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Palindrome by swapping only one character
In a C++ environment, the palindrome is a process where a set or string remains same from the intial stage of the process to termination of that same particular process. Assume, we have a string denoted as the str[]. And the task is to check the string will be paindrome or not after performing the swapping process with only one character for once.
Here is a general example of the swapping process −
Input : ARBRDD Output : true Explanation: Swap the value A(1st index present in the list) with the R. Input : INDBAN Output : true Explanation: Swap the vlaue of I(0th element present in the index set) with N(last value of the index here) or Swap I(1st value present in the index list) with N(the second value present in the index list) Input : gcagac Output : the value is totally false
Algorithm to find the palindrome by swaping only one character
In this possible algorithm, we are going to perform the printing process of a palindrome by swapping only one character in a C++ environment. With this algorithm, we will build some C++ syntax to learn about the problem statement in an efficient manner.
Step 1 − Start the process.
Step 2 − Declare the input output stream.
Step 3 − Import the built-in classes and declared functions.
Step 4 − Declare the input length.
Step 5 − Prevent the string from being a palindrome.
Step 6 − Count the difference value.
Step 7 − Keep the record of those characters which are preventing the process.
Step 8 − Start a loop till mid point.
Step 9 − Record and store the different characters.
Step 10 − Turn on the switch case.
Step 11 − If the palindrome value found then return a true.
Step 12 − Else return the false as a notation.
Step 13 − Find difference in the mid characters.
Step 14 − Declare driver code and get the return value.
Step 15 − Terminate the process.
Syntax to find the palindrome by swaping only one character
String input = "INPUT STRING";
char[] arr = input.toCharArray();
int swap = 0;
int i = 0;
int j = arr.length-1;
char temp;
while(i<j){
if(arr[i] != arr[j]){
if(arr[i+1] == arr[j]){
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
i++;j--;
swap++;
} else if(arr[i] == arr[j-1]){
temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
i++;j--;
swap++;
} else if(arr[i+1] == arr[j-1] && i+1 != j-1){
temp = arr[j];
arr[j] = arr[j-1];
arr[j-1] = temp;
temp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = temp;
i++;j--;
swap = swap+2;
} else {
swap = -1;break;
}
}
else{
i++;j--;
}
In this possible syntax, we have performed the printing process of a palindrome by swapping only one character in a C++ environment. With these possible syntax, we are heading to build some C++ syntax to learn about the problem statement in a broad manner.
Approaches to Follow
Approach 1 − C++ program to extract a palindrome value by swapping only one character with the bool isPalindromePossible(), str.length() and two pointers method
Approach 2 − C++ program to extract a palindrome value by swapping only one character by using the Lexiocographical value and iteration method
Approach 1: Using the Bool IsPalindromePossible(), Str.length() and two Pointers Method
Use of Bool IsPalindromePossible() Method
In this possible approach, we are going to apply the boolean isPalindromePossible() method to print a palindrome value by swapping only one character from a particular string.
bool check(string s){
int n=s.length();
map<char,int> m;
for(auto i:s){
m[i]++;
}
int cnt=0;
for(auto i=m.begin();i!=m.end();i++)
128{
if(i->second%2){
cnt++;
}
}
if(n%2&&cnt==1){return true;}
if(!(n%2)&&cnt==0){return true;}
return false;
}
Example
//C++ program palindrome by swapping only one character with the bool isPalindromePossible() method
#include <bits/stdc++.h>
using namespace std;
bool isPalindromePossible(string input){
int len = input.length();
int diffCount = 0, i;
char diff[2][2];
for (i = 0; i < len / 2; i++){
if (input[i] != input[len - i - 1]){
if (diffCount == 2) return false;
diff[diffCount][0] = input[i];
diff[diffCount++][1] = input[len - i - 1];
}
}
switch (diffCount){
case 0:
return true;
case 1:{
char midChar = input[i];
if (len % 2 != 0 and
(diff[0][0] == midChar or
diff[0][1] == midChar))
return true;
}
case 2:
if ((diff[0][0] == diff[1][0] and
diff[0][1] == diff[1][1]) or
(diff[0][0] == diff[1][1] and
diff[0][1] == diff[1][0]))
return true;
}
return false;
}
int main(){
cout << boolalpha
<< isPalindromePossible("RDD") << endl;
cout << boolalpha
<< isPalindromePossible("ARBRDD") << endl;
cout << boolalpha
<< isPalindromePossible("INDBDKOLDHAKA") << endl;
return 0;
}
Output
true false false
Use of Str.length() Method
In this possible approach, we are going to apply the str.length() method to print a palindrome value by swapping only one character from a particular string.
