Program(c):-Given a String S, reverse the string without reversing its individual words.

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Given a String S, reverse the string without reversing its individual words. Words are separated by dots.

Example 1:

Input:
S = i.like.this.program.very.much
Output: much.very.program.this.like.i
Explanation: After reversing the whole
string(not individual words), the input
string becomes
much.very.program.this.like.i

Example 2:

Input:
S = pqr.mno
Output: mno.pqr
Explanation: After reversing the whole
string , the input string becomes
mno.pqr

Your Task:
You dont need to read input or print anything. Complete the function reverseWords() which takes string S as input parameter and returns a string containing the words in reversed order. Each word in the returning string should also be separated by '.' 

Expected Time Complexity: O(|S|)
Expected Auxiliary Space: O(|S|)

Constraints:
1 <= |S| <= 105

SOLUTION :-

//{ Driver Code Starts

import java.util.*;

import java.lang.*;

import java.io.*;

class GFG {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        int t = sc.nextInt();

        while (t > 0) {

            String s = sc.next();

            Solution obj = new Solution();

            System.out.println(obj.reverseWords(s));

            t--;

        }

    }

}

// } Driver Code Ends

class Solution 

{

    //Function to reverse words in a given string.

    String reverseWords(String S)

    {

        int l=S.length();

        

        String sa ="";

        String s1="";

        int p=0;

        for(int i=l-1;i>=0;i--)

        {

            if(S.charAt(i)!='.')

            {

                s1=S.charAt(i)+s1;

            }

            else

            {

             sa=sa+s1+".";

             s1="";

             p++;

            }

        }

        sa=sa+s1;

        return sa;

    }

}

PROGRAM(C):- To find the greatest and least elements from an array.

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Given an array A of size N of integers. Your task is to find the minimum and maximum elements in the array.

 

Example 1:

Input:
N = 6
A[] = {3, 2, 1, 56, 10000, 167}
Output: 1 10000
Explanation: minimum and maximum elements of array are 1 and 10000.

 

Example 2:

Input:
N = 5
A[]  = {1, 345, 234, 21, 56789}
Output: 1 56789
Explanation: minimum and maximum element of array are 1 and 56789.

 

Your Task:  
You don't need to read input or print anything. Your task is to complete the function getMinMax() which takes the array A[] and its size N as inputs and returns the minimum and maximum element of the array.

 

Expected Time Complexity: O(N)
Expected Auxiliary Space: O(1)

 

Constraints:
1 <= N <= 105
1 <= Ai <=1012

SOLUTION:-

#include <stdio.h>

struct pair {

    long long int min;

    long long int max;

};

struct pair getMinMax(long long int arr[], long long int n) ;

int main() {

    long long int t, n, a[100002], i;

    struct pair minmax;

    scanf("%lld", &t);

    while (t--) {

        scanf("%lld", &n);

        for (i = 0; i < n; i++) scanf("%lld", &a[i]);

        minmax = getMinMax(a, n);

        printf("%lld %lld\n", minmax.min, minmax.max);

    }

    return 0;

}

// } Driver Code Ends

// User function Template for C

struct pair getMinMax(long long int arr[], long long int n) {

    struct pair minMax;

    int s=arr[0];

    int g=arr[0];

    for(int i=1;i<n;i++)

    {

        if(g<arr[i])

          g=arr[i];

        if(s>arr[i])

          s=arr[i];

    }

    minMax.min=s;

    minMax.max=g;

    return minMax;

    

}

PROGRAM(PYTHON) :-TO CHECK WEATHER THE NUMBER FORMES BY LAST DIGIT OF AN ARRAY IS DIVISIBLE BY 10 OR NOT??

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Problem

You are provided an array $�$ of size $�$ that contains non-negative integers. Your task is to determine whether the number that is formed by selecting the last digit of all the N numbers is divisible by $10$.

Note: View the sample explanation section for more clarification.

Input format

• First line: A single integer $�$ denoting the size of array $�$
• Second line: $�$ space-separated integers.

Output format

If the number is divisible by $10$, then print $���$. Otherwise, print $��$.

Constraints
$$\begin{gathered} 1\le �\le {10}^5 \\ 0\le �[�]\le {10}^5 \end{gathered}$$

Sample Input

5
85 25 65 21 84

Sample Output

No

Time Limit: 1

Memory Limit: 256

Source Limit:

Explanation

Last digit of $85$ is $5$.
Last digit of $25$ is $5$.
Last digit of $65$ is $5$.
Last digit of $21$ is $1$.
Last digit of $84$ is $4$.
Therefore the number formed is $55514$ which is not divisible by $10$.

SOLUTION:-

num=0;

d=0;

N = int(input())

data = [int(x) for x in input().split()]

for i in range(N):

    d=data[i]

    num=num*10+d

if num%10==0:

    ans="Yes"

else:

    ans="No"

# Write your code here

# ans =

print(ans)

C Program to print a 2d array in snake form :-

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Q:-Given a matrix of size N x N. Print the elements of the matrix in the snake like pattern depicted below.

Example 1:

Input:
N = 3 
matrix[][] = {{45, 48, 54},
             {21, 89, 87}
             {70, 78, 15}}
Output: 
45 48 54 87 89 21 70 78 15 
Explanation:
Matrix is as below:
45 48 54
21 89 87
70 78 15
Printing it in snake pattern will lead to 
the output as 45 48 54 87 89 21 70 78 15.

Example 2:

Input:
N = 2
matrix[][] = {{1, 2},
              {3, 4}}
Output: 
1 2 4 3
Explanation:
Matrix is as below:
1 2 
3 4
Printing it in snake pattern will 
give output as 1 2 4 3.

Your Task:
You dont need to read input or print anything. Complete the function snakePattern() that takes matrix as input parameter and returns a list of integers in order of the values visited in the snake pattern. 

Expected Time Complexity: O(N * N)
Expected Auxiliary Space: O(N * N) for the resultant list only.

Constraints:
1 <= N <= 103
1 <= mat[i][j] <= 109

Solution:-

//{ Driver Code Starts

//Initial Template for C

#include <stdio.h>

// } Driver Code Ends

//User function Template for C

//Function to return list of integers visited in snake pattern in matrix.

int* snakePattern(int n, int matrix[][n])

{   

    int i,j,k,x,t[n];

    

    for(i=0;i<n;i++)

    {

        if((i%2)==0)

        {  

            

            for(j=0;j<n;j++)

            {

               

            matrix[i][j] =matrix[i][j];

                

            }

          

        }

        else

        {   x=0;

            for(k=n-1;k>=0;k--)

            {     

                t[x] =matrix[i][k];

                x++;

            }

            

             for(j=0;j<n;j++)

            {

               

               matrix[i][j]=t[j];

                

            }

            

            

        }

         

    }

    return matrix;

}

//{ Driver Code Starts.

int main()

{

    int tc;

scanf("%d", &tc);

while(tc--){

int n;

scanf("%d", &n);

int matrix[n][n];

 

for(int i = 0; i < n; i++){

for(int j = 0; j < n; j++){

scanf("%d", &matrix[i][j]);

}

}

int *result = snakePattern(n, matrix);

for(int i = 0; i < n*n; i++)

printf("%d ", result[i]);

printf("\n");

}

return 0;

}

// } Driver Code Ends