write a function to check given string matches with given pattern

write a function to check given string matches with given pattern
Condition: only one wildcard used in the pattern, that is '*', but can be used in the pattern more than once.
Example: 

pattern: *abc*def*.doc*
string: adsfabcxyzdefgh.docx
output : Matching


pattern: *abc*def*.doc*
string: adsfabcxyzdefgh.do1cx
output : Not Matching

public class isMatchingWithWildcard {

public static void main(String [] args){

String pattern = "*abc*def*.doc*";

String str = "adsfabcxyzdefgh.do1docx";

if(isMatching(str, pattern))

System.out.print("Matching");

else

System.out.print("Not Matching");

}

public static Boolean isMatching(String str, String pattern){

int l = pattern.length();

if(pattern.lastIndexOf("*") == l-1)

pattern= (String) pattern.subSequence(0, l-1);

if(pattern.charAt(0) == '*')

pattern= (String) pattern.subSequence(1, pattern.length());

pattern = pattern.replace("*", "__");

String [] patternArray = pattern.split("__");

String sorttenString =str;

for(String aPattern:patternArray){

String [] temp = sorttenString.split(aPattern);

if(temp.length==1 && (aPattern == patternArray[patternArray.length-1] 

&& !sorttenString.toLowerCase().contains(aPattern.toLowerCase())))

return false;

str.substring(temp[0].length()+aPattern.length());

}

return true;

}

}

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Finding Pythagorean triplets in an array.

Finding Pythagorean triplets in an array.
Input :     3 9 4 8 6 11 12
Output :  No such solution exist


Input :    3 5 4 8 6 11 10
Output : 5.0, 3.0, 4.0    10.0, 6.0, 8.0

//   WORKING code in java with  time O(n^2)

import java.util.Arrays;

public class PythagoreanTripletsInArray {

public static double [] getSquare(String [] args){

double [] array = new double[args.length];

for(int i=0;i<args.length;i++){

double j=Integer.parseInt(args[i]);

array[i]=j*j;

}

return array;

}

public static void main(String [] args ){

double [] array =getSquare(args);

Arrays.sort(array);

Boolean flage = true;

for(double aInt : array){

for(int i=0,j=(array.length-1);i<j;){

double sum = array[i]+array[j]; 

if(aInt == sum){

System.out.println(Math.sqrt(aInt)+", "+Math.sqrt(array[i])+", "+Math.sqrt(array[j]));

flage = false;

break;

}

else if(aInt > sum)

i++;

else

j--;

}

}

if(flage)

System.out.print("No such solution exist");

}

}

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Re-arrange the odd/even to odd/even places

Given an array of integers [3,1,4,5,7,6,10,8],
re arrange elements such that either even numbers in even locations or odd numbers in odd locations.
in O(n)

output :  6 1 4 5 8 3 10 7

#include<stdio.h>

void Swap (int A[], int i, int j)
{
    int temp = A[i];
    A[i]=A[j];
    A[j]=temp;
}

int main()
{
    int A[] = {3,1,4,5,7,6,10,8};
    int i=0,j=1,n;
    n=sizeof(A)/sizeof(int);
    while(i<n)
    {
        while(i<n && A[i]%2 == 0)
            i+=2;
        while(j<n && A[j]%2 == 1)
            j+=2;
        if(i<n && j<n)
            Swap(A,i,j);
    }// end of while

    for(i=0;i<n;i++)
        printf(" %d",A[i]);

return 0;
}

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Sort an Array of 0 1& 2 in one pass

 Given an array int a[] = {0, 1, 1 ,0, 2, 2, 1, 0 ,1, 2, 1,0 }
to be converted into {0, 0, 0, 0, 1, 1, 1, 1, 1 ,2 ,2 ,2}
inplace and single pass

Input :  int A[] = {2,2,2,1,1,1,0,0,0,1,2,0};
Output:  0 0 0 0 1 1 1 1 2 2 2 2

#include<stdio.h>

int main()
{
    int A[] = {2,2,2,1,1,1,0,0,0,1,2,0};
    int j,i,x,y,n,t=A[0];
    n=sizeof(A)/sizeof(A[0]);
printf("Size of array : %d\n",n);
    for(x=i=0,y=j=n-1;i<=j;)
    {
        if(t==0)
        {
            A[x]=0;
            x++;
            i++;
            t=A[i];
        }
        else if(t==2)
        {
            t=A[y];
            A[y]=2;
            y--;
            j--;
        }
        else
        {
            i++;
            t=A[i];
        }
    }

    while(x<=y)
    {
        A[x++]=1;
    }

    for(i=0;i<n;i++)
    {
        printf(" %d",A[i]);
    }
return 0;
}

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Given a 2–d matrix , which has only 1’s and 0’s in it. Find the total number of connected sets in that matrix.

