Find Longest Palindrome in a string : O(n*n) C code

  Given a string S, find the longest palindromic substring in S.

For example,
S = “caba".

The longest palindromic substring is “aba”.

Algorithm:

1. Palindrome mirrors around center and there are 2N-1 such centers in string. The reason is center of palindrome can be in between two letters(for even length string) or the letter itself (for odd length string).
2. Expand a palindrome around its center in both direction and look for maximum possible palindrome (O(n) time).
3. If the length of string is odd, the center would be letter, therefore repeat the step 2 for each letter and update maximum palindrome on the way.
   
4. If the length of string is even, the center would be between two letters, therefore repeat the step 2 for each such center and update the maximum palindrome on the way.
5. Since there are O(N) such centers, time complexity would be O(n²).

// O(n*n) time complexity and O(1) space algorithm
#include<stdio.h>
#include<string.h>

void longestpalindrome(char *str);

int main()
{
char *str = "abacba";
int i;
longestpalindrome(str);
scanf("%d", &i);
}

void longestpalindrome(char *str) {

int length = strlen(str);
printf("%d\n", length);
int i,j,k;
int start,end,max=0,curr;

if(length == 0) {printf("string null"); return;}

if(length%2!=0) { //if string is of odd length
    for(i=0;i<length-1;i++) {
        j=k=i;
        while(j>0 && k<length-1 && str[--j]==str[++k]); 
        curr = k-j+1;
        if(curr > max) {
           max = curr;
           start = j;
           end = k;
        }
    }
    for(i=start;i<=end;i++) printf("%c", str[i]);
    printf("\n");
}
else { //if string is of even length
    for(i=0;i<length-1;i++) {
        if(str[i]==str[i+1]) {
            j=i;
            k=i+1;
            while(j>=0 && k<=length-1 && str[j--]==str[k++]);
                    curr = k-j-1;
                    if(max < curr) {
                       max = curr;
                       start = j+1;
                       end = k-1;
                    }
        }
        for(i=start;i<=end;i++) printf("%c", str[i]);
        printf("\n");
    }
}
}
  
                         
           

// Time Complexity: O(n²)

// Space Complexity: O(1)