spiral matrix for odd number of rows

#include<stdio.h>
int k=0,res[100]={0},flag=0,a[100][100]={0},n,m;

void spiral(){
int i,row;
row=0;
while(row<=n/2){
for(i=row;i<m-1-row;i++){
 res[k]=a[row][i];
 printf(" %d",res[k]);
 k++;
}
for(i=row;i<n-1-row;i++){
 res[k]=a[i][m-1-row];
 printf(" %d",res[k]);
 k++;
}
for(i=m-1-row;i>row && row<n/2;i--){
 res[k]=a[n-1-row][i];
 printf(" %d",res[k]);
 k++;
}
for(i=n-1-row;i>row;i--){
 res[k]=a[i][row];
 printf(" %d",res[k]);
 k++;
}
row++;
}
if(flag==1 && (n==m)){
printf(" %d",a[n/2][n/2]);
}
if(flag==1 && (n!=m))
printf(" %d",a[n-row][m-row]);
}
// DRIVER PROGRAM
void main(){
int i,j;
//clrscr();
scanf("%d %d",&n,&m);
for(i=0;i<n;i++){
for(j=0;j<m;j++){
scanf("%d",&a[i][j]);
}
}
// CHECK IF NUMBER OF ROW IS ODD OR EVEN
if(n%2!=0)
flag=1;
else
flag=0;
spiral();

//getch();
}

To find the given sting is palindrome or circular palindrome or not a palindrome

#include<stdio.h>

#include<conio.h>

int first,last,len,palindrome=0,true,flag=0;

char s[100];

// TO CHECK WHETHER THE GIVEN STRING IS PALINDROME 

int ispal(){

last=len-1;

for(first=0;first<len/2;first++){

if(s[first]!=s[last--]){

return 0;

}

}

return 1;

}

// TO CHECK WHETHER THE GIVEN STRING IS CIRCULAR PALINDROME

int pal(int f,int l){

int count=0;

for(;count<len;){

if(f==len) // IF FIRST REACHES THE LAST CHARACTER OF THE             f=0;          STRING..

if(l==-1)  // IF LAST REACHES THE FIRST CHARACTER OF THE

l=len-1;      STRING

if(s[f++]!=s[l--]){

return 0;

}

count++;

}

return 1;

}

//TO PRINT THE CIRCULAR PALINDROME STRING

void print(int ii){

int k;

for(k=0;k<len;k++){

if(ii==len)

ii=0;

printf("%c",s[ii++]);

}

}

//DRIVER PROGRAM

void main(){

int i,j;

clrscr();

scanf("%s",s);

for(len=0;s[len]!='\0';len++);

palindrome=ispal();

if(palindrome==1){

printf("palindrome");

}

else{

j=0;

for(i=1;i<len&&true==0;i++,j++){

true=pal(i,j);

if(true==1){

first=i;

last=j;

flag=1;

}

}

if(flag==1){

printf("%s is a circular palindrome of ",s);

print(first);

}

else

printf("\nNot palindrome");

}

getch();

}

Anagram Substring Search (Or Search for all permutations)

Given a text txt[0..n-1] and a pattern pat[0..m-1], write a function search(char pat[], char txt[]) that prints all occurrences of pat[] and its permutations (or anagrams) in txt[]. You may assume that n > m. 
Expected time complexity is O(n)

Examples:

1) Input:  txt[] = "BACDGABCDA"  pat[] = "ABCD"
   Output:   Found at Index 0
             Found at Index 5
             Found at Index 6
2) Input: txt[] =  "AAABABAA" pat[] = "AABA"
   Output:   Found at Index 0
             Found at Index 1
             Found at Index 4

// C++ program to search all anagrams of a pattern in a text #include<iostream> #include<cstring> #define MAX 256 using namespace std; // This function returns true if contents of arr1[] and arr2[] // are same, otherwise false. bool compare(char arr1[], char arr2[]) {     for (int i=0; i<MAX; i++)         if (arr1[i] != arr2[i])             return false;     return true; } // This function search for all permutations of pat[] in txt[] void search(char *pat, char *txt) {     int M = strlen(pat), N = strlen(txt);     // countP[]:  Store count of all characters of pattern     // countTW[]: Store count of current window of text     char countP[MAX] = {0}, countTW[MAX] = {0};     for (int i = 0; i < M; i++)     {         (countP[pat[i]])++;         (countTW[txt[i]])++;     }     // Traverse through remaining characters of pattern     for (int i = M; i < N; i++)     {         // Compare counts of current window of text with         // counts of pattern[]         if (compare(countP, countTW))             cout << "Found at Index " << (i - M) << endl;         // Add current character to current window         (countTW[txt[i]])++;         // Remove the first character of previous window         countTW[txt[i-M]]--;     }     // Check for the last window in text     if (compare(countP, countTW))         cout << "Found at Index " << (N - M) << endl; } /* Driver program to test above function */ int main() {     char txt[] = "BACDGABCDA";     char pat[] = "ABCD";     search(pat, txt);     return 0; }

Output:

Found at Index 0
Found at Index 5
Found at Index 6