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3 coding questions to be solved in 60 minutes




knightPosition: {3,4}
targetPosition: {2,1}

The knight can move from position (3,4) to positions (1,3), (2,2) and (4,2). Position (4,2) is selected and the ‘stepCount’ becomes 1. From position (4,2), the knight can directly jump to the position (2,1) which is the target point and ‘stepCount’ becomes 2 which is the final answer.
1. The coordinates are 1 indexed. So, the bottom left square is (1,1) and the top right square is (N, N).
2. The knight can make 8 possible moves as given in figure 1.
3. A Knight moves 2 squares in one direction and 1 square in the perpendicular direction (or vice-versa).
This problem can be seen as the shortest path in an unweighted graph. Therefore we use BFS to solve this problem. We try all 8 possible positions that a Knight can reach from its position. If the reachable position is not already visited and is inside the board, we push this state into the queue with a distance of 1 more than its parent state. Finally, we return the distance of the target position, when it gets popped out from the queue



The naive solution for this problem is to generate all subsequences of the given sequence and find the longest palindromic subsequence.



• The left subtree of a node contains only nodes with data less than the node’s data.
• The right subtree of a node contains only nodes with data greater than the node’s data.
• Both the left and right subtrees must also be binary search trees.

Level 1:
All the nodes in the left subtree of 4 (2, 1, 3) are smaller
than 4, all the nodes in the right subtree of the 4 (5) are
larger than 4.
Level 2 :
For node 2:
All the nodes in the left subtree of 2 (1) are smaller than
2, all the nodes in the right subtree of the 2 (3) are larger than 2.
For node 5:
The left and right subtrees for node 5 are empty.
Level 3:
For node 1:
The left and right subtrees for node 1 are empty.
For node 3:
The left and right subtrees for node 3 are empty.
Because all the nodes follow the property of a binary search tree, the above tree is a binary search tree.
Find the first greater value on right side of current node.
Let the index of this node be j. Return true if following
conditions hold. Else return false
(i) All values after the above found greater value are
greater than current node.
(ii) Recursive calls for the subarrays pre[i+1..j-1] and
pre[j+1..n-1] also return true.
Firstly interviewer asked me about myself. Then 2 coding questions were asked.



For the binary trees in the image below.

The left tree in the image is not a complete binary tree that’s why it is invalid and the right tree in the image is a valid complete binary tree which contains total 6 nodes.
The idea is to first traverse the tree. Since we need to see the maximum value in the given path, the pre-order traversal is used to traverse the given binary tree. While traversing the tree, we need to keep the track of the maximum value of the node that we have seen so far. If the current node is greater or equal to the max value, then increment the count of the visible node and update the max value with the current node value.



1. Constructor:
It initializes the data members(queues) as required.
2. push(data) :
This function should take one argument of type integer. It pushes the element into the stack and returns nothing.
3. pop() :
It pops the element from the top of the stack and, in turn, returns the element being popped or deleted. In case the stack is empty, it returns -1.
4. top :
It returns the element being kept at the top of the stack. In case the stack is empty, it returns -1.
5. size() :
It returns the size of the stack at any given instance of time.
6. isEmpty() :
It returns a boolean value indicating whether the stack is empty or not.
Query-1(Denoted by an integer 1): Pushes an integer data to the stack. (push function)
Query-2(Denoted by an integer 2): Pops the data kept at the top of the stack and returns it to the caller. (pop function)
Query-3(Denoted by an integer 3): Fetches and returns the data being kept at the top of the stack but doesn't remove it, unlike the pop function. (top function)
Query-4(Denoted by an integer 4): Returns the current size of the stack. (size function)
Query-5(Denoted by an integer 5): Returns a boolean value denoting whether the stack is empty or not. (isEmpty function)
Operations:
1 5
1 10
2
3
4
Enqueue operation 1 5: We insert 5 at the back of the queue.
Queue: [5]
Enqueue operation 1 10: We insert 10 at the back of the queue.
Queue: [5, 10]
Dequeue operation 2: We remove the element from the front of the queue, which is 5, and print it.
Output: 5
Queue: [10]
Peek operation 3: We return the element present at the front of the queue, which is 10, without removing it.
Output: 10
Queue: [10]
IsEmpty operation 4: We check if the queue is empty.
Output: False
Queue: [10]

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
How do you remove whitespace from the start of a string?