## Introduction to Algorithm Interview Questions and Answers

Preparing for a job interview in Algorithm. I am sure you want to know the most common 2020 Algorithm Interview Questions and Answers that will help you crack the Algorithm Interview with ease. Below is the list of top Algorithm Interview Questions and answers at your rescue.

Below is the list of 2020 Algorithm Interview Questions and answers, which can be asked during an Interview for fresher and experience. These top interview questions are divided into two parts are as follows:

### Part 1 – Algorithm Interview Questions and Answers (Basic)

This first part covers basic Interview Questions and Answers.

#### 1. Write an algorithm to reverse a string. For example, if my string is “vahbunA” then my result will be “Anubhav”.

**Answer:**

**Step 1:**Start**Step 2:**Take two variable I and j.**Step 3:**j is positioned on last character (Technically we can do this by length(string)-1)**Step 4:**I is positioned on first Character (we can do this by string[0])**Step 5:**String[i] is interchanged String[j]**Step 6:**Increment I by 1**Step 7:**Increment J by 1**Step 8:**If ‘I’ > ‘j’ then go to step 3**Step 9:**Stop

#### 2. Write an algorithm to insert a node in the linked list assuming the linked list is sorted already?

**Answer:**

**Case 1:** If linked list is empty then make the node as head and return it.

**Code: **

`New_node-> Next = head;`

head = New_node

**Case 2:** Insert node in middle

**Code: **

`While (P!= insert_position)`

{

P= p->Next;

}

Store_next= P->Next;

P->Next = New_Node;

New_Node->Next = Store_next;

**Case 3:** Insert a node at the end

**Code:**

`While (P->next!= null)`

{

P= P->Next;

}

P->Next = New_Node;

New_Node->Next = null;

#### 3.Write an algorithm for bubble sort?

**Answer:**We are going to implement the bubble sort algorithm through C language.

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**Step 1:** Repeat Step 2 and step 3

`for I = 1 to 10`

**Step 2:**

`Set j = 1`

**Step 3:** Repeat

`while j <= n (Where n is number of elements in the array)`

{ If a[i] < a[j]
Then interchange a[i] and a[j]
[End of if]
}

Set j = j + 1

[End of inner loop]
[End of step 1 outer loop]

**Step 4:**

`Exit`

#### 4.Write an algorithm for Heapsort?

**Answer:**

**Step 1:**Since the tree satisfies the max-Heap property, then the largest item is stored at the root node.**Step 2:**Remove the root element and put at the end of the array (nth position) put the last item of the tree (heap) at the vacant place.**Step 3:**Reduce the size of the heap by 1 and heap the root element again so that we have the highest element at the root.**Step 4:**The process is repeated until all the items of the list is sorted.

### Part 2 – Algorithm Interview Questions and Answers (Advanced)

Let us now have a look at the advanced Interview Questions.

#### 5. Write an algorithm for the Fibonacci search?

**Answer:**

**Step 1:** A is

`sorted_int_array;`

**Step 2:** take one variable c

**Step 3:**

`Fib2 = 1, Fib1 = 1 and fib = 2`

**Step 4:**

`While fib < n do (where n is number of element in list)`

**Step 5:** Assign the variable

`Fib2 = Fib1`

Fib1 = Fib

Fib = Fib1 + Fib2

End while

**Step 6:** Assign the value to temporary variable

`I = 0, offset = 0;`

**Step 7:**

`While Fib > 1 do`

I = min (offset + Fib2, n)

If c < A[i] then

Fib = Fib2

Fib1 = Fib1 – Fib2

Fib2 = Fib – Fib1

Else if c > A[i] then

Fib = Fib1;

Fib1 = Fib2;

Fib2 = Fib – Fib1;

Offset = I;

Else

Return true

End if

End while

Return false

#### 6. Write an algorithm of push and pop operation in the stack?

**Answer:**

For Push Operation

Procedure Add( Item, Stack, N, Top)

(Insert ‘Item’ into the ‘stack’ of maximum size ‘n’, top is the number of elements currently in ‘Stack’)

**Step 1:** Check Stack is overflow?

`If (Top >= N)`

Stack is overflow

Exit

**Step 2:** If the stack does not overflow, increment the loop

`Top = Top + 1`

**Step 3:** Insert the element

`Stack [Top] = Item`

**Step 4:**

`Exit`

**For POP Operation:**

**Step 1:** Check Stack is underflow means empty

`If (Top <= 0)`

Stack is empty

Exit

**Step 2:** If stack is not underflow, then deleting the element

`Item = stack[top]`

**Step 3:** decrementing the top value

`Top = Top – 1`

**Step 4:**

`Exit`

#### 7. Write an algorithm for insert and delete operation in Queue?

**Answer:**

For insertion operation

Procedure add(queue, F, R, N, item)

(This will insert ‘item’ in the ‘queue’ after ‘R’(rare) where ‘n’ is the size of the array.)

**Step 1:** Check Queue is overflow means the queue is full

`If (R >= N)`

Queue is full

Exit

**Step 2:** If the Queue is not overflowed, then increment the loop

`R = R + 1`

**Step 3:** Insert an element into the queue

`Queue[R] = item`

**Step 4:** Setting the ‘F’(Front) Pointer

`If ( F = 0)`

F = 1

Exit

For Deletion operation in Queue

Procedure delete(queue, F, R, item)

(Delete ‘item’ from the ‘stack’, ‘F’ is the Front-end pointer and ‘R’ is the rare end pointer.

**Step 1:** Check Queue is underflow means empty

`If ( R < = 0)`

Queue is empty

Exit

**Step 2:** Deleting an element from the queue

`Item = queue [F]`

**Step 3:** Incrementing the value of F

`F = F+1`

**Step 4:** Checking the empty queue

`If (F > R)`

Then F = R = 0

Exit

#### 8. Write an algorithm to find the minimum depth of a binary tree?

**Answer:**

Let “node” be the pointer to the root node of a subtree.

**Step 1:**If the node is equal to Null, return 0**Step 2:**If the node is a leaf node return 1.**Step 3:**Recursively, find the minimum depth of left and right sub-tree, let it be left Min Depth and right min depth respectively.**Step 4:**To get the minimum height of tree rooted at the node, we will take a minimum of left min depth and right min depth and 1 for the root node.

**Program:**

Procedure minDepth ( Node )

**Step 1:**

`if ( root = null)`

Return 0

**Step 2:**

`if (root -> Left = Null and root -> right = Null )`

Return 1

**Step 3:**

`if (root -> left is not null)`

Return minDepth (root -> right ) + 1;

**Step 4:**

`If (root -> Right is not null )`

Return minDepth ( root -> left) + 1;

**Step 5:**

`return min(minDepth (root -> left), minDepth (root -> right)) + 1`

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