Tip 1 : For Cisco, try solving DSA problems in C language. Cisco's codebase is in C. Hence, they focus more on C language in coding round and in interviews.
Tip 2 : They ask more from pointers, bit manipulations in C.
Tip 1 : Keep your resume precise and concise, one pager. Do not put false things on resume. Hiring Manager goes through resume and tries to cover each part of it in interview.
Tip 2 : Don't put too many projects and mention only relevant ones.
Tip 3 : Mention you C & Python skills are these are most used languages in Cisco
Timing: Late Night around 8 PM
Environment: At home, online on Hackerrank platform
Round consist of 2 DSA questions and 15 MCQs.



Here, sorted paths mean that the expected output should be in alphabetical order.
Given a square matrix of size 4*4 (i.e. here 'N' = 4):
1 0 0 0
1 1 0 0
1 1 0 0
0 1 1 1
Expected Output:
DDRDRR DRDDRR
i.e. Path-1: DDRDRR and Path-2: DRDDRR
The rat can reach the destination at (3, 3) from (0, 0) by two paths, i.e. DRDDRR and DDRDRR when printed in sorted order, we get DDRDRR DRDDRR.
Since the constraints are not too high, a backtracking-based solution can be used. A recursive relation can be formed, which will try to cover all the valid paths from source to destination, and the base case will be:
1. If it is a valid path, add this path to the answer
2. Else, backtrack and try out a different path.
Time Complexity: O(4 ^ (n ^ 2))
Space Complexity: O(n*n)
where n = dimension of the maze
Timing: 10 AM Morning
Environment: Cisco Webex & Coderpad
The interviewer was a helpful and cool person



The Linked Lists, where a1, a2, c1, c2, c3 is the first linked list and b1, b2, b3, c1, c2, c3 is the second linked list, merging at node c1.

Solution Steps:
1. Get the count of the nodes in the first list, let the count be c1.
2. Get the count of the nodes in the second list, let the count be c2.
3. Get the difference of counts d = abs(c1 – c2)
4. Now traverse the bigger list from the first node till d nodes so that from here onwards both the lists have an equal no of nodes
5. Then we can traverse both the lists in parallel till we come across a common node. (Note that getting a common node is done by comparing the address of the nodes)



If A = “aab”, 'B' = “abc”, 'C' = “aaabbc”
Here 'C' is an interleaving string of 'A' and 'B'. because 'C' contains all the characters of 'A' and 'B' and the order of all these characters is also the same in all three strings.

If 'A' = “abc”, 'B' = “def”, 'C' = “abcdefg”
Here 'C' is not an interleaving string of 'A' and 'B'. 'B'ecause neither A nor 'B' contains the character ‘g’.
Solution Steps:
1 . Create a DP array (matrix) of size M*N, where m is the size of the first string, and n is the size of the second string. 2. Initialize the matrix to false.
3. If the sum of sizes of smaller strings is not equal to the length of the larger string, then return false and break the
array as they cant be interleaved to form the larger string.
4. Run a nested loop, the outer loop from 0 to m and the inner loop from 0 to n. Loop counters are i and j.
5. If the values of i and j are both zeroes then mark dp[i][j] as true. If the value of i is zero and j is non zero and the j-1
character of B is equal to j-1 character of C the assign dp[i][j] as dp[i][j-1] and similarly if j is 0 then match i-1 the
character of C and A and if it matches then assign dp[i][j] as dp[i-1][j].
6. Take three characters x, y, z as (i-1)th character of A and (j-1)th character of B and (i + j – 1)th character of C.
7. if x matches with z and y does not match with z then assign dp[i][j] as dp[i-1][j] similarly if x is not equal to z and y is
equal to z then assign dp[i][j] as dp[i][j-1]
8. if x is equal to y and y is equal to z then assign dp[i][j] as bitwise OR of dp[i][j-1] and dp[i-1][j].
9. return value of dp[m][n].
This was a Hiring Manager round and was mostly based on the resume. The interviewer covered all the sections of my resume in detail. He asked briefly about all of my projects, my future goals, and some situation-based questions.
For example: What will you do, if you are tucked at something and you are not getting any help from seniors or peers?
Timing: 12 PM
Environment: Cisco Webex
The interviewer was a Software Development Manager at Cisco and was an experienced person.

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