Example
//C++ program palindrome by swapping only one character with the possible method to convert the string into palindrome string by changing only one character
#include<bits/stdc++.h>
using namespace std;
bool checkPalindrome(string str){
int n = str.length();
int count = 0;
for (int i = 0; i < n/2; ++i)
if (str[i] != str[n - i - 1])
++count;
return (count <= 1);
}
int main(){
string str = "ARBRDDINDDBDKOLKATADHAKA";
if (checkPalindrome(str))
cout << "Yes - The Process Done " << endl;
else
cout << "No - The Process Is Not Done" << endl;
return 0;
}
Output
No - The Process Is Not Done
Use of two Pointer Approach
In this possible approach, we are going to apply the method by using two pointers to print a palindrome value by swapping only one character from a particular string.
Example
//C++ program palindrome by swapping only one character with the possible method by using two pointer approach
#include <bits/stdc++.h>
using namespace std;
int countSwap(string s){
int n = s.length();
int count = 0;
for (int i = 0; i < n / 2; i++){
int left = i;
int right = n - left - 1;
while (left < right){
if (s[left] == s[right]){
break;
} else{
right--;
}
}
if (left == right){
return -1;
}
for (int j = right; j < n - left - 1; j++){
swap(s[j], s[j + 1]);
count++;
}
}
return count;
}
int main(){
string s = "ARBRDDKOLDHKAINDBD";
int ans1 = countSwap(s);
reverse(s.begin(), s.end());
int ans2 = countSwap(s);
cout << max(ans1, ans2);
return 0;
}
Output
-1
Approach 2: Using the Lexiocographical Value and Iteration Method
Use of the Lexicographical Method
Lexicographical comparison is a comparing function which accepts two arguments for a process. In this possible approach, we are going to apply the lexiocographical method to print a palindrome value by swapping only one character from a particular string.
int main(){
string a;
while(cin>>a){
if(a[0]=='0'){
break;
}
string s;s=a;
int n=s.length();
int cnt=0;
bool ini=false;
if(n%2){ini=true;}
if(check(s)){
for(int i=0;i<n/2;i++){
bool fl=false;
int j=0;
for(j=n-1-i;j>i;j--){
if(s[j]==s[i]){
fl=true;
for(int k=j;k<n-1-i;k++){
swap(s[k],s[k+1]);
cnt++;
cout<<cnt<<endl<<flush;
}
cout<<" "<<i<<" "<<cnt<<endl<<flush;
break;
}
}
if(!fl&&ini){
for(int k=i;k<n/2;k++){
swap(s[k],s[k+1]);
cnt++;
}
cout<<cnt<<" "<<i<<" "<<endl<<flush;
}
}
cout<<cnt<<endl;
}
else{
cout<<"OUTPUT NOTATION PROCESS"<<endl;
Example
//C++ program check whether is it possible to make string A lexicographically smaller than string B to perform the palindrome swaping with one character
#include <bits/stdc++.h>
using namespace std;
void swap(char& x, char& y){
char temp = x;
x = y;
y = temp;
}
bool IsLexicographicallySmaller(
string A, string B){
if (A < B){
return true;
}
string temp = A;
sort(temp.begin(), temp.end());
int index = -1;
for (int i = 0; i < A.length(); i++){
if (A[i] != temp[i]) {
index = i;
break;
}
}
if (index == -1){
return false;
}
int j;
for (int i = 0; i < A.length(); i++){
if (A[i] == temp[index])
j = i;
}
swap(A[index], A[j]);
if (A < B){
return true;
} else {
return false;
}
}
int main(){
string A = "ABONI DAS";
string B = "RUDRA BISWAS";
if (IsLexicographicallySmaller(A, B)) {
cout << "Yes"
<< "\n";
} else {
cout << "No"
<< "\n";
}
return 0;
}
Output
Yes
Use of the Iteration Method
In this possible approach, we are going to apply the iteration method to print a palindrome value by swapping only one character from a particular string loop in the range of (0, N/2). Here we are going to perform the palindrome swaping with one character after removal or neglecting character c.
Example
//C++ program to perform the palindrome swaping with one character after removal or neglecting character c and iterate a loop in the range of (0, N/2)
#include <bits/stdc++.h>
using namespace std;
bool check_palindrome(string str, char c){
int n = str.length(), i = 0, j = n - 1;
while (i < j) {
if (str[i] == c)
i++;
else if (str[j] == c)
j--;
else if (str[i] != str[j])
return 0;
else
i++, j--;
}
return 1;
}
string make_palindrome(string str){
bool is_palindrome = 1;
int n = str.length();
if (n == 1 || n == 2) {
return "YES";
}
for (int i = 0; i < n / 2; ++i) {
if (str[i] != str[n - 1 - i]) {
is_palindrome
= check_palindrome(str, str[i])
|| check_palindrome(
str, str[n - 1 - i]);
break;
}
}
if (is_palindrome)
return "Yes";
else
return "No";
}
int main(){
string S = "ARBRDDKOLKATADHAKAINDBANGLADESH";
string res = make_palindrome(S);
cout << (res);
return 0;
}
Output
No
Conclusion
Today in this article we have learned about how to implement the process to construct and apply the various methods to print a palindrome value set by swapping only one character in a C++ environment. With the above mentioned logics, syntax and algoritm; we have tried to build some C++ codes to solve the problem statement in an efficient manner.
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