MIND TREE

1.
1 1 1 1 2
 3 2 2 2 2
 3 3 3 3 4
 5 4 4 4 4
 5 5 5 5 6
 7 6 6 6 6
 7 7 7 7 8
 9 8 8 8 8
 9 9 9 9 10
#include <stdio.h>

 int main()
{
int i,j;
int n = 9;
for(i=1;i<=n;i++)
{
    for(j=1;j<=5;j++)
    {

         if(i%2==0)
        {
            if(j==1)
            {
                printf(" %d",i+1);
            }
            else
                printf(" %d",i);
        }
        else
        {
            if(j==5)
            {
                printf(" %d",i+1);
                break;
            }
            else
            {
                printf(" %d",i);
            }
        }
    }
    printf("\n");
}
return 0;
}

 2.
input :
abc      cba
1

 ab  aa

 -1

 #include<stdio.h>
#include<string.h>
int test(char*str1,char *str2)
{
    int ctr1=0,ctr2=0,i,j;
for(j=0;str1[j]!=0;j++)
{
    for(i=0;i<strlen(str1);i++)
    if(str1[i]==str1[0])
    ctr1++;
   
    for(i=0;i<strlen(str2);i++)
    if(str2[i]==str1[0])
    ctr2++;
   

     if(ctr1==ctr2)
    {
        j++;
   
    continue;
    }
    else
    return -1;
   
}
return 1;
}

 int main() {
   char*s1="ab";
   char*s2="aa";
   printf("%d",test(s1,s2));
}

 input :
abc      cba
1

 ab  aa
-1

 3.Sub String occurance in another string with out case sensitive Mind Tree
public class GFG {
  
    static int stringOccuranceCount(String pat, String txt) {       
        int M = pat.length();       
        int N = txt.length();       
        int res = 0;
              String pat1=pat.toUpperCase();
              String txt1=txt.toUpperCase();
            
        /* A loop to slide pat[] one by one */
        for (int i = 0; i <= N - M; i++) {
            /* For current index i, check for 
        pattern match */
            int j;           
            for (j = 0; j < M; j++) {
              
                if (txt1.charAt(i + j) != pat1.charAt(j)) {
                  
                    break;
                }
            }
             // if pat[0...M-1] = txt[i, i+1, ...i+M-1] 
            if (j == M) {               
                res++;               
                j = 0;               
            }           
        }       
        return res;       
    } 

 4. input : 3 2 0 1
output: 2 3 1 0

 #include<stdio.h>

 int main() 
{
int n;
int a[20], res[20];
printf("Enter n: ");
scanf("%d", &n);
printf("Enter array elements: ");
for(int i = 0; i < n; i++)
{
scanf("%d", &a[i]);
}
for(int i = 0; i < n; i++)
{
res[a[i]] = i;
}
for(int i = 0; i < n; i++)
{
printf("%d ", res[i]);
}

     return 0;

 }

 5.* C Program - Delete Vowels from String */
  
#include<stdio.h>
#include<conio.h>
#include<string.h>
void main()
{
 clrscr();
 char str[20];
 int len, i, j;
 printf("Enter a string : ");
 gets(str);
 len=strlen(str);
 for(i=0; i<len; i++)
 {
  if(str[i]=='a' || str[i]=='e' || str[i]=='i' ||
  str[i]=='o' || str[i]=='u' || str[i]=='A' ||
  str[i]=='E' || str[i]=='I' || str[i]=='O' ||
  str[i]=='U')
  {
   for(j=i; j<len; j++)
   {
    str[j]=str[j+1];
   }
  len--;
  }
 }
 printf("After deleting the vowels, the string will be : %s",str);
 getch();
}
5.o/p:

1
2*2
3*3*3
4*4*4*4
4*4*4*4
3*3*3
2*2
1




#include<stdio.h>

int main() {
 
 int i,j,n=4;
 for(i=1;i<=n;i++)
 {
     for(j=1;j<=i;j++)
     {
         printf("%d",i);
         if(j!=i)
         printf("*");
     }
     printf("\n");
 }
 for(i=n;i>=1;i--)
 {
     for(j=1;j<=i;j++)
     {
         printf("%d",i);
         if(j!=i)
         printf("*");
     }
     printf("\n");
 }
}




6. write an algorithm to figure out the max no of calls required special function to obtain an array ascending order
i/p
teat case 1)
5 1 3 11  
size 4
o/p
23

test case2 )
5,1,2,3,8,0,12
7
o/p
5039

#include<stdio.h>

int fact(int n)
{
    int f=1,i;
    for(i=1;i<=n;i++)
    {
        f=f*i;
    }
    return f;
}
int maxOperations(int*list,int len)
{
    
    
    return fact(len)-1;
}

int main()
{
    int no[]={5,1,2,3,8,0,12};
    printf("%d\n\n",maxOperations(no,7));
    

}


7.sort elements and display alternative
#include<stdio.h>
int findAlt(int arr[],int len)
{
    int i,j,tmp;
    
