DSA & Coding Interview Mock Tests, Syllabus & Complete Exam Guide

What is the output of: def rotate(a, k): n = len(a) k = k % n a[:] = a[n - k:] + a[:n - k] return a print(rotate([1, 2, 3, 4, 5], 2))

• A.[1, 2, 4, 5, 3]
• B.[4, 5, 1, 2, 3]
• C.[3, 4, 5, 1, 2]
• D.[2, 3, 4, 5, 1]

Solution: Right-rotating by k=2 moves the last 2 elements to the front. a[n-k:] is a[3:] = [4, 5] and a[:n-k] is a[:3] = [1, 2, 3], so the result is [4, 5] + [1, 2, 3] = [4, 5, 1, 2, 3].

Which technique detects a cycle in a singly linked list using only O(1) extra space?

• A.Storing every visited node in a hash set (this needs O(n) space)
• B.Sorting the node addresses and scanning for duplicates
• C.Floyd's tortoise-and-hare: advance one pointer by 1 and another by 2 until they meet
• D.Reversing the list twice and comparing

Solution: Floyd's cycle detection moves a slow pointer one step and a fast pointer two steps; if a cycle exists they eventually meet, all in O(1) extra space. A hash set also detects cycles but uses O(n) space, so it fails the constraint.

In a binary search tree (BST), the left subtree of any node contains only values that are:

• A.Greater than the node's value
• B.Less than the node's value
• C.Equal to the node's value
• D.Unordered relative to the node

Solution: The BST invariant says every key in a node's left subtree is smaller than the node, and every key in its right subtree is larger. This ordering is what makes searching, inserting, and deleting efficient.

What is the output of: def fib(n): if n < 2: return n return fib(n - 1) + fib(n - 2) print(fib(6))

• A.6
• B.13
• C.8
• D.5

Solution: The Fibonacci sequence starting at fib(0)=0 is 0, 1, 1, 2, 3, 5, 8, so fib(6) = 8. (This naive recursion recomputes subproblems, which is why memoization matters.)