Which of the following is not the application of Divide and Conquer technique?
Show Hint
- Quick Sort
- Strassen's Matrix Multiplication
- Linear Search
- Binary Search
The Correct Option is C
Solution and Explanation
Step 1: Grasp the Divide and Conquer paradigm.
The Divide and Conquer technique decomposes a problem into smaller, manageable subproblems. These subproblems are solved individually, and their solutions are then integrated to address the original problem.
Step 2: Assess each presented option.
- Quick Sort: This sorting algorithm employs the Divide and Conquer strategy by dividing the array into smaller partitions.
- Strassen's Matrix Multiplication: This algorithm leverages Divide and Conquer for efficient matrix multiplication.
- Linear Search: This search algorithm sequentially examines elements and does not utilize the Divide and Conquer approach.
- Binary Search: This technique employs Divide and Conquer by reducing the search space by half in each iteration.
Step 3: Determination.
The accurate choice is (3) Linear Search because it does not incorporate the Divide and Conquer methodology.
Top Questions on Algorithms
- Consider an unordered list of \(N\) distinct integers. What is the minimum number of element comparisons required to find an integer in the list that is NOT the largest in the list?
- An array of 25 distinct elements is to be sorted using quicksort. Assume that the pivot element is chosen uniformly at random. The probability that the pivot element gets placed in the worst possible location in the first round of partitioning (rounded off to 2 decimal places) is __________
Let 𝐺(𝑉, 𝐸) be a simple, undirected, edge-weighted graph with unique edge weights.
Which of the following statements about the minimum spanning trees (MST)
of 𝐺 is/are true?
Let 𝐺(𝑉, 𝐸) be an undirected, edge-weighted graph with integer weights. The weight
of a path is the sum of the weights of the edges in that path. The length of a path is
the number of edges in that path.
Let 𝑠∈𝑉 be a vertex in 𝐺. For every 𝑢∈𝑉 and for every 𝑘 ≥0, let 𝑑𝑘(𝑢) denote
the weight of a shortest path (in terms of weight) from 𝑠 to 𝑢 of length at most 𝑘. If
there is no path from 𝑠 to 𝑢 of length at most 𝑘, then 𝑑𝑘(𝑢) = ∞.
Consider the statements:
S1:
For every 𝑘 ≥0 and 𝑢 ∈𝑉, 𝑑𝑘+1(𝑢) ≤𝑑𝑘(𝑢).
S2:
For every (𝑢, 𝑣) ∈𝐸, if (𝑢, 𝑣) is part of a shortest path (in terms of
weight) from 𝑠 to 𝑣, then for every 𝑘≥ 0, 𝑑𝑘(𝑢) ≤𝑑𝑘(𝑣).
Which one of the following options is correct?- Want to practice more? Try solving extra ecology questions todayView All Questions