You are given a string s (only lowercase alphabet) with length n. Print the position of every character of the string it must acquire so that it will form a palindromic string.
Examples:
Input : c b b a a Output : 3 1 5 2 4 To make string palindrome 'c' must be at position 3, 'b' at 1 and 5, 'a' at 2 and 4. Input : a b c Output : Not Possible Any permutation of string cannot form palindrome.
The idea is to create an array of vectors (or dynamic size array) that stores all positions of every character. After storing positions, we check if the count of odd characters is more than one. If yes, we return “Not Possible”. Otherwise, we first print the first half positions from the array, then one position of the odd character (if it is present), and finally the second half positions.
Implementation:
C++
// CPP program to print original// positions of characters in a// string after rearranging and// forming a palindrome#include <bits/stdc++.h>using namespace std;// Maximum number of charactersconst int MAX = 256;void printPalindromePos(string &str){ // Insert all positions of every // character in the given string. vector<int> pos[MAX]; int n = str.length(); for (int i = 0; i < n; i++) pos[str[i]].push_back(i+1); /* find the number of odd elements. Takes O(n) */ int oddCount = 0; char oddChar; for (int i=0; i<MAX; i++) { if (pos[i].size() % 2 != 0) { oddCount++; oddChar = i; } } /* A palindrome cannot contain more than 1 odd characters */ if (oddCount > 1) cout << "NO PALINDROME"; /* Print positions in first half of palindrome */ for (int i=0; i<MAX; i++) { int mid = pos[i].size()/2; for (int j=0; j<mid; j++) cout << pos[i][j] << " "; } // Consider one instance odd character if (oddCount > 0) { int last = pos[oddChar].size() - 1; cout << pos[oddChar][last] << " "; pos[oddChar].pop_back(); } /* Print positions in second half of palindrome */ for (int i=MAX-1; i>=0; i--) { int count = pos[i].size(); for (int j=count/2; j<count; j++) cout << pos[i][j] << " "; }}// Driver codeint main(){ string s = "neveropengk"; printPalindromePos(s); return 0;} |
Java
// JAVA program to print original// positions of characters in a// String after rearranging and// forming a palindromeimport java.util.*;class GFG{// Maximum number of charactersstatic int MAX = 256;static void printPalindromePos(String str){ // Insert all positions of every // character in the given String. Vector<Integer> []pos = new Vector[MAX]; for (int i = 0; i < MAX; i++) pos[i] = new Vector<Integer>(); int n = str.length(); for (int i = 0; i < n; i++) pos[str.charAt(i)].add(i + 1); /* find the number of odd elements. Takes O(n) */ int oddCount = 0; char oddChar = 0; for (int i = 0; i < MAX; i++) { if (pos[i].size() % 2 != 0) { oddCount++; oddChar = (char) i; } } /* A palindrome cannot contain more than 1 odd characters */ if (oddCount > 1) System.out.print("NO PALINDROME"); /* Print positions in first half of palindrome */ for (int i = 0; i < MAX; i++) { int mid = pos[i].size() / 2; for (int j = 0; j < mid; j++) System.out.print(pos[i].get(j) + " "); } // Consider one instance odd character if (oddCount > 0) { int last = pos[oddChar].size() - 1; System.out.print(pos[oddChar].get(last) + " "); pos[oddChar].remove(pos[oddChar].size() - 1); } /* Print positions in second half of palindrome */ for (int i = MAX - 1; i >= 0; i--) { int count = pos[i].size(); for (int j = count / 2; j < count; j++) System.out.print(pos[i].get(j) + " "); }}// Driver codepublic static void main(String[] args){ String s = "neveropengk"; printPalindromePos(s);}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to print original# positions of characters in a# string after rearranging and# forming a palindrome# Maximum number of charactersMAX = 256def printPalindromePos(Str): global MAX # Insert