Explanation:

Connected set can be defined as group of cell(s) which has 1 mentioned on it and have at least one other cell in that set with which they share the neighbor relationship. A cell with 1 in it and no surrounding neighbor having 1 in it can be considered as a set with one cell in it. Neighbors can be defined as all the cells adjacent to the given cell in 8 possible directions ( i.e N , W , E , S , NE , NW , SE , SW direction ). A cell is not a neighbor of itself.

Input format :

First line of the input contains T , number of test-cases.

Then follow T testcases. Each testcase has given format.

N [ representing the dimension of the matrix N X N ].

Followed by N lines , with N numbers on each line.

Ouput format :

For each test case print one line ,  number of connected component it has.

Sample Input :

4

4

0 0 1 0

1 0 1 0

0 1 0 0

1 1 1 1

4

1 0 0 1

0 0 0 0

0 1 1 0

1 0 0 1

5

1 0 0 1 1

0 0 1 0 0

0 0 0 0 0

1 1 1 1 1

0 0 0 0 0

8

0 0 1 0 0 1 0 0

1 0 0 0 0 0 0 1

0 0 1 0 0 1 0 1

0 1 0 0 0 1 0 0

1 0 0 0 0 0 0 0

0 0 1 1 0 1 1 0

1 0 1 1 0 1 1 0

0 0 0 0 0 0 0 0

Sample output :

1

3

3

9

Constraint :

0 < T < 6 

0 < N < 1009 

=================WORKING CODE=================

#include<stdio.h>

int count_connected_set(int A[1015][1015],int i,int j,int n)

{

    A[i][j]=2;

    if(A[i-1][j-1]==1 && i>0 && j>0 && i<=n && j<=n )     //NW

        count_connected_set(A,i-1,j-1,n);

    if(A[i-1][j]==1 && i>0 && j>0 && i<=n && j<=n)       //N

        count_connected_set(A,i-1,j,n);

    if(A[i-1][j+1]==1 && i>0 && j>0 && i<=n && j<=n)//   //NE

        count_connected_set(A,i-1,j+1,n);

    if(A[i][j-1]==1 && i>0 && j>0 && i<=n && j<=n)       //W

        count_connected_set(A,i,j-1,n);

    if(A[i][j+1]==1 && i>0 && j>0 && i<=n && j<=n)       //E

        count_connected_set(A,i,j+1,n);

    if(A[i+1][j-1]==1 && i>0 && j>0 && i<=n && j<=n)     //SW

        count_connected_set(A,i+1,j-1,n);

    if(A[i+1][j]==1 && i>0 && j>0 && i<=n && j<=n)       //S

        count_connected_set(A,i+1,j,n);

    if(A[i+1][j+1]==1 && i>0 && j>0 && i<=n && j<=n)     //SE

        count_connected_set(A,i+1,j+1,n);

return 1;

}// end of count_connected_set()

int main()

{

    int i,j,t,k,n,A[1015][1015],count;

    scanf("%d",&t);

    while(t--)

    {

        scanf("%d",&n);

        for(i=1;i<=n;i++)

            for(j=1;j<=n;j++)

                scanf("%d",&A[i][j]); // geting input a test case;

        count =0;

        for(i=1;i<=n;i++)

        {

            for(j=1;j<=n;j++)

                if(A[i][j]==1)

                    if(count_connected_set(A,i,j,n))

                        count++;

        }// of for

        printf("%d\n",count);

    }// end of while

return 0;

}

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REVERSE a string and if any charecter is repeating then keep that one comes last

    INPUT           OUTPUT
    ABC             CBA
    ABBC            CBA
    NHHGFDGS        SGDFHN
    ABBCC           CBA

