# Check if given string can be made Palindrome by removing only single type of character

• Difficulty Level : Easy
• Last Updated : 23 Nov, 2021

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:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: S = “abczdzacb”
Output: Yes
Explanation: Remove first and second occurrence of character ‘a’, string S becomes “bczdzcb”, which is a palindrome .

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 ``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 st = ``new` `HashSet();``    ` `    ``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

 ``

Output:
`Yes`

Time Complexity: O(n*26)
Auxiliary Space:  O(n)

My Personal Notes arrow_drop_up