Checking whether a String is a Lapindrome or not [duplicate] - c++

Question

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How to find whether the string is a Lapindrome? [closed]
(2 answers)
Closed 2 years ago.
The question is to check whether a given string is a lapindrome or not(CodeChef). According to the question, Lapindrome is defined as a string which when split in the middle, gives two halves having the same characters and same frequency of each character.
I have tried solving the problem using C++ with the code below
#include <iostream>
#include<cstring>
using namespace std;
bool lapindrome(char s[],int len){
int firstHalf=0,secondHalf=0;
char c;
for(int i=0,j=len-1;i<j;i++,j--){
firstHalf += int(s[i]);
secondHalf += int(s[j]);
}
if(firstHalf == secondHalf){
return true;
}
else
return false;
}
int main() {
// your code goes here
int t,len;
bool result;
char s[1000];
cin>>t;
while(t){
cin>>s;
len = strlen(s);
result = lapindrome(s,len);
if(result == true)
cout<<"YES"<<endl;
else
cout<<"NO"<<endl;
--t;
}
return 0;
}
I have taken two count variables which will store the sum of ascii code of characters from first half and second half. Then those two variables are compared to check whether both the halves are equal or not.
I have tried the code on a couple of custom inputs and it works fine. But after I submit the code, the solution seems to be wrong.

Answer 1

Replace the lapindrome function to this one:
bool isLapindrome(std::string str)
{
int val1[MAX] = {0};
int val2[MAX] = {0};
int n = str.length();
if (n == 1)
return true;
for (int i = 0, j = n - 1; i < j; i++, j--)
{
val1[str[i] - 'a']++;
val2[str[j] - 'a']++;
}
for (int i = 0; i < MAX; i++)
if (val1[i] != val2[i])
return false;
return true;
}
Example Output
Input a string here: asdfsasd
The string is NOT a lapindrome.
---
Input a string here: asdfsdaf
The string is a lapindrome.
Enjoy!

Answer 2

You're not counting frequencies of the characters, only their sum. You could simply split the string into halves, create two maps for character frequencies of both sides e.g. std::map containing the count for each character. Then You can compare both maps with something like std::equal to check the complete equality of the maps (to see whether the halves are the same in terms of character frequency).

Answer 3

Instead of counting the frequency of characters (in the two halfs of input string) in two arrays or maps, it's actually sufficient to count them in one as well.
For this, negative counts have to be allowed.
Sample code:
#include <iostream>
#include <string>
#include <unordered_map>
bool isLapindrome(const std::string &text)
{
std::unordered_map<unsigned char, int> freq;
// iterate until index (growing from begin) and
// 2nd index (shrinking from end) cross over
for (size_t i = 0, j = text.size(); i < j--; ++i) {
++freq[(unsigned char)text[i]]; // count characters of 1st half positive
--freq[(unsigned char)text[j]]; // count characters of 2nd half negative
}
// check whether positive and negative counts didn't result in 0
// for at least one counted char
for (const std::pair<unsigned char, int> &entry : freq) {
if (entry.second != 0) return false;
}
// Otherwise, the frequencies were balanced.
return true;
}
int main()
{
auto check = [](const std::string &text) {
std::cout << '\'' << text << "': "
<< (isLapindrome(text) ? "yes" : "no")
<< '\n';
};
check("");
check("abaaab");
check("gaga");
check("abccab");
check("rotor");
check("xyzxy");
check("abbaab");
}
Output:
'': yes
'abaaab': yes
'gaga': yes
'abccab': yes
'rotor': yes
'xyzxy': yes
'abbaab': no
Live Demo on coliru
Note:
About the empty input string, I was a bit uncertain. If it's required to not to count as Lapindrome then an additional check is needed in isLapindrome(). This could be achieved with changing the final
return true;
to
return !text.empty(); // Empty input is considered as false.

Answer 4

The problem with your code was, that you only compare the sum of the characters. What's meant by frequency is that you have to count the occurrence of each character. Instead of counting frequencies in maps like in the other solutions here, you can simply sort and compare the two strings.
#include <iostream>
#include <string>
#include <algorithm>
bool lapindrome(const std::string& s) {
// true if size = 1, false if size = 0
if(s.size() <= 1) return (s.size());
std::string first_half = s.substr(0, s.size() / 2);
std::sort(first_half.begin(), first_half.end());
std::string second_half = s.substr(s.size() / 2 + s.size() % 2);
std::sort(second_half.begin(), second_half.end());
return first_half == second_half;
}
// here's a shorter hacky alternative:
bool lapindrome_short(std::string s) {
if (s.size() <= 1) return (s.size());
int half = s.size() / 2;
std::sort(s.begin(), s.begin() + half);
std::sort(s.rbegin(), s.rbegin() + half); // reverse half
return std::equal(s.begin(), s.begin() + half, s.rbegin());
}
int main() {
int count;
std::string input;
std::cin >> count;
while(count--) {
std::cin >> input;
std::cout << input << ": "
<< (lapindrome(input) ? "YES" : "NO") << std::endl;
}
return 0;
}
Live Demo