#include<stdio.h>
#include<string.h>
int main()
{
    char S[20],S1[20];
    int i,x,j,l,A[26]={0},count=0;
    gets(S);            // UPPER CASE ONLY
    l=strlen(S);
    for(i=0;i<l;i++)
    {
        x=S[i]-'A'+1;
        if(!A[x])
            count++;
        A[x]++;
    }
    S1[count]=NULL;
    for(j=l-1,i=0;count>i;)
    {
        x=S[j]-'A'+1;
        if(A[x]>0)
        {
            S1[i++]=S[j];
            A[x]=0;
        }
        j--;
    }
    printf("\n> %s",S1);
return 0;
}

WAP to find the position of a given word in a given text

WAP to find the position of a given word in a given text.
  INPUT  :
        Text : my name is kumar name
        Word : name
 OUTPUT :
         Position  : 3 17

#include<stdio.h>
#include<string.h>
int main()
{
    char S[100],W[20];
    int i,j,l,m,flage=0;
    printf("Enter Text : ");
    gets(S);
    printf("\nEnter Word : ");
    gets(W);
    l=strlen(S);
    m=strlen(W);
    for(i=0;i<l;i++)
    {
        if(S[i]==W[0])
        {
            for(j=1;j<m;j++)
            {
                if(S[i+j]!=W[j])
                    break ;
            }
            if(j==m)
            {
                if(!flage)
                    printf("\nPosition : ");
                flage =1;
                printf(" %d",i);
                //flage=
            }
        }
    }
    if(!flage)
        printf("\nNot found \n");
return 0;
}

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Which type of triangle is ?

Write a function that receives three integer inputs for the lengths of the sides of
a triangle and returns one of four values to determine the triangle type
(1=scalene, 2=isosceles, 3=equilateral, 4=error).

#include<stdio.h>
int which_triangle(int a,int b,int c)
{
    if( !(a&&b&&c&&(a+b)>c&&(a+c)>b&&(b+c)>a) )
        return 4;
    else if(a==b && a==c)
        return 3;
    else if(a==b || a==c || b==c)
        return 2;
    else
        return 1;
}

int main()
{
    int a,b,c,x;
    printf("enter 3 side length : ");
    scanf("%d %d %d",&a,&b,&c);
    x=which_triangle(a,b,c);
    switch(x){
    case 1:
        printf("\nScalene Triangle\n");
        break ;
    case 2:
        printf("\nIsoscalene Triangle\n");
        break ;
    case 3:
        printf("\nEquilateral Triangle\n");
        break ;
    case 4:
        printf("\nError\n");
        break ;
    }// end of switch
return 0;
}

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Find the number of substrings of a string that are palindromes

// INPUT : w3asd1dsaf
// OUTPUT
asd1dsa
sd1ds
d1d
total 3 palindrom in given string

#include<stdio.h>
#include<string.h>

int is_palindrom(char S[20],int f,int l)
{
    for(;f<=l;f++,l--)
    {
        if(S[f]!=S[l])
            return 0;
    }
return 1;
}
int main()
{
    char S[20];
    int i,l,j,k,count=0;
    gets(S);
    l=strlen(S);
    for(i=0;i<l;i++)
    {
        for(j=i+1;j<l;j++)
        {
            if(S[i]==S[j])
                if(is_palindrom(S,i,j))
                {
                    printf("\n");
                    count++;
                    for(k=i;k<=j;k++)
                        printf("%c",S[k]);
                }
        }
    }
    printf("\n\nTotal << %d >> no of palindrom in given string\n",count);
return 0;
}

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To copy the contents of an array A to contents of Array B without using loops and any standard string copy functions.

