SDE - 1

1). First make a queue of pair of coordinates
2). Insert all those coordinates where value is 2
3). Run BFS on 2D Grid 
4). Then run a while loop and correspondingly make the movement from one cell to another only if value at next cell is 1 and the movement doesn't cross matrix boundaries and then make that value to 2, and corresponding increase the time value by 1 in each loop.
5). At the end, if we are left with atleast cell where value is 1 then return -1, else return time.

1). I used recursion here to generate all root to left paths of the given Binary Tree.
2). The base condition for this recursive function was when we hit the leaf node, we need to check whether the sum equals to targetSum, if yes then insert the path vector into the answer.
3). Once you traverse the whole binary tree, you'll get your resultant root-to-leaf nodes.

1). Here, l used a stack for checking the validity of parentheses, and later remove the indexes of invalid parentheses from the string s. 

2). First, iterate the string s and mark the index of those characters which need to be removed to make it parentheses string using a special symbol '#'.

Here, a stack is used for finding the valid pair of parentheses, and while doing so also mark the indexes of invalid parentheses in s.

3). Finally, iterate s again and append non-marked symbol (#) to ans.

1). Here, we hash the indices of all elements in a hashMap. In case of repeated elements, the last occurrence index would be stored in hashMap.

2). Here also we fix a number (num[i]), by traversing the loop. But the loop traversal here for fixing numbers would leave the last two indices. These last two indices would be covered by the nested loop.

3). If the number fixed is +ve, break there because we can't make it zero by searching after it.
Make a nested loop to fix a number after the first fixed number. (num[j])

4). To make sum 0, we would require the -ve sum of both fixed numbers. Let us say this required.
Now, we will find the this required number in hashMap. 

5). If it exists in hashmap and its last occurrence index > 2nd fixed index, we found our triplet. Push it in answer vector.

6). Update j to last occurence of 2nd fixed number to avoid duplicate triplets.
7). Update i to last occurence of 1st fixed number to avoid duplicate triplets.
8)> Return answer vector.

1). I used the Recursive here.
2). If current node value is greater than low, and less than right, then update the left and right child node of cur node by calling the recursion function for their left and right child individually.
3). If current node value is less than low, then call recursive function for right child.
4). Else call recursive function for left child.

Tip 1 : Understand the problem and think rationally taking into consideration the trade-offs that you would be making in choosing the one than the other.
Tip 2 : Always speak it out loud and make sure that you and the interviewer are on the same page.
Tip 3 : Always asked and clarify the requirements from the interviewer beforehand.