Explain the rotation operations used in AVL trees and their significance in maintaining balance.
- Primary and secondary rotations; Primary rotations adjust immediate subtrees, while secondary rotations modify distant subtrees.
- Simple and complex rotations; Simple rotations involve basic adjustments, while complex rotations involve intricate reconfigurations.
- Single and double rotations; Single rotations involve left or right rotations, while double rotations involve combinations of single rotations.
- Triple and quadruple rotations; Triple rotations involve three consecutive rotations, while quadruple rotations involve four rotations simultaneously.
Rotation operations used in AVL trees are single and double rotations. Single rotations include left rotations and right rotations, which help maintain balance by adjusting the heights of subtrees. Double rotations are combinations of single rotations performed to restore balance in specific cases, such as the double rotation involving left-right or right-left rotations.
Selection sort's time complexity can be improved to _______ by implementing certain optimizations.
- O(log n)
- O(n log n)
- O(n)
- O(n^2)
Selection sort's time complexity can be improved to O(n log n) by implementing certain optimizations, such as using more advanced data structures or algorithms to perform the selection in a more efficient manner.
Merge sort is a _______ sorting algorithm that follows the _______ strategy.
- Bubble
- Divide and Conquer
- Dynamic Programming
- Greedy
Merge sort is a Divide and Conquer sorting algorithm that follows the Divide and Conquer strategy. It recursively divides the array into two halves, sorts them, and then merges them back together.
Explain why binary search is more efficient than linear search for large datasets.
- Binary search always finds the element in the first comparison
- Binary search can only be used with small datasets
- Binary search divides the search space in half at each step, reducing the time complexity to O(log n)
- Linear search has a time complexity of O(n^2)
Binary search is more efficient for large datasets because it divides the search space in half at each step, resulting in a time complexity of O(log n), which is significantly faster than linear search (O(n)).
BFS guarantees finding the shortest path in an unweighted graph because it explores nodes in _______ order.
- Increasing
- Lexicographical
- Non-decreasing
- Non-increasing
BFS guarantees finding the shortest path in an unweighted graph because it explores nodes in increasing order. As it systematically traverses nodes level by level, the first time a node is encountered, it is reached through the shortest path.
How can the longest common substring problem be extended to handle multiple strings?
- Apply the algorithm separately to each pair of strings and combine the results.
- Extend dynamic programming to a multidimensional array to account for multiple strings.
- Longest common substring problem cannot be extended to handle multiple strings.
- Utilize greedy algorithms to find common substrings among multiple strings.
To handle multiple strings in the longest common substring problem, dynamic programming can be extended to a multidimensional array. This array helps store the common substrings for each pair of strings, and the results can then be combined.
The Longest Increasing Subsequence problem can be efficiently solved using _______.
- Binary Search
- Bubble Sort
- Depth-First Search
- QuickSort
The Longest Increasing Subsequence (LIS) problem can be efficiently solved using Binary Search. The binary search approach allows us to find the length of the LIS in an optimized way, reducing the time complexity.
Can LCS be applied to strings of different lengths? Why or why not?
- No, because it can only be applied to arrays, not strings.
- No, because it only works on strings of equal lengths.
- Yes, as long as the algorithm is modified to handle different lengths.
- Yes, without any modification.
Yes, the longest common subsequence (LCS) algorithm can be applied to strings of different lengths. It involves modifying the dynamic programming approach to handle the differences in lengths by considering all possible pairs of substrings and building the LCS table accordingly.
In the Fractional Knapsack Problem, items can be divided to fit into the knapsack partially, whereas in the 0/1 Knapsack Problem, items must be chosen _______.
- Arbitrarily
- Completely
- Exponentially
- Sequentially
In the 0/1 Knapsack Problem, items must be chosen completely, meaning either an item is included in its entirety or not at all. On the other hand, the Fractional Knapsack Problem allows items to be divided and included partially.
The time complexity of radix sort is _______ in most scenarios.
- O(k * n)
- O(n * log n)
- O(n + k)
- O(n^2)
The time complexity of radix sort is O(k * n), where 'k' is the number of digits or components in the keys, and 'n' is the number of elements. It is linear and often more efficient.