#include<stdio.h>
int Copy_Array(int A[],int B[], int n,int i)
{
    if(i==n)
        return 0;
    B[i]=A[i];
    Copy_Array(A,B,n,i+1);
return 0;
}

int main()
{
    int A[10],B[10],i,n;
    printf("Enter size : ");
    scanf("%d",&n);
    printf("Enter Array : ");
    for(i=0;i<n;i++)
        scanf("%d",&A[i]);
    printf("\nArray A : ");
    for(i=0;i<n;i++)
        printf("%d ",A[i]);
    Copy_Array(A,B,n,0);
    printf("\nArray B : ");
    for(i=0;i<n;i++)
        printf("%d ",B[i]);
return 0;
}

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Verify if the given password is valid or not

Verify if the given password is valid or not.
for a valid password it should follow these 2 condition
1. must be 5-12 characters long
2. must contain atleast one number and one lowercase & one upper character and a symbol



#include<stdio.h>
#include<string.h>
int main()
{
    char S[20],c;
    int i=0,l,L=0,U=0,N=0,Sy=0;
    gets(S);
    l=strlen(S);
    if(l>12 || l<5)
    {
        printf("\n\Invalid password \n");
        return 0;
    }
    while(S[i++]!=NULL)
        if(S[i-1]>='a'&&S[i-1]<='z')
            L=1;
        else if(S[i-1]>='A'&&S[i-1]<='Z')
            U=1;
        else if(S[i-1]>='0'&&S[i-1]<='9')
            N=1;
        else Sy=1;
    if(L&&U&&N&&Sy)
        printf("\nvalid \n");
    else printf("\n\Invalid password \n");
return 0;
}

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Find the seed of a number.

Eg : 1716 = 143*1*4*3 =1716so 143 is the seed of 1716.
find all possible seed for a given number.

// Input   ||  OUTPUT 
//  1250        125       
//  174720      1456   

#include<stdio.h>
int main()
{
    int n,i,j,m,f=0;
    scanf("%d",&n);
    printf("\n");
    for(i=0;i<n;i++)
    {
        m=i;
        j=i;
        while(j)
        {
            m=m*(j%10);
            j=j/10;
        }
        if(m==n)
        {
            printf("\n%d",i);
            f=1;
        }
    }
    if(!f)
        printf("\nNo such number exit\n\n");
return 0;
}

Find duplicates in O(n) time and O(1) extra space

Given an array of n elements which contains elements from 0 to n-1, with any of these numbers appearing any number of times.Find these repeating numbers in O(n) and using only constant memory space.

For example, let n be 7 and array be {1, 2, 3, 1, 3, 0, 6}, the answer should be 1 & 3.

#include<stdio.h>
#include<math.h>
int main()
{
    int n,A[15],i,flage,cycle=0,j,temp;
    printf("\nEnter size of array : ");
    scanf("%d",&n);
    printf("\nEnter elements of array( %d space saperated  integer) : \n",n );
    for(i=0;i<n;i++)
      scanf("%d",&A[i]);
    flage=0;j=0;
    printf("\n OUTPUT : ");

    for(i=0;i<n;i++)
    {
        if( A[abs(A[i])] >= 0)
           A[abs(A[i])]=-A[abs(A[i])];
        else
            printf(" %d",abs(A[i]) );
    }
return 0;
}

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Kahan We Met

Geet is travelling to meet her boyfriend in Mumbai.

Aditya is travelling to Mumbai for his business meeting.

Geet starts from Delhi while Aditya starts from Bangalore. Both of them are travelling on bus passing from town to town.

They meet at a nondescript city and end up travelling together to Mumbai. By the end of the journey they fall in love and settle down.

Enough of Bollywood stories, The problem at hand is to find Kahan they met?

For all interested programmers, Represent this as a programming problem(Data Structure) and find the location where Geet and Aditya would have met.

P.S. No need to actually find any city on the map where they met.

*/

//   ===========================================================================

//      Input Formate :

//

//            12 13 14 21 22            (these are 5 city pincode(2 disit integer pincoede for each city they pass through))

//

//            31 32 33 42 20 21 22      (these are 7 city pincode )

//

//     Output : 21                      (pincode of the city where they will met)

