Implement Segment Tree with without recursion

Question

Implement a Segment Tree data structure that supports insert, delete, and lookup operations in O(log n) time.

Codemia · Accepted Answer

The Segment Tree is a tree data structure that allows for efficient range queries and updates on an array. Here, we require a non-recursive implementation that supports insert, delete, and lookup operations in O(log n) time. This data structure is particularly useful for problems involving dynamic data where we need to perform multiple range queries and updates. The pattern we will use here is a binary tree representation, allowing us to divide the input array into segments, which will enable efficient querying and updating. 1. **Initialization**: Create a Segment Tree where each node represents a segment of the array. The tree will be built based on the size of the input array, and the tree's height will be log(n). 2. **Insert Operation**: To insert a value, we will find the correct position in the tree and update the corresponding segment. For example, if inserting a value at index 3, we navigate through the tree, adjusting the relevant nodes' values to reflect this insertion. 3. **Delete Operation**: To delete a value, similarly navigate through the tree to find the node corresponding to the value and remove it, updating the necessary segments. 4. **Lookup Operation**: For lookup, traverse the tree to find the value in the required segment and return it. 5. **Example**: Given an array [1, 3, 5, 7, 9, 11], inserting 6 would lead to adjusting nodes corresponding to segments that include the index of 6. ```python class SegmentTree: def __init__(self, data): self.n = len(data) self.tree = [0] * (2 * self.n) self.build(data) def build(self, data): # Build the tree for i in range(self.n): self.tree[self.n + i] = data[i] for i in range(self.n - 1, 0, -1): self.tree[i] = self.tree[i << 1] + self.tree[i << 1 | 1] def update(self, index, value): # Update the value at index pos = index + self.n self.tree[pos] = value while pos > 1: pos >>= 1 self.tree[pos] = self.tree[pos << 1] + self.tree[pos << 1 | 1] def delete(self, index): # For simplicity, we will set the value to 0 to represent deletion self.update(index, 0) def lookup(self, left, right): # Lookup the sum in range [left, right] left += self.n right += self.n result = 0 while left < right: if left & 1: result += self.tree[left] left += 1 if right & 1: right -= 1 result += self.tree[right] left >>= 1 right >>= 1 return result ``` The time complexity for the insert, delete, and lookup operations is O(log n) due to the tree structure, where each operation involves traversing the height of the tree. The space complexity is O(n), as we need to store the entire segment tree. Given that the segment tree is represented as an array of size 2n (for a complete binary tree), we ensure that our operations remain efficient even as the data size scales. 1. **Empty Tree**: Inserting into an empty Segment Tree should initialize the structure correctly. 2. **Single Element**: When the tree has only one element, both insert and delete operations should handle this gracefully. 3. **Duplicate Values**: Inserting duplicate values should be handled correctly, allowing for multiple entries. 4. **Range Queries on Empty Segments**: For the lookup operation, if the specified range is empty or invalid, the implementation should return 0 or handle the case without errors.

Implement Segment Tree with without recursion

Slack

What the Interviewer Expects

Key Topics to Cover

How to Approach This

Possible Follow-up Questions

Sharpen Your Skills on Codemia

Sample Answer

Problem Analysis

Approach

Submit Your Answer

Slack Software Engineer Interview Guide

Related Questions

Binary Search on graph

Count kth largest element in array

Find connected components in string

Merge Three Sorted strings

Binary Search on string