Infor Global Solution
ABB GISL Interview Questions and Answers
Q1. Explain OOPS concept and how you apply it in your implementation
OOPS is a programming paradigm based on the concept of objects, which can contain data in the form of fields and code in the form of procedures.
OOPS focuses on the concept of classes and objects
Encapsulation: bundling data and methods that operate on the data within a single unit
Inheritance: allows a class to inherit properties and behavior from another class
Polymorphism: ability to present the same interface for different data types
Example: Creating a class 'Car' with proper...read more
Q2. Difference between Comparable and Comparator. Difference between Abstract Class and Interface
Comparable is an interface used for natural ordering, Comparator is an interface used for custom ordering. Abstract class can have method implementations, Interface cannot.
Comparable interface is used to define the natural ordering of objects. Example: String class implements Comparable interface for natural ordering based on alphabetical order.
Comparator interface is used to define custom ordering of objects. Example: Sorting a list of objects based on a specific attribute u...read more
Q3. Given a string. Find the number of occurrences of each character
Count occurrences of each character in a given string
Create an array to store the count of each character
Iterate through the string and increment the count of each character in the array
Return the array with counts for each character
Q4. Given an array. Remove the number of duplicates
Remove duplicates from an array of strings
Iterate through the array and store each element in a set to keep track of unique values
Create a new array with the unique values from the set
Q5. Container with most water
The container with most water problem involves finding the maximum area that can be formed between two vertical lines by selecting two points on a line.
Start with two pointers at the beginning and end of the array.
Calculate the area between the two pointers using the formula min(height[left], height[right]) * (right - left).
Move the pointer with the smaller height towards the center to potentially increase the area.
Repeat until the pointers meet in the middle.
Q6. Maximum subarray with 0 sum
Find the maximum subarray with a sum of 0 in an array.
Iterate through the array and keep track of the running sum.
Store the running sum in a hashmap along with the index.
If the running sum is seen again, the subarray between the two occurrences has a sum of 0.
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