//   ===================================================================


#include<stdio.h>


int kahan_they_met(int A[20],int G[20],int g,int a)
{
    int i_g=0,i_a=0;
    if(g>a)
          i_g+=g-a;
    else  i_a+=a-g;


    while(i_g <= g)
    {
        if(A[i_a]==G[i_g])
            return A[i_a];
        i_g++;
        i_a++;
    }
return 0;
}


int convert_input(int K[],char string[200])
{
    int j=0,i=0,temp=0;
    while(string[i]!='\0')
    {
        temp=0;
        while(string[i]!=' ' && string[i]!='\0')
            temp=temp*10 + (string[i++]-'0') ;
        if(string[i]==' ')
            i++;
        K[j++]=temp;
    }


return j-1;
}




int main()
{
    char string[200];
    int g,a,i,G[20],A[20],met;
    printf("\nEnter city beetween D and M for GEET (eg: 12 13 14 21 22 ) : ");
    gets(string);
    g=convert_input(G,string);


    printf("\n-----------------\nEnter city beetween B and M for  for Aditya visit : ");
    gets(string);
    a=convert_input(A,string);


//    for(i=0;i<=g;i++)
//        printf("\n%d>>%d",i,G[i]);


   met=kahan_they_met(A,G,g,a);


   if(met)
        printf("\n\nGeet and Aditya will met at : %d\n",met);
    else
        printf("\n\nGeet and Aditya will met at Mumbai\n");


return 0;


}

Given an image represented by an NxN matrix, where each pixel in the image is 4 bytes, write a method to rotate the image by 90 degrees. Can you do this in place?

INPUT  >   OUTPUT        INPUT    >  >    OUTPUT  
4                                        4                                         
1 2 3 4        1 1 1 1           11 12 13 14      41 31 21 11
1 2 3 4        2 2 2 2           21 22 23 24      42 32 22 12
1 2 3 4        3 3 3 3           31 32 33 34      43 33 23 13
1 2 3 4        4 4 4 4           41 42 43 44      44 34 24 14



#include<stdio.h>
void rotate(int I[10][10],int O[10][10],int j,int i)
{
    int k,l,temp;
    for(k=0;k<j;k++)
    {
        temp=I[k+i][i];              // copy of Left in temp
        O[k+i][i]=I[i+j-1][i+k];     // copy bottom to left
        O[i+j-1][i+k]=I[i+j-k-1][i+j-1];   // copy right to bottom
        O[i+j-k-1][i+j-1]=I[i][i+j-k-1];   // copy top to right
        O[i][i+j-k-1]=temp;            // copy temp(left) to top
    }
}// end of rotate

int main()
{
    int I[10][10],O[10][10],i,j,N;
    scanf("%d",&N);
    for(i=0;i<N;i++)
    {
        for(j=0;j<N;j++)
            scanf("%d",&I[i][j]);
    }
    i=0;
    j=N;
    while(j>0)
    {
        rotate(I,O,j,i);  // i is base,  j is size
        j=j-2;
        i++;
    }
    printf("\n\n");
    for(i=0;i<N;i++)
    {
        printf("\n");
        for(j=0;j<N;j++)
            printf("%d ",O[i][j]);
    }
return 0;
}

Reduce aabcccccaaa to a2b1c5a3 for new string is of smaller size otherwise live it original

#include<stdio.h>
#include<string.h>
int main()
{
     char s[20],o[20],temp;
     int i=0,j=0,c=0;
     gets(s);
     while(s[i]!=NULL)
     {
         temp=s[i];
         while(s[i]== temp)
         {
             i++;
             c++;
         }
         o[j++]=temp;
         o[j++]=c+48;// ASCII of 0 is 48
         c=0;
     }// end of while
     if(strlen(o)<strlen(s))
        puts(o);
    else
        puts(s);
return 0;
}

Write a method to decide if two strings are premutation of each other or not.

#include<stdio.h>
#include<string.h>
int check_permutation(char s1[], char s2[])
{
    int c[27],i;
    for(i=0;i<27;i++)
        c[i]=0;
    if(strlen(s1)!=strlen(s2))
        return 0;
    for(i=0;s1[i]!=NULL;i++)
    {
        c[0]= s1[i];
        if(c[0]>96)
             c[0]=c[0]-96;
        else c[0]=c[0]-64;
        c[c[0]]++;

        c[0]= s2[i];
        if(c[0]>96)
             c[0]=c[0]-96;
        else c[0]=c[0]-64;
        c[c[0]]--;
    }
    for(i=1;i<27;i++)
    {
        if(c[i]!=0)
            return 0;
    }
    return 1;
}

int main()
{
    char s1[20],s2[20];
    scanf("%s%s",&s1,&s2);
    if( check_permutation(s1,s2) )
       printf("\n Permutation of each other");
    else
        printf("\n NOT a Permutation ");
return 0;
}

Implement an algorithm to determine if a string has all unique characters.

