In binary heap, whenever the root is removed then the rightmost element of last level is replaced by the root. Why?

Data Structures Algorithms - Quiz(MCQ)

It is the easiest possible way.

Because left and right subtree might be missing.

To make sure that it is still complete binary tree.

None of the above

Correct Answer : Option (C) : To make sure that it is still complete binary tree.

Explanation : A binary heap (whether max or min) has to satisfy the property of complete binary tree at all times.

Recently Updated in Data Structures Algorithms Questions

Consider the following code snippet:

void solve(vector<int> &a) { int queries; cin >> queries; while(queries--) { int type; cin >> type; if(type == 1) { int index, value; cin >> index >> value; update(a, index, value); } else { int l, r; cin >> l >> r; cout << getXOR(a, l, r) << endl; } } }

The update() function updates the element at the given index in the array to some given value. The getXOR() function returns the XOR of the elements in the array a, in the range [l, r]. Which of the following data structures can perform the above tasks optimally?

Tries.

Stacks

Prefix XOR Arrays.

Segment Trees.

Correct Answer : Option (D) : Segment Trees.

Consider we have a function, getLCA(), which returns us the Lowest Common Ancestor between 2 nodes of a tree. Using this getLCA() function, how can we calculate the distance between 2 nodes, given that distance from the root, to each node is calculated?

dist(u) + dist(v)

dist(u) + dist(v) - 2 * dist(getLCA(u, v))

dist(u) + dist(v) - dist(getLCA(u, v))

dist(u) + dist(v) + 2 * dist(getLCA(u, v))

Correct Answer : Option (B) : dist(u) + dist(v) - 2 * dist(getLCA(u, v))

Explanation : The distance between 2 nodes, can be broken down into the sum of distances from the root, to each of the nodes. Observe, that in each of these paths, the path from the root to the LCA comes in each of them. So, this needs to be removed from the answer, and hence we get our formula in Option (B).

What will be the output of the following code snippet?

void solve() { deque<int> dq; for(int i = 1; i <= 5; i++) { if(i % 2 == 0) { dq.push_back(i); } else { dq.push_front(i); } } for(auto x: dq) { cout << x << " "; } cout << endl; }

1 2 3 4 5

1 3 5 2 4

5 3 1 2 4

5 4 3 2 1

Correct Answer : Option (C) : 5 3 1 2 4