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recursion.cpp
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recursion.cpp
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#include<iostream>
#include<cmath>
#define THOUSAND 1000
using namespace std;
int getArray(int arr[], int size)
{
cin >> size;
// Read the array
for(int j=0;j<size;++j)
cin >> arr[j];
return size;
}
void printArray(int arr[], int size)
{
// Print Array
cout << "The array of size " << size << " is: " << endl;
for(int i=0;i<size;++i)
cout << arr[i] << ' ';
cout << endl;
}
void print(int n){
/* Given the code to print number from 1 to n in increasing order
* recursively. But it contains few bugs, that you need to rectify such that
* all the test cases should pass.*/
if(n == 1){
cout << n << " ";
return;
}
print(n - 1);
cout << n << " ";
}
int count(int n){
/* return the number of digits present in a number recursively */
if(n == 0){
return 0;
}
int smallAns = count(n / 10);
return smallAns + 1;
}
int power(int x, int n) {
/* find x to the power n (i.e. x^n) */
if(n == 0){
return 1;
}
int smallAns = power(x,n-1);
return smallAns * x;
}
int sum(int n) {
/* Given an integer n, find and return the sum of numbers from 1 to n using
* recursion.*/
if(n == 1){
return 1;
}
int smallAns = sum(n-1);
return smallAns + n;
}
int sumArray(int input[], int n) {
/* Given an array of length N, you need to find and return the sum of all
* elements of the array.*/
if(n == 1){
return input[0];
}
int smallAns = sumArray(input, n-1);
return smallAns + input[n-1];
}
bool checkNumber(int input[], int size, int x) {
/* Given an array of length N and an integer x, you need to find if x is
* present in the array or not. Return true or false.*/
if(size == 1){
return x==input[0];
}
bool smallAns = checkNumber(input, size-1, x);
return smallAns || (x==input[size-1]);
}
int firstIndex(int input[], int size, int x) {
/* Given an array of length N and an integer x, you need to find and return
* the first index of integer x present in the array. Return -1 if it is not
* present in the array. First index means, the index of first occurrence of
* x in the input array. Do this recursively. Indexing in the array starts
* from 0. */
if(size <= 0){
return -1;
}
if(x==input[0]) return 0;
int smallAns = firstIndex(input+1, size-1, x);
if(smallAns == -1)
return -1;
else
return smallAns+1;
}
int lastIndex(int input[], int size, int x) {
/* Given an array of length N and an integer x, you need to find and return
* the last index of integer x present in the array. Return -1 if it is not
* present in the array. Last index means - if x is present multiple times in
* the array, return the index at which x comes last in the array. */
if(size <= 0){
return -1;
}
if(x==input[size-1]) return size-1;
int smallAns = lastIndex(input, size-1, x);
if(smallAns == -1)
return -1;
else
return smallAns;
}
int allIndexes(int input[], int size, int x, int output[]) {
/* Given an array of length N and an integer x, you need to find all the
* indexes where x is present in the input array. Save all the indexes in an
* array (in increasing order). Do this recursively. Indexing in the array
* starts from 0.*/
if(size <= 0) return 0;
int smallAns = allIndexes(input,size-1,x,output);
if(x==input[size-1])
output[smallAns++] = size-1;
return smallAns;
}
int multiplyNumbers(int m, int n) {
/* Given two integers m & n, calculate and return their multiplication using
* recursion. You can only use subtraction and addition for your calculation.
