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 LokeshPython3
# 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>
Output:
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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