#include<stdio.h>
#include<string.h>
int main()
{
    char input[27];
    int i,l,check[28];
    scanf("%s",&input);
    l=strlen(input);
    for(i=0;i<27;i++)
        check[i]=0;
    for(i=0;i<l;i++)
    {
        check[27]=input[i];
        if(check[27]>96)
            check[27]=check[27]-96;
        else
            check[27]=check[27]-64;
        if(!check[check[27]])
            check[check[27]]=1;
        else
        {
            printf("\nNOT");
            return 0;
        }
    }
    printf("\n unique");
    return 0;
}

Write a C program to reverse all words but not string.

// Input :   My name is Kumar

// Outout: yM eman is ramuK

#include<stdio.h>

int main()

{

    char S[50];

    int i,j,t;

    printf("\nEnter Sentence : ");

    gets(S);

    printf("\nOUTPUT : ");

    for(i=0;S[i-1]!='\0';i++)

    {

        j=i;

        while(S[j]!=' ' && S[j]!=NULL)

            j++;

        t=j;

        while(j>=i)

            printf("%c",S[j--]);

        printf(" ");

        i=t;

    }

    printf("\n");

return 0;

}

In a given String Find the alphabet Which occur most.

        //  Time  ::  O(n)
    //INPUT : asd adg s
    //OUTPUT : a

#include<stdio.h>
#include<string.h>
int main()
{
    char S[20];
    int i=0,j,A[26]={0},C[2]={0,0};
    gets(S);
    printf("%s",S);

    while(S[i]!='\0')
    {
        if(S[i++]==' ')
            continue ;
        j=S[i-1]-'a';
        A[j]++;
        if(C[1]<A[j])
        {
            C[0]=j;
            C[1]=A[j];
        }
    }
    printf("\nOUTPUT : %c\n",C[0]+'a');
return 0;
}

sort an array a[n] containing 0s and 1s with time complexity n and you cant use more than 1 data structure

//      TIME: O(n)  ||   Space : O(1)


#include<stdio.h>
int main()
{
    int A[20],i,j,n,temp;
    printf("Enter n :");
    scanf("%d",&n);
    for(i=0;i<n;i++)
        scanf("%d",&A[i]);
    printf("\n--");
    i=0;
    j=n-1;
    while(i<j)
    {
        while(A[i]==0)
            i++;
        while(A[j]==1)
            j--;
        if(i<j)
        {
            temp=A[i];
            A[i]=A[j];
            A[j]=temp;
            i++; j--;
        }
    }// end of while
    printf("\n");
    for(i=0;i<n;i++)
        printf("%d ",A[i]);
    printf("\n");
return 0;
}

WAP to return numbered index if input is excel sheet column header name. e.g excel sheet column headers are A, B, C , D ... Z, AA, AB...AZ, BA,, etc if Input is D , output should be 4 and for AA output should be 27

INPUT                 OUTPUT

A                          1   

AA                        27   

BA                        53   

AAA                      703  



//    TIME : O(n)

#include<stdio.h>
#include<string.h>
int main()
{
    int i=0,l,x,index=0;
    char S[5];
    printf("Enter Column header : ");
    gets(S);
    l=strlen(S);
    while(i<l)
    {
        x=S[i++]-'A'+1;
        index=index*26+x;
    }
    printf("%d",index);
return 0;
}

WAP to return all anagrams from the array of strings. (all possible permutation of a String)

#include<stdio.h>
#include<string.h>
int anagram(char S[10],int l,char C[10])
{
    char D[10];
    int j,k,m=strlen(C),i;
    for(j=0;j<=m;j++)
    {
        for(k=m;k>j;k--)
            D[k]=C[k-1];
        D[k]=S[m];
        D[m+1]=NULL;
        for(k=j-1;k>=0;k--)
            D[k]=C[k];
        if((m+1)==l)
            printf("%s\n",D);
        else
            anagram(S,l,D);
    }
return 0;

}
int main()
{
    char S[10],C[10];
    C[0]=NULL;
    gets(S);
    printf("\n");
    anagram(S,strlen(S),C);
return 0;
}

Consider you have a grid of size m x n. There are stones placed randomly in some of the squares of this grid. Design a way to find out minimum rectangular area which covers all the stones in this grid.

