Given a string of lowercase letters. Find minimum characters to be inserted in the string so that it can become palindrome. We can change the positions of characters in the string.
Examples:
Input : neveropen Output : 2 neveropen can be changed as: neveropenroforskeeg neveropenorfroskeeg and many more Input : aabbc Output : 0 aabbc can be changed as: abcba bacab
Method 1: A palindromic string can have one odd character only when the length of the string is odd otherwise all characters occur an even number of times. So, we have to find characters that occur at odd times in a string.
The idea is to count the occurrence of each character in a string. As a palindromic string can have one character which occurs odd times, so the number of insertion will be one less than the count of characters that occur at odd times. And if the string is already palindrome, we do not need to add any character, so the result will be 0.
Implementation:
C++
// CPP program to find minimum number // of insertions to make a string // palindrome #include <bits/stdc++.h> using namespace std; // Function will return number of // characters to be added int minInsertion(string str) { // To store string length int n = str.length(); // To store number of characters // occurring odd number of times int res = 0; // To store count of each // character int count[26] = { 0 }; // To store occurrence of each // character for (int i = 0; i < n; i++) count[str[i] - 'a']++; // To count characters with odd // occurrence for (int i = 0; i < 26; i++) if (count[i] % 2 == 1) res++; // As one character can be odd return // res - 1 but if string is already // palindrome return 0 return (res == 0) ? 0 : res - 1; } // Driver program int main() { string str = "neveropen"; cout << minInsertion(str); return 0; } |
Java
// Java program to find minimum number // of insertions to make a string // palindrome public class Palindrome { // Function will return number of // characters to be added static int minInsertion(String str) { // To store string length int n = str.length(); // To store number of characters // occurring odd number of times int res = 0; // To store count of each // character int[] count = new int[26]; // To store occurrence of each // character for (int i = 0; i < n; i++) count[str.charAt(i) - 'a']++; // To count characters with odd // occurrence for (int i = 0; i < 26; i++) { if (count[i] % 2 == 1) res++; } // As one character can be odd return // res - 1 but if string is already // palindrome return 0 return (res == 0) ? 0 : res - 1; } // Driver program public static void main(String[] args) { String str = "neveropen"; System.out.println(minInsertion(str)); } } |
Python3
# Python3 program to find minimum number # of insertions to make a string # palindrome import math as mt # Function will return number of # characters to be added def minInsertion(tr1): # To store string length n = len(str1) # To store number of characters # occurring odd number of times res = 0 # To store count of each # character count = [0 for i in range(26)] # To store occurrence of each # character for i in range(n): count[ord(str1[i]) - ord('a')] += 1 # To count characters with odd # occurrence for i in range(26): if (count[i] % 2 == 1): res += 1 # As one character can be odd return # res - 1 but if string is already # palindrome return 0 if (res == 0): return 0 else: return res - 1 # Driver Code str1 = "neveropen"print(minInsertion(str1)) # This code is contributed by # Mohit kumar 29 |
C#
// C# program to find minimum number // of insertions to make a string // palindrome using System; public class GFG { // Function will return number of // characters to be added static int minInsertion(String str) { // To store string length int n = str.Length; // To store number of characters // occurring odd number of times int res = 0; // To store count of each // character int[] count = new int[26]; // To store occurrence of each // character for (int i = 0; i < n; i++) count[str[i] - 'a']++; // To count characters with odd // occurrence for (int i = 0; i < 26; i++) { if (count[i] % 2 == 1) res++; } // As one character can be odd // return res - 1 but if string // is already palindrome // return 0 return (res == 0) ? 0 : res - 1; } // Driver program public static void Main() { string str = "neveropen"; Console.WriteLine(minInsertion(str)); } } // This code is contributed by vt_m. |
PHP