* No other operators are allowed.*/
if(m==0 || n==0) return 0;
if(n>0)
{
int smallAns = multiplyNumbers(m,n-1);
return smallAns + m;
}
else
{
int smallAns = multiplyNumbers(m,n+1);
return smallAns - m;
}
}
int countZeros(int n){
/* Given an integer n, count and return the number of zeros that are present
* in the given integer using recursion.*/
if(n<0) n *= -1; // Make n positive
if(n<10)
{
if(n == 0) return 1;
return 0;
}
int smallAns = countZeros(n / 10);
if(n%10==0)
smallAns++;
return smallAns;
}
double geometricSum(int k){
/* Given k, find the geometric sum i.e. 1 + 1/2 + 1/4 + 1/8 + ... + 1/(2^k)*/
if(k == 0){
return 1;
}
double smallAns = geometricSum(k-1);
return smallAns + 1/pow(2,k);
}
bool RcheckPalindrome(char input[],int size) {
/* Check if a given String S is palindrome or not (using recursion). Return
* true or false.*/
if(size <= 1){
return true;
}
if(input[0]!=input[size-1]) return false;
return RcheckPalindrome(input+1,size-2);
}
bool checkPalindrome(char input[]) {
int size;
if(input==nullptr) return false;
for(size=0;input[size]!='\0';size++);
return RcheckPalindrome(input,size);
}
int sumOfDigits(int n){
/* Write a recursive function that returns the sum of the digits of a given
* integer. */
if(n == 0){
return 0;
}
int smallAns = sumOfDigits(n / 10);
return smallAns + n%10;
}
void replacePi(char input[]) {
/* Given a string, compute recursively a new string where all appearances of
* "pi" have been replaced by "3.14".*/
if(input[0] == '\0')
return;
else if(input[0]!='p')
replacePi(input+1);
else if(input[1]=='\0')
return;
else if(input[1]=='i')
{
int size;
// Calculate Size
for(size=2;input[size]!='\0';size++);
for(int i=size+2;i>=4;i--)
input[i] = input[i-2];
input[0] = '3';
input[1] = '.';
input[2] = '1';
input[3] = '4';
replacePi(input+4);
}
else
replacePi(input+1);
}
void removeX(char input[]) {
/* Given a string, compute recursively a new string where all 'x' chars have
* been removed.*/
if(input[0] == '\0')
return;
else if(input[0]!='x')
removeX(input+1);
else
{
int i;
for(i=0;input[i]!='\0';i++)
input[i] = input[i+1];
input[i] = '\0';
removeX(input);
}
}
int stringToNumberNonRecuresive(char input[])
{
/* Write a recursive function to convert a given string into the number it
* represents. That is input will be a numeric string that contains only
* numbers, you need to convert the string into corresponding integer and
* return the answer.*/
int i, number=0;
for(i=0;input[i]!='\0';i++)
{
int digit = input[i]-'0';
number = (number*10) + digit;
}
return number;
}
int stringToNumber(char input[])
{
/* Write a recursive function to convert a given string into the number it
* represents. That is input will be a numeric string that contains only
* numbers, you need to convert the string into corresponding integer and
* return the answer.*/
if(input[0] == '\0')
return 0;
int smallAns = stringToNumber(input+1);
int power=1;
for(int i=1;input[i]!='\0';i++)
power *= 10;
int digit = input[0]-'0';
return digit*power + smallAns;
}
void pairStar(char input[])
{
/* Given a string, compute recursively a new string where identical chars
* that are adjacent in the original string are separated from each other by
* a "*".*/
if(input[0] == '\0')
return;
pairStar(input+1);
if(input[0]==input[1])
{
int size;
// Calculate Size
for(size=2;input[size]!='\0';size++);
for(int i=size+1;i>=2;i--)
input[i] = input[i-1];
input[1] = '*';
}
}
void towerOfHanoi(int n, char source, char auxiliary, char destination) {
/* Tower of Hanoi is a mathematical puzzle where we have three rods and
* n disks. The objective of the puzzle is to move all disks from source rod
* to destination rod using third rod (say auxiliary). The rules are:
* 1) Only one disk can be moved at a time.
* 2) A disk can be moved only if it is on the top of a rod.
* 3) No disk can be placed on the top of a smaller disk.
* Print the steps required to move n disks from source rod to destination
* rod. Source Rod is named as 'a', auxiliary rod as 'b' and destination rod
* as 'c'.*/
if(n == 0) return;
if(n == 1)
{
cout << source << ' ' << destination << endl;
return;
}
towerOfHanoi(n-1,source,destination,auxiliary);
cout << source << ' ' << destination << endl;
towerOfHanoi(n-1,auxiliary,source,destination);
}
int recursion()
{
int n,x,val;
/*
int arr[THOUSAND];
int output[THOUSAND];
n=getArray(arr,THOUSAND);
cin >> x;
int count = allIndexes(arr, n, x, output);
printArray(output,count);
*/
cin >> x; cout << geometricSum(x) << endl;
return 0;
}