Source One Management Services Software Engineer Interview Questions and Answers

Round duration - 126 minutes
Round difficulty - Hard

This round consist of coding questions as well as mcqs based on aptitude as well as some domain sections. This round is hard.

Q1. 
Maximum Path Sum in a Matrix
Given an N*M matrix filled with integer numbers, determine the maximum sum that can be obtained from a path starting from any cell in the first row to any cell in the last row...read more
Ans. 
Find the maximum sum path in a matrix from top row to bottom row by moving down or diagonally.
• Use dynamic programming to keep track of maximum sum at each cell.

• At each cell, consider the maximum sum from the cell above, left diagonal, and right diagonal.

• Add the current cell value to the maximum of the three possibilities.

• Repeat this process for all cells in the matrix to find the maximum sum path.
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Q2. 
Boolean Matrix Transformation Challenge
Given a 2-dimensional boolean matrix mat of size N x M, your task is to modify the matrix such that if any element is 1, set its entire row and column to 1. Specifi...read more
Ans. 
Modify a boolean matrix such that if any element is 1, set its entire row and column to 1.
• Iterate through the matrix to find elements with value 1

• Store the row and column indices of these elements

• Update the entire row and column for each element found to be 1
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Round duration - 45 minutes
Round difficulty - Easy

In this round the interviewer ask few questions related to domain and at last he ask some behavioral questions.

Q1. 
Intersection of Linked List Problem
You are provided with two singly linked lists containing integers, where both lists converge at some node belonging to a third linked list.
Your task is to determine t...read more
Ans. 
Find the node where two linked lists merge, return -1 if no merging occurs.
• Traverse both lists to find the lengths and the last nodes

• Align the starting points of the lists by adjusting the pointers

• Traverse both lists simultaneously until a common node is found
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