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., ![S[i] != S[N-i-1]](https://www.geeksforgeeks.org/wp-content/ql-cache/quicklatex.com-d4c5cf11941c2c2702b654d35d214d6a_l3.png "Rendered by QuickLaTeX.com")
. 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.
- If the length is even then we can always insert any letter in the middle.
- If the length is odd then we can insert the letter which is in middle, to the left or right to it.
- 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.
- If
is a palindrome, then we can insert* at any position before
until
, K in range
.(*letter = S[N-i-1])
- 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.
Implementation:
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 palindromebool 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 codeint 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 palindromeimport 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 palindromedef isPalindrome(s, i, j): p = j for k in range(i, p + 1): if (s[k] != s[p]): return False p -= 1 return Truedef 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 codeif __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 palindromefunction 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 palindromefunction 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 codelet s = "abca";document.write( countWays(s)); // This code is contributed by avanitrachhadiya2155</script> |
Output
2
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