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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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., $S[i] != S[N-i-1]$ . 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 $S[i] != S[N-1-i]$ , So we increase the count and check for the cases if insertion at any other position makes it a palindrome.
1) If $S[i]...S[N-i-2]$ is a palindrome, then we can insert* at any position before $S[i]$ until $S[K] != S[N-i-1]$ , K in range $[i-1, 0]$ .(*letter = S[N-i-1])
2.)If $S[i+1]...S[N-i-1]$ is a palindrome, then we can insert* at any position after $S[n-i-1]$ until $S[K] != S[i]$ , K in range $[N-i, N-1]$ .(*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 
?> 

Output:

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python 3

C#

PHP

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