Genc Next

Step 0 : I solved it using backtracking algorithm.
Step 1 : Fix a character in the first position and swap the rest of the character with the first character. Like in ABC, in the first iteration three strings are formed: ABC, BAC, and CBA by swapping A with A, B and C respectively.
Step 2 : Repeat step 1 for the rest of the characters like fixing second character B and so on.
Step 3 : Now swap again to go back to the previous position. E.g., from ABC, we formed ABC by fixing B again, and we backtrack to the previous position and swap B with C. So, now we got ABC and ACB.
Step 4 : Repeat these steps for BAC and CBA, to get all the permutations.

Algorithm :
1) Define a string.
2) Fix a character and swap the rest of the characters.
3) Call the generatePermutation() for rest of the characters.
4) Backtrack and swap the characters again.

Step 0 : To find out the maximum width of the binary tree. The width of the binary tree is the number of nodes present in any level. So, the level which has the maximum number of nodes will be the maximum width of the binary tree. To solve this problem, traverse the tree level-wise and count the nodes in each level.
Step 1 : Define Node class which has three attributes namely: data left and right. Here, left represents the left child of the node and right represents the right child of the node.
Step2 : When a node is created, data will pass to data attribute of the node and both left and right will be set to null.
Step 3 : Define another class which has an attribute root.
Root represents the root node of the tree and initializes it to null.
Step 4 : findMaximumWidth() will find out the maximum width of the given binary tree:
     4.1)Variable maxWidth will store the maximum number of nodes present in any level.
     4.2)The queue is used for traversing binary tree level-wise.
     4.3)It checks whether the root is null, which means the tree is empty.
     4.4)f not, add the root node to queue. Variable nodesInLevel keeps track of the number of nodes in each level.
     4.5)If nodesInLevel > 0, remove the node from the front of the queue and add its left and right child to the queue. For the first iteration,               node 1 will be removed and its children nodes 2 and 3 will be added to the queue. In the second iteration, node 2 will be removed, its               children 4 and 5 will be added to the queue and so on.
     4.6)MaxWidth will store max(maxWidth, nodesInLevel). So, at any given point of time, it will represent the maximum number of nodes.
     4.7)This will continue till all the levels of the tree is traversed.

Tip 1 : Try to give relevant examples as well.
Tip 2 : Study and go through basics of OS.

Tip 1 : Study and go through basics of DBMS.

I first used the iterative method to solve.
Iterative way :
1) Initialize start and end indexes as start = 0, end = n-1 
2) In a loop, swap arr[start] with arr[end] and change start and end as follows : 
start = start +1, end = end – 1
Then interviewer asked to code recursively. Then i used recursive method
Recursive Way :
1) Initialize start and end indexes as start = 0, end = n-1 
2) Swap arr[start] with arr[end] 
3) Recursively call reverse for rest of the array.

I used the partioning of array . it goes like :
I use 0 as a pivot element and whenever I see a non zero element I will swap it with the pivot element. So all the non zero element will come at the beginning.

Tip 1 : Be relaxed and try to speak your mind loud.
Tip 2 : Don't panic and remember for some behavioural sometimes there's no right or wrong. So, don't panic. They just want to check your thought process.