Intermediate.txt

Example : 
Solution :
# Python  
age=input(int("Age : "))
print(age);
                                                                                                                                               Maintainer : Gaurav-2803
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1. Write a guessing game where the user has to guess a secret number. After every guess the program tells the user whether  their number was too large or too small. At the end the number of tries needed should be printed. It counts only as one try if they input the same number multiple times consecutively.
Solution : 

2. Write function that reverses a array, preferably in place.
Solution : 
#Java
static void reverseArray(int[] intArray,int size){
        int i, temp; 
        for (i = 0; i < size/ 2; i++) { 
            
            //store first index value at temp
            temp = intArray[i]; 
            
            //store lastIndex-i-1 value in i'th Index
            intArray[i] = intArray[size - i - 1]; 
            
            //store temp value at lastIndex-i-1 
            intArray[size - i - 1] = temp; 

            //Loop will run till the mid position of given array
        } 
        //printing the reversed Array
        System.out.println(Arrays.toString(intArray)); 
}
3. Write a function that tests whether a string is a palindrome.
Solution : 
class Main{
public static void main(String[] args){
String str="abba";
String str1="";
StringBuilder ans=new StringBuilder();
ans.append(str);
ans.reverse;
str1=ans.toString();
if(str==str){
System.out.print("Palindromic");
}
Systme.out.print("Not palindromic");
}
}

#CPP
class Solution{
public:	
	int isPalindrome(string S)
	{
	       int ch=0;
         string st;
         while(S[ch]!='\0'){
            st+=S[ch];
            ch++;
          }
          string newst;
          for(int i=ch-1;i>=0;i--){
          newst+=S[i];
          }
          if(newst==st){
            return 1;
          }
          return 0;
	}

};

4. Write a function that tests whether a number is a palindrome.
Solution : 
#Python
# Python program to check if number is Palindrome

def isPalindrome(n):  #user-defined function
    # calculate reverse of number
    temp=n
    reverse = 0
    reminder = 0
    while(n != 0):
        remainder = n % 10
        reverse = reverse * 10 + remainder
        n = int(n / 10)
    if(temp== reverse):
        return 1
    else:
        return 0


5. Write a function on_all that applies a function to every element of a list. Use it to print the first twenty perfect squares. The perfect squares can be found by multiplying each natural number with itself. The first few perfect squares are 1*1= 1, 2*2=4, 3*3=9, 4*4=16. Twelve for example is not a perfect square because there is no natural number m so that m*m=12.
Solution : 

6. Write a function that combines two arrays by alternatingly taking elements, e.g. [a,b,c], [1,2,3] → [a,1,b,2,c,3].
Solution : 
#CPP
vector<int> mergeArr(int arr1[],int arr2[],m,n){
	vector<int> ans(m+n);
	int a1=0;
	int a2=0;
	for(int i=0;i<(m+n);i++){
		if(i%2==0){
			ans.push_back(arr1[a1];
			a1++;
		}else{
			ans.push_back(arr2[a2]);
			a2++;
		}
	}
	return ans;
} 

7. Write a function that rotates a array by k elements. For example [1,2,3,4,5,6] rotated by two becomes [3,4,5,6,1,2]. Try solving this without creating a copy of the array. How many swap or move operations do you need?
Solution : 

 
class Main
{
    public static void rightRotateByOne(int[] A)
    {
        int last = A[A.length - 1];
        for (int i = A.length - 2; i >= 0; i--) {
            A[i + 1] = A[i];
        }
 
        A[0] = last;
    }
 
    public static void rightRotate(int[] A, int k)
    {
        if (k < 0 || k >= A.length) {
            return;
        }
 
        for (int i = 0; i < k; i++) {
            rightRotateByOne(A);
        }
    }
 
    public static void main(String[] args)
    {
        int[] A = { 1, 2, 3, 4, 5, 6, 7 };
        int k = 3;
 
        rightRotate(A, k);
 
        System.out.println(Arrays.toString(A));
    }
}

8. Write a function that takes a number and returns a array of its digits. So for 2342 it should return [2,3,4,2].
Solution : 
#Java
  static void getDigit(int number){
        ArrayList<Integer> arr=new ArrayList<Integer>();
        // Printing the last digit of the number
        while (number > 0) {
            // Finding the remainder (Last Digit)
            int remainder = number % 10;
            // Printing the remainder/current last digit
            arr.add(remainder);
            // Removing the last digit/current last digit
            number = number / 10;
    }
    Collections.reverse(arr);
    System.out.println(arr);
}


9. Write functions that add, subtract, and multiply digits of two numbers.
Solution : 
#python

def add(a,b):
    return a+b;
def sub(a,b):
    return a-b;
def multiply(a,b):
    return a*b;

f1=add(5,5.2)
f2=sub(5,4.2)
f3=multiply(5,5.2)
print("Addition: ", f1)
print("substraction: ", f2)
print("multiplication: ", f3)


10. Implement binary search.
Solution : 
//java
class Solution {
    int binarysearch(int arr[], int n, int k){
        int start=0;
        int end=n-1;
        while(start<=end){
            int mid =start +(end-start)/2;
            if(arr[mid]<k){
                start=mid+1;
            }
            if(arr[mid]>k){
                end=mid-1;
            }
            if(arr[mid]==k) {
              return mid;
            }
        }
        return -1;
    }
}


