Tip 1: You should be good at logic building and problem-solving.
Tip 2: Prepare JavaScript in depth, at least to a strong working level.
Tip 1: Your resume should be well documented and clearly structured.
Tip 2: Aim for an ATS score of at least 70 to improve your chances of shortlisting.
The test consisted of 30 MCQs on HTML, CSS, and JavaScript, along with two coding questions.



Input: 'arr' = [1, 2, 7, -4, 3, 2, -10, 9, 1]
Output: 11
Explanation: The subarray yielding the maximum sum is [1, 2, 7, -4, 3, 2].
The Intuition behind the code is to find the maximum sum of a contiguous subarray within the given array nums. It does this by scanning through the array and keeping track of the current sum of the subarray. Whenever the current sum becomes greater than the maximum sum encountered so far, it updates the maximum sum. If the current sum becomes negative, it resets the sum to 0 and starts a new subarray. By the end of the loop, the code returns the maximum sum found.



An array is called good if the sum of elements in odd indexes is equal to the sum of elements in even indexes.
In array A= [1 2 4 3 6], if we delete A[4]=6, we will get new array B= [1 2 4 3], where B[0] + B[2] = B[1] + B[3] = 5, which means array B is good.
One pass on A,
For each different number a in A,
we need to count its frequency and it first occurrence index.
If a has the maximum frequency,
update the degree = count[a] and res = i - first[A[i]] + 1.
If a is one of the numbers that has the maximum frequency,
update the res = min(res, i - first[A[i]] + 1)



For ‘N’ = 3,
All possible combinations are:
((()))
(()())
(())()
()(())
()()()
The idea is to add ')' only after valid '('
We use two integer variables left & right to see how many '(' & ')' are in the current string
If left < n then we can add '(' to the current string
If right < left then we can add ')' to the current string


The width of each bar is the same and is equal to 1.
Input: ‘n’ = 6, ‘arr’ = [3, 0, 0, 2, 0, 4].
Output: 10
Explanation: Refer to the image for better comprehension:

You don't need to print anything. It has already been taken care of. Just implement the given function.
Maintain two pointers left and right.
Track leftMax and rightMax.
If height[left] < height[right], then water trapped at left depends only on leftMax.
If height[left] >= leftMax, update leftMax.
Else, add leftMax - height[left].
Move left++.
Else do the same for right using rightMax.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
Which data structure is used to implement a DFS?