Number of positions where a letter can be inserted such that a string becomes palindrome
Last Updated :
20 Jul, 2022
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] . 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 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.
- 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])
- 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.
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 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 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 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# 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 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
?>
<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>
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