all positions of every # character in the given string. pos = [[] for i in range(MAX)] n = len(Str) for i in range(n): pos[ord(Str[i])].append(i+1) # find the number of odd elements. # Takes O(n) oddCount = 0 for i in range(MAX): if (len(pos[i])% 2 != 0): oddCount += 1 oddChar = i # A palindrome cannot contain more than 1 # odd characters if (oddCount > 1): print("NO PALINDROME") # Print positions in first half # of palindrome for i in range(MAX): mid = len(pos[i]) // 2 for j in range(mid): print(pos[i][j],end = " ") # Consider one instance odd character if (oddCount > 0): last = len(pos[oddChar]) - 1 print(pos[oddChar][last],end = " ") pos[oddChar].pop() # Print positions in second half # of palindrome for i in range(MAX-1,-1,-1): count = len(pos[i]) for j in range(count//2,count): print(pos[i][j],end = " ") # Driver codes = "neveropengk"printPalindromePos(s)# This code is contributed by shinjanpatra. |
C#
// C# program to print original// positions of characters in a// String after rearranging and// forming a palindromeusing System;using System.Collections.Generic;class GFG{// Maximum number of charactersstatic int MAX = 256;static void printPalindromePos(String str){ // Insert all positions of every // character in the given String. List<int> []pos = new List<int>[MAX]; for (int i = 0; i < MAX; i++) pos[i] = new List<int>(); int n = str.Length; for (int i = 0; i < n; i++) pos[str[i]].Add(i + 1); /* find the number of odd elements. Takes O(n) */ int oddCount = 0; char oddChar = (char)0; for (int i = 0; i < MAX; i++) { if (pos[i].Count % 2 != 0) { oddCount++; oddChar = (char) i; } } /* A palindrome cannot contain more than 1 odd characters */ if (oddCount > 1) Console.Write("NO PALINDROME"); /* Print positions in first half of palindrome */ for (int i = 0; i < MAX; i++) { int mid = pos[i].Count / 2; for (int j = 0; j < mid; j++) Console.Write(pos[i][j] + " "); } // Consider one instance odd character if (oddCount > 0) { int last = pos[oddChar].Count - 1; Console.Write(pos[oddChar][last] + " "); pos[oddChar].RemoveAt(pos[oddChar].Count - 1); } /* Print positions in second half of palindrome */ for (int i = MAX - 1; i >= 0; i--) { int count = pos[i].Count; for (int j = count / 2; j < count; j++) Console.Write(pos[i][j] + " "); }}// Driver codepublic static void Main(String[] args){ String s = "neveropengk"; printPalindromePos(s);}}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript program to print original// positions of characters in a// string after rearranging and// forming a palindrome// Maximum number of charactersvar MAX = 256;function printPalindromePos(str){ // Insert all positions of every // character in the given string. var pos = Array.from(Array(MAX), ()=>Array()); var n = str.length; for (var i = 0; i < n; i++) pos[str[i].charCodeAt(0)].push(i+1); /* find the number of odd elements. Takes O(n) */ var oddCount = 0; var oddChar; for (var i=0; i<MAX; i++) { if (pos[i].length % 2 != 0) { oddCount++; oddChar = i; } } /* A palindrome cannot contain more than 1 odd characters */ if (oddCount > 1) document.write( "NO PALINDROME"); /* Print positions in first half of palindrome */ for (var i=0; i<MAX; i++) { var mid = pos[i].length/2; for (var j=0; j<mid; j++) document.write( pos[i][j] + " "); } // Consider one instance odd character if (oddCount > 0) { var last = pos[oddChar].length - 1; document.write( pos[oddChar][last] + " "); pos[oddChar].pop(); } /* Print positions in second half of palindrome */ for (var i=MAX-1; i>=0; i--) { var count = pos[i].length; for (var j=count/2; j<count; j++) document.write( pos[i][j] + " "); }}// Driver codevar s = "neveropengk";printPalindromePos(s);</script> |
Output:
2 1 4 5 7 6 3
Time Complexity : O ( n )
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