Check if a string can be rearranged to form special palindrome
Given a string str, the task is to check if it can be rearranged to get a special palindromic string. If we can make it print YES else print NO.
A string is called special Palindrome that contains an uppercase and a lower case character of the same character in the palindrome position.
Example: ABCcba is a special palindrome but ABCCBA is not a special palindrome.
Examples:
Input: str = “ABCcba”
Output: YESInput: str = “ABCCBA”
Output: NO
Approach: Check if the occurrence of the uppercase letter of a character is same as the occurrence of lower case letter of the same character. And also there should be only one odd occurring character. We increase the frequency of each uppercase character by 1 and decrease the frequency of each lowercase character by 1. After this, there should be either zero, 1 or -1 in the frequency array If anything else occurs then we directly say NO else print YES.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach #include<bits/stdc++.h> using namespace std; // Driver code int main() { string s = "ABCdcba" ; // creating a array which stores the // frequency of each character int u[26] = {0}; int n = s.length() ; for ( int i = 0; i < n ; i++) { // Checking if a character is uppercase or not if ( isupper (s[i])) { // Increasing by 1 if uppercase u[s[i] - 65] += 1; } else { // Decreasing by 1 if lower case u[s[i] - 97] -= 1 ; } } bool f1 = true ; // Storing the sum of positive // numbers in the frequency array int po = 0 ; // Storing the sum of negative // numbers in the frequency array int ne = 0 ; for ( int i = 0 ; i < 26 ; i++) { if (u[i] > 0) po += u[i] ; if (u[i] < 0) ne += u[i] ; } // If all character balances out then its Yes if (po == 0 && ne == 0) cout << ( "YES" ) << endl; // If there is only 1 character which // does not balances then also it is Yes else if (po == 1 && ne == 0) cout << ( "YES" ) << endl; else if (po == 0 && ne == -1) cout << ( "YES" ) << endl; else cout << ( "NO" ) << endl; } // This code is contributed by // Surendra_Gangwar |
Java
// Java implementation of the above approach public class Improve { public static void main(String args[]) { String s = "ABCdcba" ; // creating a array which stores the // frequency of each character int u[] = new int [ 26 ]; int n = s.length() ; for ( int i = 0 ; i < n ; i++) { // Checking if a character is uppercase or not if (Character.isUpperCase(s.charAt(i))) { // Increasing by 1 if uppercase u[s.charAt(i) - 65 ] += 1 ; } else { // Decreasing by 1 if lower case u[s.charAt(i) - 97 ] -= 1 ; } } boolean f1 = true ; // Storing the sum of positive // numbers in the frequency array int po = 0 ; // Storing the sum of negative // numbers in the frequency array int ne = 0 ; for ( int i = 0 ; i < 26 ; i++) { if (u[i] > 0 ) po += u[i] ; if (u[i] < 0 ) ne += u[i] ; } // If all character balances out then its Yes if (po == 0 && ne == 0 ) System.out.println( "YES" ) ; // If there is only 1 character which // does not balances then also it is Yes else if (po == 1 && ne == 0 ) System.out.println( "YES" ) ; else if (po == 0 && ne == - 1 ) System.out.println( "YES" ) ; else System.out.println( "NO" ) ; } // This code is contributed by ANKITRAI1 } |
Python3
# Python implementation of the above approach s = "ABCdcba" # creating a list which stores the # frequency of each character u = [ 0 ] * 26 n = len (s) for i in range (n): # Checking if a character is uppercase or not if (s[i].isupper()): # Increasing by 1 if uppercase u[ ord (s[i]) - 65 ] + = 1 else : # Decreasing by 1 if lower case u[ ord (s[i]) - 97 ] - = 1 fl = True # Storing the sum of positive # numbers in the frequency array po = 0 # Storing the sum of negative # numbers in the frequency array ne = 0 for i in range ( 26 ): if (u[i] > 0 ): po + = u[i] if (u[i] < 0 ): ne + = u[i] # If all character balances out then its Yes if (po = = 0 and ne = = 0 ): print ( "YES" ) # If there is only 1 character which # does not balances then also it is Yes elif (po = = 1 and ne = = 0 ): print ( "YES" ) elif (po = = 0 and ne = = - 1 ): print ( "YES" ) else : print ( "NO" ) |
C#
// C# implementation of the // above approach using System; class GFG { public static void Main() { string s = "ABCdcba" ; // creating a array which stores // the frequency of each character int [] u = new int [26]; int n = s.Length ; for ( int i = 0; i < n ; i++) { // Checking if a character is // uppercase or not if (Char.IsUpper(s[i])) { // Increasing by 1 if uppercase u[s[i] - 65] += 1 ; } else { // Decreasing by 1 if lower case u[s[i] - 97] -= 1 ; } } // Storing the sum of positive // numbers in the frequency array int po = 0 ; // Storing the sum of negative // numbers in the frequency array int ne = 0 ; for ( int i = 0 ; i < 26 ; i++) { if (u[i] > 0) po += u[i] ; if (u[i] < 0) ne += u[i] ; } // If all character balances // out then its Yes if (po == 0 && ne == 0) Console.Write( "YES" + "\n" ) ; // If there is only 1 character which // does not balances then also it is Yes else if (po == 1 && ne == 0) Console.Write( "YES" + "\n" ) ; else if (po == 0 && ne == -1) Console.Write( "YES" + "\n" ) ; else Console.Write( "NO" + "\n" ) ; } } // This code is contributed // by ChitraNayal |
Javascript
<script> // JavaScript implementation of the approach // driver code let s = "ABCdcba" ; // creating a array which stores the // frequency of each character let u = Array(26).fill(0); let n = s.length ; for (let i = 0; i < n ; i++) { // Checking if a character is // uppercase or not if (s[i].toUpperCase()) { // Increasing by 1 if uppercase u[s[i] - 65] += 1 ; } else { // Decreasing by 1 if lower case u[s[i] - 97] -= 1 ; } } let f1 = true ; // Storing the sum of positive // numbers in the frequency array let po = 0 ; // Storing the sum of negative // numbers in the frequency array let ne = 0 ; for (let i = 0 ; i < 26 ; i++) { if (u[i] > 0) po += u[i] ; if (u[i] < 0) ne += u[i] ; } // If all character balances out then its Yes if (po == 0 && ne == 0) document.write( "YES" ) ; // If there is only 1 character which // does not balances then also it is Yes else if (po == 1 && ne == 0) document.write( "YES" ) ; else if (po == 0 && ne == -1) document.write( "YES" ) ; else document.write( "NO" ) ; </script> |
YES
Time complexity: O(n)
Auxiliary space: O(1)
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