All palindromic contiguous substring !

This is so far what I have come up with ! I think there are better ways to do this .  : ) )
But it 's working for my problem
:-P


Here's to the code - >

//Sorry this doesn't work ! I will edit it when I'm busy :-/

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/* @_@ catGotBored @_@ */ #include<bits/stdc++.h> using namespace std; int isPalindrome[100][100]; map<string,int>isThere; void all_palindromic_conti_sub_str(string s) { int N = s.length(); //Lets find all such palindromes with length 1 . Easy , huh ? for(int idx = 0; idx<N; idx++) isPalindrome[idx][idx] = 1; //Lets find all such palindromes with length 2 . Easy again !! for(int idx = 0; idx < N -1 ; idx++) isPalindrome[idx][idx+1] = (s[idx] == s[idx+1] ); //Now,lets do it for all lengths < = N //Idea is simple : //If I know s[i...j] is a palindrome then s[i-1.....j+1] is a palindrome //if and only if s[i-1] == s[j+1] //O(N^2) for marking only for(int idx = 0 ,len = 3; idx <= N -len; idx++,len++) { for(int jdx = idx + len -1 ; jdx < N ; jdx ++ ) { isPalindrome[idx][jdx] = isPalindrome[idx+1][jdx-1] & s[idx] == s[jdx] ; } } / //Bad approach : O(N^3 * log (N)) -> Need to do better // for(int ff = 0; ff<N; ff++) for(int ss = 0; ss<N; ss++) if(isPalindrome[ff][ss]) { string temp; for(int xfsx = ff ; xfsx <= ss ; xfsx++) temp+=s[xfsx] ; // cout<<s[xfsx]; //Every palindromes non distinct ones too if(isThere[temp] == 0) { cout<<temp; isThere[temp]++; cout<<endl; } } } int main() { string str_ = "aassaa"; memset(isPalindrome,0,sizeof(isPalindrome)); // just for nothing all_palindromic_conti_sub_str(str_); return 0; }