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Articles by Sunidhi Bansal
Page 81 of 81
Print Postorder traversal from given Inorder and Preorder traversals
Given with inorder and preorder of a tree program must find the postroder traversal and print the sameInput: Inorder traversal in[] = {4, 2, 5, 1, 3, 6} Preorder traversal pre[] = {1, 2, 4, 5, 3, 6} Output: Postorder traversal post[] = {4, 5, 2, 6, 3, 1}AlgorithmSTART Step 1 -> declare function as find_value(int p, int in_order[], int n) Loop For i=0 and i declare function as postorder(int pre_order[], int in_order[], int n) Declare int variable as root = find_value(pre_order[0], in_order, n) IF root!=0 Call postorder(pre_order+1, in_order, root) End IF ...
Read MorePrint Reverse a linked list using Stack
Given with a linked list program must print the list starting from the end till the front using the stack data structureInput : 10 -> 5 -> 3 -> 1 -> 7 -> 9 Output: 9 -> 7 -> 1 -> 3 -> 5 -> 10Here the user can use the approach of poping elements from the stack pointing top at the stack[0] location and than going till stack[n] elementAlgorithmSTART Step 1 -> create structure Linked_list Declare int data Declare struct linked_list *next End Step 2 -> declare int stack[30], top = -1 Step 3 -> declare struct ...
Read MorePrint the nearest prime number formed by adding prime numbers to N
As per the question, the task is to find the nearest prime number by adding the prime number starting from 2 if the number N is not Prime.Input: N=6 Output: 11ExplanationSince 6 is not prime add first prime to 6 i.e. 2 which will result to 8 now 8 is also not prime now add next prime after 2 which is 3 which will give 8+3 = 11. Hence 11 is a prime number output will be 11.AlgorithmSTART Step 1- > declare num=15, i = num/2 Step 2 -> Loop For k=2 and k
Read MorePrint N lines of numbers such that every pair among numbers has a GCD K
GCDGCD stands for Greatest Common Divisor of two or more integers excluding 0Like, to find the greatest common divisor of 48 and 18048 = 2 × 2 × 2 × 2 × 3180 = 2 × 2 × 3 × 3 × 5Greatest common divisor = 2 × 2 × 3 = 12.In the given problem, N lines should be printed with elements have GCD as specifiedInput : N=2 GCD=2 Ouput : 2-4-6-10 14-16-18-22AlgorithmSTART Step 1 -> take input n(e.g. 2) and k(e.g. 2) as int values and i Step 2-> Loop For i to 0 and i end loop ...
Read MorePrint the kth common factor of two numbers
Given with two numbers x and y the output should contain their kth common factor.Input: x=9 y=18 k=1 Output : k common factor = 2 Factors of 9 : 1, 3, 9 Factors of 18 : 1, 2, 3, 6, 9, 18 Greatest Common Factor : 9AlgorithmSTART Step 1 -: take input as x and y lets say 3 and 21 and k as 1 Step 2 -: declare start variables as int i,num,count=1 Step 3 -: IF x
Read MorePrint n terms of Newman-Conway Sequence
Newman-Conway Sequence is used for generating following integer sequence.1 1 2 2 3 4 4 4 5 6 7 7 8 8 8 8 9 10 11 12Formula used for generating Newman-Conway sequence for n numbers is −P(n) = P(P(n - 1)) + P(n - P(n - 1)) Where, p(1) =p(2) =1AlgorithmSTART Step 1 -> Input variable n(e.g. 20) Step 2 -> start variables as i, p[n+1], p[1]=1, p[2]=1 Step 3 -> Loop For i=3 and i End Loop For STOPExample#include int main() { int n = 20,i; int p[n + 1]; p[1] = 1; p[2] = 1; printf("Newman-Conway Sequence is :"); printf("%d %d ",p[1],p[2]); for (i = 3; i
Read MorePrint n smallest elements from given array in their original order
Given with array of let’s say k elements the program must find the n smallest elements amongst them in their appearing order.Input : arr[] = {1, 2, 4, 3, 6, 7, 8}, k=3 Ouput : 1, 2, 3 Input k is 3 it means 3 shortest elements among the set needs to be displayed in original order like 1 than 2 and than 3AlgorithmSTART Step 1 -> start variables as int i, max, pos, j, k=4 and size for array size Step 2 -> Loop For i=k and i=0 and j-- If arr[j]>max ...
Read MorePrint values of ‘a’ in equation (a+b) <= n and a+b is divisible by x
Given with equation program must find the value of ‘a’ where a+b Declare start variables b=10, x=9, n=40 and flag=0, divisible Step 2 -> Loop For divisible = (b / x + 1 ) * x and divisible = 1 Print divisible-1 Set flag=1 End END STOPExample#include int main(int argc, char const *argv[]) { int b=10, x=9, n=40, flag = 0; int divisible; for (divisible = (b / x + 1 ) * x ; divisible = 1) { printf("%d ", divisible - b ); flag = 1; } } return 0; }Outputif we run above program then it will generate following output8 17 26
Read MorePrint the given pattern recursively
Here, as per the given problem pattern needs to be displayed using recursive approach.Recursive function is the one that calls itself n number of times. There can be ‘n’ number of recursive function in a program. The problem working with recursive function is their complexity.AlgorithmSTART Step 1 -> function int printpattern(int n) If n>0 Printpattern(n-1) Print * End IF End Step 2 -> function int pattern(int n) If n>0 pattern(n-1) End IF Printpattern(n) Print End STOPExample#include int printpattern(int n) { if(n>0) { ...
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