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Articles by Sunidhi Bansal
Page 59 of 81
Reverse actual bits of the given number in Java
Given an integer n that is not negative. The goal is to reverse the bits of n and report the number that results from doing so. While reversing the bits, the actual binary form of the integer is used; no leading 0s are taken into account.Let us see various input output scenarios for thisInput − 13Output − Reverse actual bits of the given number 11(13)10 = (1101)2. After reversing the bits, we get: (1011)2 = (11)10.Explanation − The binary bits are obtained from the input number which is then reversed and finally converted to decimal format which is returned as output.Input − 18Output − ...
Read MoreRecursive sum of digits of a number is prime or no in C++
Given an integer variable number as input. The goal is to calculate the sum of digits of the input number and check if that sum is prime or not. Do this till the obtained number with sum of digits becomes a single digit number. Check if that number is prime or not. If the input number is 123 than the sum of digits will be 1+2+3=6 which is non-prime and also 6 is a single digit number.Let us see various input output scenarios for thisInput − number=12341Output − Recursive sum of digits of a number is PRIMEExplanation −1+2+3+4+1=111+1=22 is a prime number.Input − ...
Read MoreRecursive sum of digits of a number formed by repeated appends in C++
Given two integers ‘number’ and ‘repeat’ as input. The goal is to calculate the sum of digits of the input number repeated ‘repeat’ number of times until the sum becomes a single digit. Do this till the obtained number with sum of digits becomes a single digit number. If the input number is 123 and repeat=2 than the sum of digits of 123123 will be 1+2+3+1+2+3=12 which is not a single digit number. Now the sum of digits of 12 is 1+2=3. Output will be 3Let us see various input output scenarios for thisInput − number=32 repeat=3Output − Recursive sum of digits ...
Read MoreRefactorable number in C++
We are given an integer type value, let's say, number. The task is to check whether the given number is Refactorable or not. If yes print that the number is a refactorable number else print not possible.What is a Refactorable Number?A number is refactorable when it is divisible by its total count of factors available. For example, number 9 is refactorable as it has a total number of factors i.e. 3(1, 3, 9) and 9 is divisible by 3 hence its a refactorable number.Let us see various input output scenarios for this −Input − int number = 9Output − It is a ...
Read MoreRearrange characters in a string such that no two adjacent are same in C++
We are given a string, let's say, str of any given length. The task is to rearrange the given string in such a manner that there won't be the same adjacent characters arranged together in the resultant string.Let us see various input output scenarios for this −Input − string str = "itinn"Output − Rearrangement of characters in a string such that no two adjacent are same is: initn.Explanation − We are given a string type variable let’s say, str. Now we will rearrange the characters of an input string in such a manner that no two same characters occur at the same position ...
Read MoreRectangle with minimum possible difference between the length and the width in C++
Given an area of rectangle as input. The goal is to find the sides of a rectangle such that the difference between the length and breadth is minimum.Area of Rectangle = length * breadth.ExamplesInput − Area = 100Output −Side of Rectangle with minimum difference:Length = 10, Breadth = 10Explanation − Sides with area = 100.2 - 50, 4 - 25, 5 - 20, 10 - 10. Sides with minimum difference are 10-10 with difference = 0. As we know tha square is a rectangle with equal length on all sides.Input − Area = 254Output − Side of Rectangle with minimum difference :Length = 127, Breadth ...
Read MoreRecursive program to insert a star between pair of identical characters in C++
Given a string str1 as input. The goal is to insert a ‘*’ between a pair of identical characters in the input string and return the resultant string using a recursive approach.If the input string is str1= "wellness" then output will be "wel*lnes*s"ExamplesInput − str1="happiness"Output − String after adding * : hap*pines*sExplanation − Adding * between pairs pp and ss will gives us resultant string hap*pines*sInput − str1=”swimmmmingggg pooool”Output − String after adding * : swim*m*m*ming*g*g*g po*o*o*olExplanation − Adding * between pairs mm, gg and oo will gives us resultant string swim*m*m*ming*g*g*g po*o*o*olApproach used in the below program is as followsIn this approach take string ...
Read MoreRecursive Approach to find nth node from the end in the linked list in C++
Given a Singly linked list and positive integer N as input. The goal is to find the Nth node from the end in the given list using recursion. If the input list has nodes a → b → c → d → e → f and N is 4 then the 4th node from the end will be c.We will first traverse till the last node in the list and while returning from recursion (backtracking) increment count. When count is equal to N then return pointer to current node as result.Let us see various input output scenarios for this −Input − ...
Read MoreRecursive approach for alternating split of Linked List in C++
Given a Singly linked list as input. The goal is to split the list into two singly linked lists that have alternate nodes of the original list. If the input list has nodes a → b → c → d → e → f then after the split, two sub-lists will be a → c → e and b → d → f.We will take two pointers N1 and N2 one pointing to the head of the original list and another pointing to the head → next. Now move both pointers to the next of next node and create sublists.ExamplesInput − ...
Read MoreRecursive insertion and traversal linked list in C++
We are given with integer values that will be used to form a linked list. The task is to firstly insert and then traverse a singly linked list using a recursive approach.Recursive addition of nodes at the endIf head is NULL → add node to headElse add to head( head → next )Recursive traversal of nodesIf head is NULL → exitElse print( head → next )ExamplesInput − 1 - 2 - 7 - 9 - 10Output − Linked list : 1 → 2 → 7 → 9 → 10 → NULLInput − 12 - 21 - 17 - 94 - 18Output − Linked list ...
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