Smallest Rectangle

Input :(place 1 as rock other wise 0)
5 4
0 0 0 0
0 1 0 0
0 0 0 1
1 0 0 0
0 0 0 0

Output:   (1,0), (1,3), (3,0), (3,3)
*/

#include<stdio.h>
#define MAX(a,b) (a>b?a:b)
#define MIN(a,b) (a<b?a:b)
int main()
{
    int m,n,A[10][10],i,j,i1=10,i2=0,j1=10,j2=0;
    printf("enter m & n (m*n) :");
    scanf("%d %d",&m,&n);
    printf("entet array (place 1 as rock other wise 0) :\n");
    for(i=0;i<m;i++)
        for(j=0;j<n;j++)
            scanf("%d",&A[i][j]);
    for(i=0;i<m;i++)
        for(j=0;j<n;j++)
            if(A[i][j]==1)
            {
                i1=MIN(i1,i);
                i2=MAX(i2,i);
                j1=MIN(j1,j);
                j2=MAX(j2,j);
            }
    printf("\n(%d,%d), (%d,%d), (%d,%d), (%d,%d)\n",i1,j1,i1,j2,i2,j1,i2,j2);
return 0;
}

The character 'a' to 'z' are encoded as 1 - 26. Given a string of digits, compute the number of valid decodings of the string. For example, both 'aa' and 'k' can be encoded as '11'. Hence num_valid_encodings('11') = 2.

      recursive 

#include <stdio.h>
#include <string.h>
int count_decode(char S[10],int n)
{
    int x1=S[n-1]-'0',x2=S[n-2]-'0',z;
    if(!n)
        return 0;
    else if(n==1)
    {
        return 1;
    }
    else
    {
        z=x2*10+x1;
      //  printf("z=%d ",z);
        if(z>26)
            return (count_decode(S,n-1));
        else if(n==2)
            return (count_decode(S,n-1) + count_decode(S,n-2)+1);
        else
            return (count_decode(S,n-1) + count_decode(S,n-2));
    }
}
int main()
{
    int i,j,n;
    char S[10];
    gets(S);
    n=strlen(S);
    printf("< %d >",count_decode(S,n));
return 0;
}

atoi is a function in the C programming language that converts a string into an integer numerical representation. atoi stands for ASCII to integer. It is included in the C standard library header file stdlib.h. Its prototype is as follows:

atoi : ASCII to integer 


    //INPUT : 24 83 0 2 


    //OUTPUT : 24 83 0 2

#include<stdio.h>

#include<string.h>
int atoi(int K[10],char string[30])
{
    int j=0,i=0,temp=0;
    while(string[i]!='\0')
    {
        temp=0;
        while(string[i]!=' ' && string[i]!='\0')
            temp=temp*10 + (string[i++]-'0') ;
        if(string[i]==' ')
            i++;
        K[j++]=temp;
    }
return j;
}

int main()
{
    char string[30];
    int i,a,A[10];
    printf("\nEnter String : ");
    gets(string);
    a=atoi(A,string);
    printf("\nIntegers are : ");
    for(i=0;i<a;i++)
        printf("%d ",A[i]);
    printf("\n");
return 0;
}

String Masking : Given a string"S" and we have to remove a an occurance of a small string "M" from there.

Input:
KUMAR
UM

Output:
KAR



#include<stdio.h>
#include<string.h>
int str_com(char K[20],int i,int m,char M[10])
{
    int k=0;
    for(k=0;k<m;k++)
        if(K[i+k]!=M[k])
            break;
    if(k==m)
        return 1;
    return 0;
}

int main()
{
    char S[20],M[10];
    int i,j=0,l,m,k,flage;
    gets(S);
    gets(M);
    l=strlen(S);
    m=strlen(M);
    for(i=0;i<l;i++)
    {
        if(S[i]!=M[0])
            S[j++]=S[i];
        else
            if( str_com(S,i,m,M) )
                i+=m-1;
            else S[j++]=S[i];
    }
    S[j]=NULL;
    printf("< %s >",S);
return 0;
}