:
      return zero_flips

Big(O) Analysis

Time Complexity

O(n) – The algorithm processes each node of the binary tree exactly once in a bottom-up manner. At each node, a constant amount of work is performed to calculate the minimum flips required to achieve 0 and 1 based on the node's type (leaf, AND, OR) and its children's costs. Since there are n nodes in the binary tree, and the work done at each node is constant, the total time complexity is directly proportional to the number of nodes. Therefore, the time complexity is O(n).

Space Complexity

O(N) – The dominant space complexity comes from the recursion stack used during the traversal of the binary tree. In the worst-case scenario, the tree can be skewed (like a linked list), requiring a recursive call for each node, resulting in a recursion depth of N, where N is the number of nodes in the tree. We store intermediate cost to achieve 0 or 1 for each node which could potentially utilize space proportional to the number of nodes as well. Thus the algorithm's auxiliary space is O(N).

Edge Cases

Case	How to Handle
Null or Empty Tree	Return 0 if the tree is null or empty, as no flips are needed.
Single Node Tree	If the single node's value equals the desired result, return 0, otherwise return 1 if flips are allowed, and -1 or error if flips are not allowed in this case.
Tree with only AND nodes and desired result is 1, but all leaves are 0	Return -1 or throw an exception indicating that the target result is unreachable given the tree structure and node values.
Tree with only OR nodes and desired result is 0, but all leaves are 1	Return -1 or throw an exception indicating that the target result is unreachable given the tree structure and node values.
Deeply nested tree exceeding recursion depth limits	Consider iterative solutions or memoization to avoid stack overflow errors with deep trees.
Tree with a very large number of nodes (scalability)	Ensure the chosen approach (e.g., memoization, dynamic programming) has reasonable time and space complexity with respect to the tree size.
Desired result cannot be achieved with given number of flips available	Return -1 or throw exception when the desired value is not reachable based on the value of nodes and logic gates.
Integer overflow during intermediate calculations	Consider using larger data types or modular arithmetic if the number of flips or intermediate calculations could potentially overflow.