Number of positions where a letter can be inserted such that a string becomes palindrome
Given a string str, we need to find the no. of positions where a letter(lowercase) can be inserted so that string becomes a palindrome.
Examples:
Input : str = "abca" Output : possible palindromic strings: 1) acbca (at position 2) 2) abcba (at position 4) Hence, the output is 2. Input : str = "aaa" Output : possible palindromic strings: 1) aaaa 2) aaaa 3) aaaa 4) aaaa Hence, the output is 4.
Naive Approach:This approach is to insert all 26 alphabets at every position possible i.e., N+1 positions and check at every position if this insertion makes it a palindrome and increase the count.
Efficient Approach:
First you have to observe that we have to make insertion only at the point when the character at that point violates the palindrome condition i.e., . Now, there will be two cases based on the above fact:
Case I: What if the given string is already a palindrome
Then we can only insert at the position such that the insertion does not violate the palindrome.
1) If the length is even then we can always insert any letter in the middle.
2) If the length is odd then we can insert the letter which is in middle, to the left or right to it.
3) In both the cases we can insert the letter which is in middle(let it be ‘CH’), at positions equals to:
(no.of consecutive CH’s to the left of middle letter)*2.
Case II:If it is not a palindrome
As mentioned above we should start inserting at position where , So we increase the count and check for the cases if insertion at any other position makes it a palindrome.
1) If is a palindrome, then we can insert* at any position before
until
, K in range
.(*letter = S[N-i-1])
2.)If is a palindrome, then we can insert* at any position after
until
, K in range
.(*letter = S[i])
In all the cases we keep increasing the count.
C++
// CPP code to find the no.of positions where a // letter can be inserted to make it a palindrome #include <bits/stdc++.h> using namespace std; // Function to check if the string is palindrome bool isPalindrome(string &s, int i, int j) { int p = j; for ( int k = i; k <= p; k++) { if (s[k] != s[p]) return false ; p--; } return true ; } int countWays(string &s) { // to know the length of string int n = s.length(); int count = 0; // if the given string is a palindrome(Case-I) if (isPalindrome(s, 0, n - 1)) { // Sub-case-III) for ( int i = n / 2; i < n; i++) { if (s[i] == s[i + 1]) count++; else break ; } if (n % 2 == 0) // if the length is even { count++; count = 2 * count + 1; // sub-case-I } else count = 2 * count + 2; // sub-case-II } else { for ( int i = 0; i < n / 2; i++) { // insertion point if (s[i] != s[n - 1 - i]) { int j = n - 1 - i; // Case-I if (isPalindrome(s, i, n - 2 - i)) { for ( int k = i - 1; k >= 0; k--) { if (s[k] != s[j]) break ; count++; } count++; } // Case-II if (isPalindrome(s, i + 1, n - 1 - i)) { for ( int k = n - i; k < n; k++) { if (s[k] != s[i]) break ; count++; } count++; } break ; } } } return count; } // Driver code int main() { string s = "abca" ; cout << countWays(s) << endl; return 0; } |
Java
// Java code to find the no.of positions where a // letter can be inserted to make it a palindrome import java.io.*; class GFG { // Function to check if the string is palindrome static boolean isPalindrome(String s, int i, int j) { int p = j; for ( int k = i; k <= p; k++) { if (s.charAt(k) != s.charAt(p)) return false ; p--; } return true ; } static int countWays(String s) { // to know the length of string int n = s.length(); int count = 0 ; // if the given string is a palindrome(Case-I) if (isPalindrome(s, 0 , n - 1 )) { // Sub-case-III) for ( int i = n / 2 ; i < n; i++) { if (s.charAt(i) == s.charAt(i + 1 )) count++; else break ; } if (n % 2 == 0 ) // if the length is even { count++; count = 2 * count + 1 ; // sub-case-I } else count = 2 * count + 2 ; // sub-case-II } else { for ( int i = 0 ; i < n / 2 ; i++) { // insertion point if (s.charAt(i) != s.charAt(n - 1 - i)) { int j = n - 1 - i; // Case-I if (isPalindrome(s, i, n - 2 - i)) { for ( int k = i - 1 ; k >= 0 ; k--) { if (s.charAt(k) != s.charAt(j)) break ; count++; } count++; } // Case-II if (isPalindrome(s, i + 1 , n - 1 - i)) { for ( int k = n - i; k < n; k++) { if (s.charAt(k) != s.charAt(i)) break ; count++; } count++; } break ; } } } return count; } // Driver code public static void main(String[] args) { String s = "abca" ; System.out.println(countWays(s)); } } // This code is contributed by vt_m. |
Python 3
