Tip 1: Solve problems to gain hands-on confidence.
Tip 2: Solve high-quality questions.
Tip 3: Try solving questions yourself instead of relying on YouTube videos.
Tip 1: Include projects related to the company's current endeavors, such as adding Spring Boot.
Tip 2: Highlight relevant skills; since fintech mostly works with Java, I added Java to my repertoire.



1. Horizontally as 1x2 tile
2. Vertically as 2x1 tile
The number of ways might be large so output your answer modulo 10^9 + 7.

I have done this problem earlier so got the DP based approach during the test and this approach passed all the test cases.



Example: String "aabbbcdcdcd" will be encrypted as "a2b3cd3".
Input string will always be lowercase characters without any spaces.
If the count of a substring is 1 then also it will be followed by Integer '1'.
Example: "aabcdee" will be Encrypted as "a2bcd1e2"
This means it's guaranteed that each substring is followed by some Integer.
Also, the frequency of encrypted substring can be of more than one digit. For example, in "ab12c3", ab is repeated 12 times. No leading 0 is present in the frequency of substring.
The frequency of a repeated substring can also be in parts.
Example: "aaaabbbb" can also have "a2a2b3b1" as Encrypted String.
I simply decrypt the string by reading substring and their frequency and append current substring to the decrypted string and after the end of traversal of given string our answer will be kth element of the decrypted string.



The pair consists of equal absolute values, one being positive and another negative.
Return an empty array, if no such pair exists.
I asked for some clarifications whether I should print all distinct x‘s or if I should print an x if a pair of +x and -x is encountered. The first approach I told was to use a map and I was keeping a flag for +x and -x if it’s found once. Later he asked me to print all pairs, so I stored the frequencies of all the elements in the map and iterated through the negative elements and for each element x , I would print x min(count[-x],count[+x]) times. He said he can’t afford that much space and he wanted me to optimise space further. So I told him a 2 pointer approach where I sort the array once and then keep two pointers to the start and end. I would move the start pointer forward if the sum is less than 0 and I’ll move the end pointer backward if the sum is greater than 0. He was fine with the solution and asked me to code it in a paper. I wrote the code and walked him through it.



Input: ‘N’ = 4, ‘arr’ = [3, 1, 2, 4], 'K' = 6
Output: 2
Explanation: The subarrays that sum up to '6' are: [3, 1, 2], and [2, 4].
Your task is to find the total number of subarrays of the given array whose sum of elements is equal to k.

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
What is recursion?