WAP to Find all Possible Palindromes in a String Efficiently

This is not the regular palindrome finder. Given a string, we should be able to print all the possible palindromes. Let me quickly give a brief example.

For the following text : abccbab

All the possible palindromes are,

bab

abccba

bccb

cc

However, there is even a simpler solution available. First do a parse for odd occurring palindromes followed by even palindromes.For odd palindromes run through each character from the text. For each character, see if there the pre and post occuring characters are equal, if they are equal print them and do the same for the next levels. In the following example shown below, assume you are at 'y' and see the previous and next characters are equal. If they are see further more until the palindrome functionality ceases. Print all of them whilst this time.

java Program

class FindAllPalindromes {
 public static void main(String[] args){
  FindAllPalindromes finder = new FindAllPalindromes();
  finder.printAllPalindromes("abcddcbaABCDEDCBA");
 }
  
 public void printAllPalindromes(String inputText){
  if(inputText==null){
   System.out.println("Input cannot be null!");
   return;
  }
  if(inputText.length()<=2){
   System.out.println("Minimum three characters should be present");
  }
  //ODD Occuring Palindromes
  int len = inputText.length();
  for(int i=1;i<len-1;i++){
   for(int j=i-1,k=i+1;j>=0&&k<len;j--,k++){
    if(inputText.charAt(j) == inputText.charAt(k)){
     System.out.println(inputText.subSequence(j,k+1));
    }else{
     break;
    }
   }
  }
  //EVEN Occuring Palindromes
  for(int i=1;i<len-1;i++){
   for(int j=i,k=i+1;j>=0&&k<len;j--,k++){
    if(inputText.charAt(j) == inputText.charAt(k)){
     System.out.println(inputText.subSequence(j,k+1));
    }else{
     break;
    }
   }
  }
 
 }
}

/*
Sample Output:
DED
CDEDC
BCDEDCB
ABCDEDCBA
dd
cddc
bcddcb
abcddcba
*/


Time Complexity O(n)
Space Complexity O(n)

Run Here https://ideone.com/weINb

C++program

#include<string.h>
#include <iostream>
using namespace std;
 
void printAllPalindromes(char*);
 
char* subSequence(char*,int,int);
 
int main() {
 char s[] = "abcddcbaABCDEDCBA";
 printAllPalindromes(s);
 return 0;
}
 
char* subSequence(char* mainSequence, int from, int to){
 char * tgt = new char[to-from+1];
 for(int i=0;i<(to-from);i++){
  tgt[i] = mainSequence[i+from];
 }
 tgt[to-from] = '\0';
 return tgt;
}
 
void printAllPalindromes(char* inputText) {
 if(!inputText) {
  printf("Input cannot be null!");
  return;
 }
 if(strlen(inputText)<=2) {
  printf("Minimum three characters should be present\n");
 }
 //ODD Occuring Palindromes
 int len = strlen(inputText);
 for(int i=1;i<len-1;i++) {
  for(int j=i-1,k=i+1;j>=0&&k<len;j--,k++) {
   if(inputText[j] == inputText[k]) {
    char* subSeq = subSequence(inputText,j,k+1);
    cout<<subSeq<<endl;
    delete subSeq;
   } else {
    break;
   }
  }
 }
 //EVEN Occuring Palindromes
 for(int i=0;i<len-1;i++) {
  for(int j=i,k=i+1;j>=0&&k<len;j--,k++) {
   if(inputText[j] == inputText[k]) {
    char* subSeq = subSequence(inputText,j,k+1);
    cout<<subSeq<<endl;
    delete subSeq;
   } else {
    break;
   }
  }
 }
 
}

/*
Sample Output:
DED
CDEDC
BCDEDCB
ABCDEDCBA
dd
cddc
bcddcb
abcddcba
*/


Time Complexity O(n)
Space Complexity O(n)
Run Here https://ideone.com/WhZzz