# Python 3 code to find the no.of positions # where a letter can be inserted to make it # a palindrome # Function to check if the string # is palindrome def isPalindrome(s, i, j): p = j for k in range (i, p + 1 ): if (s[k] ! = s[p]): return False p - = 1 return True def countWays(s): # to know the length of string n = len (s) count = 0 # if the given string is a palindrome(Case-I) if (isPalindrome(s, 0 , n - 1 )) : # Sub-case-III) for i in range (n / / 2 , n): if (s[i] = = s[i + 1 ]): count + = 1 else : break if (n % 2 = = 0 ): # if the length is even count + = 1 count = 2 * count + 1 # sub-case-I else : count = 2 * count + 2 # sub-case-II else : for i in range (n / / 2 ) : # insertion point if (s[i] ! = s[n - 1 - i]) : j = n - 1 - i # Case-I if (isPalindrome(s, i, n - 2 - i)) : for k in range (i - 1 , - 1 , - 1 ): if (s[k] ! = s[j]): break count + = 1 count + = 1 # Case-II if (isPalindrome(s, i + 1 , n - 1 - i)) : for k in range (n - i, n) : if (s[k] ! = s[i]): break count + = 1 count + = 1 break return count # Driver code if __name__ = = "__main__" : s = "abca" print (countWays(s)) # This code is contributed by ita_c |
C#
// C# code to find the no. of positions // where a letter can be inserted // to make it a palindrome. using System; class GFG { // Function to check if the // string is palindrome static bool isPalindrome(String s, int i, int j) { int p = j; for ( int k = i; k <= p; k++) { if (s[k] != s[p]) return false ; p--; } return true ; } static int countWays(String s) { // to know the length of string int n = s.Length; int count = 0; // if the given string is // a palindrome(Case-I) if (isPalindrome(s, 0, n - 1)) { // Sub-case-III) for ( int i = n / 2; i < n; i++) { if (s[i] == s[i + 1]) count++; else break ; } // if the length is even if (n % 2 == 0) { count++; // sub-case-I count = 2 * count + 1; } else // sub-case-II count = 2 * count + 2; } else { for ( int i = 0; i < n / 2; i++) { // insertion point if (s[i] != s[n - 1 - i]) { int j = n - 1 - i; // Case-I if (isPalindrome(s, i, n - 2 - i)) { for ( int k = i - 1; k >= 0; k--) { if (s[k] != s[j]) break ; count++; } count++; } // Case-II if (isPalindrome(s, i + 1, n - 1 - i)) { for ( int k = n - i; k < n; k++) { if (s[k] != s[i]) break ; count++; } count++; } break ; } } } return count; } // Driver code public static void Main() { String s = "abca" ; Console.Write(countWays(s)); } } // This code is contributed by nitin mittal |
PHP
<?php // PHP code to find the no. of // positions where a letter can // be inserted to make it a palindrome // Function to check if the // string is palindrome function isPalindrome( $s , $i , $j ) { $p = $j ; for ( $k = $i ; $k <= $p ; $k ++) { if ( $s [ $k ] != $s [ $p ]) return false; $p --; } return true; } function countWays( $s ) { // to know the length of string $n = strlen ( $s ); $count = 0; // if the given string is // a palindrome(Case-I) if (isPalindrome( $s , 0, $n - 1)) { // Sub-case-III) for ( $i = $n / 2; $i < $n ; $i ++) { if ( $s [ $i ] == $s [ $i + 1]) $count ++; else break ; } // if the length is even if ( $n % 2 == 0) { $count ++; // sub-case-I $count = 2 * $count + 1; } else // sub-case-II $count = 2 * $count + 2; } else { for ( $i = 0; $i < $n / 2; $i ++) { // insertion point if ( $s [ $i ] != $s [ $n - 1 - $i ]) { $j = $n - 1 - $i ; // Case-I if (isPalindrome( $s , $i , $n - 2 - $i )) { for ( $k = $i - 1; $k >= 0; $k --) { if ( $s [ $k ] != $s [ $j ]) break ; $count ++; } $count ++; } // Case-II if (isPalindrome( $s , $i + 1, $n - 1 - $i )) { for ( $k = $n - $i ; $k < $n ; $k ++) { if ( $s [ $k ] != $s [ $i ]) break ; $count ++; } $count ++; } break ; } } } return $count ; } // Driver code $s = "abca" ; echo countWays( $s ) ; // This code is contributed by nitin mittal ?> |
Javascript
<script> // Javascript code to find the no.of positions where a // letter can be inserted to make it a palindrome function isPalindrome(s,i,j) { let p = j; for (let k = i; k <= p; k++) { if (s[k] != s[p]) return false ; p--; } return true ; } // Function to check if the string is palindrome function countWays(s) { // to know the length of string let n = s.length; let count = 0; // if the given string is a palindrome(Case-I) if (isPalindrome(s, 0, n - 1)) { // Sub-case-III) for (let i = n / 2; i < n; i++) { if (s[i] == s[i + 1]) count++; else break ; } if (n % 2 == 0) // if the length is even { count++; count = 2 * count + 1; // sub-case-I } else count = 2 * count + 2; // sub-case-II } else { for (let i = 0; i < n / 2; i++) { // insertion point if (s[i] != s[n - 1 - i]) { let j = n - 1 - i; // Case-I if (isPalindrome(s, i, n - 2 - i)) { for (let k = i - 1; k >= 0; k--) { if (s[k] != s[j]) break ; count++; } count++; } // Case-II if (isPalindrome(s, i + 1, n - 1 - i)) { for (let k = n - i; k < n; k++) { if (s[k] != s[i]) break ; count++; } count++; } break ; } } } return count; } // Driver code let s = "abca" ; document.write( countWays(s)); // This code is contributed by avanitrachhadiya2155 </script> |
Output:
2
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