<?php // PHP program to find minimum number // of insertions to make a string // palindrome // Function will return number of // characters to be added function minInsertion($str) { // To store string length $n = strlen($str); // To store number of characters // occurring odd number of times $res = 0; // To store count of each // character $count = array(26); // To store occurrence of each // character for ($i = 0; $i < $n; $i++) $count[ord($str[$i]) - ord('a')]++; // To count characters with odd // occurrence for ($i = 0; $i < 26; $i++) { if ($count[$i] % 2 == 1) $res++; } // As one character can be odd return // res - 1 but if string is already // palindrome return 0 return ($res == 0) ? 0 : $res - 1; } // Driver program $str = "neveropen"; echo(minInsertion($str)); // This code is contributed // by Mukul Singh ?> |
Javascript
<script> // JavaScript program to find minimum number // of insertions to make a string // palindrome // Function will return number of // characters to be added function minInsertion(str) { // To store string length let n = str.length; // To store number of characters // occurring odd number of times let res = 0; // To store count of each // character let count = new Array(26); for(let i=0;i<count.length;i++) { count[i]=0; } // To store occurrence of each // character for (let i = 0; i < n; i++) count[str[i].charCodeAt(0) - 'a'.charCodeAt(0)]++; // To count characters with odd // occurrence for (let i = 0; i < 26; i++) { if (count[i] % 2 == 1) res++; } // As one character can be odd return // res - 1 but if string is already // palindrome return 0 return (res == 0) ? 0 : res - 1; } // Driver program let str = "neveropen"; document.write(minInsertion(str)); // This code is contributed by unknown2108 </script> |
Output
2
Time Complexity: O(n)
Auxiliary Space: O(1)
Method 2: An approach using Bit manipulation:
- Create a mask and initialise it to zero.
- For each character in string str, toggle the bit into the mask with its corresponding position in the alphabet.
- Check if mask is equal to zero, and return 0.
- Otherwise, return number of setbit in mask – 1.
Below is the implementation of the above approach:
C++
// CPP program to find minimum number // of insertions to make a string // palindrome #include <bits/stdc++.h> using namespace std; // Function will return number of // characters to be added int minInsertion(string str) { long long mask = 0; for (auto c : str) mask ^= (1 << (c - 'a')); if (mask == 0) return 0; int count = 0; while (mask) { count += mask & 1; mask = mask >> 1; } return count - 1; } // Driver program int main() { string str = "neveropen"; cout << minInsertion(str); return 0; } // This code is contributed by hkdass001 |
Java
// Java program to find minimum number // of insertions to make a string // palindrome import java.util.*; public class GFG { // Function will return number of // characters to be added static int minInsertion(String str) { long mask = 0; for (char c : str.toCharArray()) mask ^= (1 << (c - 'a')); if (mask == 0) return 0; int count = 0; while (mask != 0) { count += mask & 1; mask = mask >> 1; } return count - 1; } // Driver program public static void main(String[] args) { String str = "neveropen"; System.out.println(minInsertion(str)); } } // This code is contributed by Karandeep1234 |
Python3
# Python program to find minimum number # of insertions to make a string # palindrome # Function will return number of # characters to be added def minInsertion(str): mask = 0 for c in str: mask ^= (1 << (ord(c) - ord('a'))) if mask == 0: return 0 count = 0 while mask: count += mask & 1 mask = mask >> 1 return count - 1 str = "neveropen"print(minInsertion(str)) # This code is contributed by ishankhandelwals. |
C#
// C# program to find minimum number // of insertions to make a string // palindrome using System; using System.Collections.Generic; class GFG { // Function will return number of // characters to be added static int minInsertion(string str) { int mask = 0; foreach(char c in str) mask ^= (1 << (c - 'a')); if (mask == 0) return 0; int count = 0; while (mask>0) { count += mask & 1; mask = mask >> 1; } return count - 1; } // Driver program static void Main(string[] args) { string str = "neveropen"; Console.Write(minInsertion(str)); } } // This code is contributed by ratiagrawal. |
Javascript
// JavaScript program to find minimum number of insertions to make a string palindrome // Function will return number of characters to be added function minInsertion(str) { let mask = 0; // loop through the string for (let c of str) { // XOR the ascii value of character with the mask mask ^= (1 << (c.charCodeAt(0) - 'a'.charCodeAt(0))); } // return 0 if mask is 0 if (mask === 0) { return 0; } let count = 0; // loop through the mask while (mask) { count += mask & 1; mask = mask >> 1; } // return the count minus 1 return count - 1; } // Driver program let str = "neveropen"; console.log(minInsertion(str)); |
Output
2
Time Complexity: O(n)
Auxiliary Space: O(1)
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