=======
#Java
 class binarySearch {
  public int search(int[] nums, int target) {
    int key, start = 0, end = nums.length - 1;
    while (start <= end) {
      key = (start+end)/2  
      if (nums[key] == target){
        return key;
      } 
      if (target < nums[key]){
        end = key - 1;
     }
      else {
        left = key + 1;
        }
    }
    return -1;
  }
 public static void main(String args[]){  
        int arr[] = {10,20,30,40,50};  
        binarySearch(arr,30);     
 }   
}
def binary_search(list1, n):  
    low = 0  
    high = len(list1) - 1  
    mid = 0  
  
    while low <= high:  
        mid = (high + low) // 2  
        if list1[mid] < n:  
            low = mid + 1  
        elif list1[mid] > n:  
            high = mid - 1  
        else:  
            return mid  
    return -1  
 
# Initial list1  
list1 = [12, 24, 32, 39, 45, 50, 54]  
n = 45  
  
# Function call   
result = binary_search(list1, n)  
  
if result != -1:  
    print("Element is present at index", str(result))  
else:  
    print("Element is not present in list1") 



11. Write a function that takes a array of strings an prints them, one per line, in a rectangular frame. For example the list ["Hello", "World", "in", "a", "frame"] gets printed as:

*********
* Hello *
* World *
* in    *
* a     *
* frame *
*********

Solution : 

12. Print the following pattern without using print in-built function.
i.      ii.     iii.    iv.
*       ***       *     ***
**      **       **      **
***     *       ***       *
v.      ii.     iii.    iv.
1       123       1     123
23      45       23      45
456     5       456       6

Solution : 

13. Convert Age -> Days (Consider leap years in between) take year of birth as input also.
Solution : 

14. Create a function that finds the maximum range of a triangle's third edge, where the side lengths are all integers. Note. (side1 + side2) - 1 = maximum range of third edge.
Solution : // C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Function to find the minimum and the
// maximum possible length of the third
// side of the given triangle
void find_length(int s1, int s2)
{
 
    // Not a valid triangle
    if (s1 <= 0 || s2 <= 0) {
        cout << -1;
        return;
    }
    int max_length = s1 + s2 - 1;
    int min_length = max(s1, s2) - min(s1, s2) + 1;
 
    // Not a valid triangle
    if (min_length > max_length) {
        cout << -1;
        return;
    }
 
    cout << "Max = " << max_length << endl;
    cout << "Min = " << min_length;
}
 
// Driver code
int main()
{
    int s1 = 8, s2 = 5;
    find_length(s1, s2);
 
    return 0;
}

15. Write a function that takes number K where is smaller than size of array, we need to find the Kth smallest and largest elementin the given array.It is given that all array elements are not distinct and If not present return -1. 
Solution : 

16. Write a function that sort array of 0s, 1s & 2s (inbuilt sort functions are not allowed)
Solution : # Python program to sort an array with
# 0, 1 and 2 in a single pass
 
# Function to sort array
 
 
def sort012(a, arr_size):
    lo = 0
    hi = arr_size - 1
    mid = 0
    # Iterate till all the elements
    # are sorted
    while mid <= hi:
        # If the element is 0
        if a[mid] == 0:
            a[lo], a[mid] = a[mid], a[lo]
            lo = lo + 1
            mid = mid + 1
        # If the element is 1
        elif a[mid] == 1:
            mid = mid + 1
        # If the element is 2
        else:
            a[mid], a[hi] = a[hi], a[mid]
            hi = hi - 1
    return a
 
# Function to print array
 
 
def printArray(a):
    for k in a:
        print(k, end=' ')
 
 
# Driver Program
arr = [0, 1, 1, 0, 1, 2, 1, 2, 0, 0, 0, 1]
arr_size = len(arr)
arr = sort012(arr, arr_size)
printArray(arr)

17. Given Unsorted array check whether elements are in Arithmetic Progression (Differnce between every 2 consecutive is same) if not return -1.
Example : 1,2,3,4,5 are in A.P
Solution : 


18. Given a sorted array of distinct integers and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order. You must write an algorithm with O(log n) runtime complexity.
Solution : 

19. Given a string S,of length N that is indexed from 0 to N-1, print its even-indexed and odd-indexed characters as 2 space-separated strings on a single line.
    I/P -> Gaurav   O/P -> Gua arv
Solution : 




20 .Sort an array of 0s, 1s and 2s (Given an array of size N containing only 0s, 1s, and 2s; sort the array in ascending order.)
/// java 
class Solution
{
    public static void sort012(int arr[], int n)
    {
        // code here 
        int zero =0;
        int one =0;
        int two =0;
        for(int i =0; i<n;i++){
            if (arr[i]==0){
                zero++;
            }
            else if(arr[i]==1){
                one++;
            }
            else{
                two++ ;
                
            }
        }
        for(int i=0;i<n;i++){
            if (zero>0){
                arr[i]=0;
                zero--;
            }
            else if(one>0){
                arr[i]=1;
                one--;
            }
            else{
                arr[i]=2;
            }
        }
    }
}