Objective: Write an algorithm to check if one string is Rotation of another string. This question has been asked in the Amazon interview.
Example:
Input Strings : 'sumitjain' and 'tjainsumi' Output : true Input String : 'Jaain' and 'ainJ' Output: false
Input: Two Strings
Output: True or false based on whether strings are rotation of each other.
Approach:
Objective: Write an algorithm to compress the given string by using the count of repeated characters and if new compressed string length is not smaller than the original string then return the original string.
Example:
Input String : ssssuuuummmmmmiiiittttttttttttt Compressed String : s4u4m6i4t13 Input String : Jaain Compressed String : Jaain (Since compressed string is length is greater than original string)
Input: A String
Output: either A compressed string or original string whichever us smaller.
Approach:
Objective: Given Two Strings, check whether one string is permutation of other
Input: Two Strings
Output: True or false based on whether strings are permutation of other or not.
Example:
"sumit" and "tiums" are permutations of each other. "abcd" and bdea" are not permutations of each other.
Approach:
Objective: Reverse the given linked list.
Input: A Linked List
Output: Reversed Linked List
Example:
Input : ->30->25->20->15->10->5 Reversed : ->5->10->15->20->25->30
NOTE : Click Reverse a Linked List – Part 2 to see the another implementation of this problem.
Approach:
Linked List- As the name suggests it’s a list which is linked.
Nodes are nothing but objects of a class and each node has data and a link to the next node.
class Node { public int data; public Node next; public Node(int data) { this.data = data; this.next = null; } }
Objective: Write an algorithm to find out whether in a given array there exists or not two numbers whose sum is exactly equals to a given number. This problem has been asked in Amazon and Microsoft interviews.
Input: An array arrA[], A number k
Output: True or false based upon we have found any two numbers in array arrA[] whose sum is equal to k
Approach:
Method 1: Using Binary Search
Objective: Write an algorithm to find out whether in a given string contains all the unique characters. This question has been asked in the Amazon and Microsoft interviews.
Input: A String
Output: True or false based upon whether string contains all the unique characters or not.
Approach:
Method 1.
Objective : Write an algorithm to find Whether Given String is palindrome or Not.
Input: A String,
Output: true or false on whether string is palindrome or not
Approach:
Example:
Jain niaJ => compare ‘J’ with ‘J’ =>returns true ain nia => compare ‘a’ with ‘a’ =>returns true in ni => compare ‘i’ with ‘i’ =>returns true n n => compare ‘n’ with ‘n’ =>returns true string length <2 => returns true
Objective : In this article we will discuss an algorithm to Find a peak element in a Given Array. We will see the recursion techniques to solve this problem.
Peak Element: peak element is the element which is greater than or equal to both of its neighbors.
Input: Array, arrA[] .
Output: A peak element and its index
Approach:
A simple approach is to do a linear scan to a array and using few variables we can find a peak element. But the Time Complexity will be O(n) but real question is, Can we do better?
The Answer is yes, by using Binary Search techniques.
Objective : Write an algorithm to find two Missing Numbers in a Sequence of Consecutive Numbers
Input: Array, arrA[] with two missing numbers and Range
Output : Two missing numbers
Approach:
Objective : You have been asked to Write an algorithm Find a Missing Number From a Sequence of Consecutive Numbers
Input: Array, arrA[] with a missing number and Range
Output : missing number
Approach:
Objective : To implement a Hash Table
Input: A set of pairs of keys and values
Approach:
Objective : Write an algorithm to find an element in an sorted array Input: A sorted array, arrA[] and an key Output : Return true if element is found, else false. Approach: The idea is to compare the middle element of array with the key, if key equal to the middle element , that’s it … Read more