    /*
    for(i=1,j=0;i<len;i+=2,j++)
    arr[j]=ar[i];
    len=j-1;
    */
    for(i=0;i<len-1;i++)
    {
        
        for(j=i+1;j<len;j++)
        {
            if(*(arr+i)>*(arr+j))
            {
                tmp=*(arr+i);
                *(arr+i)=*(arr+j);
               *(arr+j)=tmp;
            
            }
        
        }
    }
    
    }
int main()
{
    int no[]={4,2,5,3,7},i;
    findAlt(no,5);
    for(i=0;i<5;i+=2)
    printf("%d",no[i]);
}
7. Print the below pattern

Output:
1
2*3
4*5*6
7*8*9*10
7*8*9*10
4*5*6
2*3
1
import java.io.*;
public class Pattern
{
public static void main(String args[]) {
int i,j,k=1,n=4;
for (i=1;i<=n;i++) {
for (j=1;j<=i;j++) {
if(j<i) 
       System.out.print(k++ +"*"); 
       else 
       System.out.print(k++);
}
System.out.println();
}
k=k-1;
for (i=n;i>=1;i--) {
for (j=1;j<=i;j++) {
if(j<i) 
       System.out.print(k-- +"*"); 
       else 
       System.out.print(k--);
}
System.out.println();
}
}
}
8. find distict elements in an array
i/p
90,91,1,2,90,3,4,3,91
o/p
90 91 1 2 3 4 
#include<stdio.h>
int* distinctElementArray(int *arr, int len) 
{ 
    // Pick all elements one by one 
    for (int i=0; i<len; i++) 
    { 
        // Check if the picked element is already printed 
        int j; 
        for (j=0; j<i; j++) 
           if (arr[i] == arr[j]) 
               break; 
  
        // If not printed earlier, then print it 
        if (i == j) 
          printf("%d ",arr[i]); 
    } 
    return &arr[0];
} 

 int main() 
{ 
    int arr[] = {90,91,1,2,90,3,4,3,91}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    distinctElementArray(arr, n); 
    return 0; 

 }
find distict elements in an array
#include<stdio.h>
int* distinctElementArray(int *arr, int len)
{
    int i,j,k=0;
    int *ar;
    ar=(int*)malloc(sizeof(int)*len);
    for (i=0; i<len; i++)
    {

         for (j=0; j<i; j++)
           if (arr[i] == arr[j])
               break;

   if (i == j )
   ar[k++]=arr[i];
    }
    return &ar[0];
}

 int main()
{
    int *ret,i;
    int arr[] = {90,91,1,2,90,3,4,9};
    int n = sizeof(arr)/sizeof(arr[0]);
    clrscr();
    ret=distinctElementArray(arr, n);
    for(i=0;i<5;i++)
    printf("%d ",ret[i]);
    return 0;

 } 
i/p
90,91,1,2,90,3,4,3,91
o/p
90 91 1 2 3 4 
----------------------------------------------------------------------------------------------------------------------------------
pattern 
o/p
1*2*3*4
9*10*11*12
13*14*15*16

 5*6*7*8

 #include<stdio.h>
void print(int n)
 {
     int m = 1;
     int i,j;
     for( i=1; i<=n-1; i++)
     {
         if(i!=2)
         {
          for(j=1; j<=n; j++){
              if(j!=n)
           printf("%d*", m);
           else
           printf("%d", m);
           m++;
          }
          printf( "\n");
         }else{
          m = m+n;
          for(j=1; j<=n; j++){
           if(j!=n)
           printf("%d*", m);
           else
           printf("%d", m);
           m++;
          }
          printf("\n");
         }
     }
     for(i=n+1 ;i<=(n+n);i++){
      if(i!=n+n)
      printf("%d*", i);
      else
      printf("%d", i);
     }
     printf("\n");
 }

 int main() {
   print(4);

 }
--------------------------------------------------------------------------
sort elements and display alternative
#include<stdio.h>
int findAlt(int arr[],int len)
{
    int i,j,tmp;
    
    /*
    for(i=1,j=0;i<len;i+=2,j++)
    arr[j]=ar[i];
    len=j-1;
    */
    for(i=0;i<len-1;i++)
    {
        
        for(j=i+1;j<len;j++)
        {
            if(*(arr+i)>*(arr+j))
            {
                tmp=*(arr+i);
                *(arr+i)=*(arr+j);
               *(arr+j)=tmp;
            
            }
        
        }
    }
    
    }
int main()
{
    int no[]={4,2,5,3,7},i;
    findAlt(no,5);
    for(i=0;i<5;i+=2)
    printf("%d",no[i]);
}

--------------------------------------------------------------

 N=5
Output
1
3*2
4*5*6
10*9*8*7
11*12*13*14*15

 #include<stdio.h>
int main()
{
int i,j,n,count=0,k=0;
printf(“Enter N”);
scanf(“%d”,&n);
for(i=1;i<=n;i++)
{ 
count=k;
for(j=1;j<=i;j++)
{ 
if(i%2==0)
{
printf(“%d”,count+i);
count=count-1;
if(j!=i)printf(“*”);
k++;
}
else
{
count=count+1;
printf(“%d”,count);
if(j!=i)printf(“*”);
k++;
}
}
printf(“\n”);
}
return 0;
}
-----------------------------------------------------------------------------------------------------------

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