Check if given string can be made Palindrome by removing only single type of character
Given a string S, the task is to whether a string can be made palindrome after removing the occurrences of the same character, any number of times
Examples:
Input: S = “abczdzacb”
Output: Yes
Explanation: Remove first and second occurrence of character ‘a’, string S becomes “bczdzcb”, which is a palindrome .
Input: S = “madem”
Output: No
Approach: The task can be solved by iterating over each unique character in the given string, and removing its occurrences wherever there is a mismatch, if a valid palindrome is found, after removing occurrences of the same character any number of times, return “Yes” else return “No“.
Follow the below steps to solve the problem:
- Start iterating over each unique character of the string, whose occurrences are to be deleted
- Use the two-pointer technique, to check for a mismatch, Place l at the start of the string and r at the end of the string
- If S[l] == S[r], increment l, and decrement r.
- If S[l]!= S[r], check if S[l[ == char, do l++, else if S[r] == char, do r–
- If none of the conditions hold means the given can’t be converted into a palindrome
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to check if a palindrome is // possible or not string isPossible(string S) { // Stores the length of string int n = (int)S.length(); // Stores the unique characters in // the string set<char> st; for (int i = 0; i < n; i++) { st.insert(S[i]); } // Check if valid palindrome is // possible or not bool check = false; // Iterating over unique characters // of the string for (auto ele : st) { // Pointers to check the condition int low = 0, high = n - 1; bool flag = true; // Iterating over the string for (int i = 0; i < n; i++) { if (S[low] == S[high]) { // Updating low and high low++; high--; } else { if (S[low] == ele) { // Updating low low++; } else if (S[high] == ele) { // Updating high high--; } else { // It is impossible // to make palindrome // by removing // occurrences of char flag = false; break; } } } // If palindrome is formed // break the loop if (flag == true) { check = true; break; } } if (check) return "Yes"; else return "No"; } // Driver Code int main() { string S = "abczdzacb"; cout << isPossible(S); return 0; }
Java
// Java code for the above approach import java.util.*; class GFG { // Function to check if a palindrome is // possible or not static String isPossible(String S) { // Stores the length of string int n = S.length(); // Stores the unique characters in // the string Set<Character> st = new HashSet<Character>(); for (int i = 0; i < n; i++) { st.add(S.charAt(i)); } // Check if valid palindrome is // possible or not boolean check = false; // Iterating over unique characters // of the string for (Character ele : st) { // Pointers to check the condition int low = 0, high = n - 1; boolean flag = true; // Iterating over the string for (int i = 0; i < n; i++) { if (S.charAt(low) == S.charAt(high)) { // Updating low and high low++; high--; } else { if (S.charAt(low) == ele) { // Updating low low++; } else if (S.charAt(high)== ele) { // Updating high high--; } else { // It is impossible // to make palindrome // by removing // occurrences of char flag = false; break; } } } // If palindrome is formed // break the loop if (flag == true) { check = true; break; } } if (check) return "Yes"; else return "No"; } // Driver Code public static void main (String[] args) { String S = "abczdzacb"; System.out.println(isPossible(S)); } } // This code is contributed by Potta Lokesh
Python3
# python program for the above approach # Function to check if a palindrome is # possible or not def isPossible(S): # Stores the length of string n = len(S) # Stores the unique characters in # the string st = set() for i in range(0, n): st.add(S[i]) # Check if valid palindrome is # possible or not check = False # Iterating over unique characters # of the string for ele in st: # Pointers to check the condition low = 0 high = n - 1 flag = True # Iterating over the string for i in range(0, n): if (S[low] == S[high]): # Updating low and high low += 1 high -= 1 else: if (S[low] == ele): # Updating low low += 1 elif (S[high] == ele): # Updating high high -= 1 else: # It is impossible # to make palindrome # by removing # occurrences of char flag = False break # If palindrome is formed # break the loop if (flag == True): check = True break if (check): return "Yes" else: return "No" # Driver Code if __name__ == "__main__": S = "abczdzacb" print(isPossible(S)) # This code is contributed by rakeshsahni
C#
// C# code for the above approach using System; using System.Collections.Generic; public class GFG { // Function to check if a palindrome is // possible or not static String isPossible(String S) { // Stores the length of string int n = S.Length; // Stores the unique characters in // the string HashSet<char> st = new HashSet<char>(); for (int i = 0; i < n; i++) { st.Add(S[i]); } // Check if valid palindrome is // possible or not bool check = false; // Iterating over unique characters // of the string foreach (char ele in st) { // Pointers to check the condition int low = 0, high = n - 1; bool flag = true; // Iterating over the string for (int i = 0; i < n; i++) { if (S[low] == S[high]) { // Updating low and high low++; high--; } else { if (S[low] == ele) { // Updating low low++; } else if (S[high]== ele) { // Updating high high--; } else { // It is impossible // to make palindrome // by removing // occurrences of char flag = false; break; } } } // If palindrome is formed // break the loop if (flag == true) { check = true; break; } } if (check) return "Yes"; else return "No"; } // Driver Code public static void Main(String[] args) { String S = "abczdzacb"; Console.WriteLine(isPossible(S)); } } // This code is contributed by shikhasingrajput
Javascript
<script> // Javascript program for the above approach // Function to check if a palindrome is // possible or not function isPossible(S) { // Stores the length of string let n = S.length; // Stores the unique characters in // the string let st = new Set(); for (let i = 0; i < n; i++) { st.add(S[i]); } // Check if valid palindrome is // possible or not let check = false; // Iterating over unique characters // of the string for (ele of st) { // Pointers to check the condition let low = 0, high = n - 1; let flag = true; // Iterating over the string for (let i = 0; i < n; i++) { if (S[low] == S[high]) { // Updating low and high low++; high--; } else { if (S[low] == ele) { // Updating low low++; } else if (S[high] == ele) { // Updating high high--; } else { // It is impossible // to make palindrome // by removing // occurrences of char flag = false; break; } } } // If palindrome is formed // break the loop if (flag == true) { check = true; break; } } if (check) return "Yes"; else return "No"; } // Driver Code let S = "abczdzacb"; document.write(isPossible(S)); // This code is contributed by saurabh_jaiswal. </script>
Yes
Time Complexity: O(n*26)
Auxiliary Space: O(1) because the size of set cannot exceed 26 if only lowercase alphabets are considered.
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