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HackerEarth String Clash problem solution

In this HackerEarth String Clash problem solution, you are given two strings S and T of equal lengths. You need to pick some characters from the first string, some from the second string, and then form a new string by rearranging the characters you have picked. You need to find the length of the maximum string that you can make which will be a palindrome.


HackerEarth String Clash problem solution


HackerEarth String Clash problem solution.

#include<bits/stdc++.h>
#define LL long long int
#define M 1000000007
#define endl "\n"
#define eps 0.00000001
LL pow(LL a,LL b,LL m){LL x=1,y=a;while(b > 0){if(b%2 == 1){x=(x*y);if(x>m) x%=m;}y = (y*y);if(y>m) y%=m;b /= 2;}return x%m;}
LL gcd(LL a,LL b){if(b==0) return a; else return gcd(b,a%b);}
LL gen(LL start,LL end){LL diff = end-start;LL temp = rand()%start;return temp+diff;}
using namespace std;
int f[1000001];
int main() {
ios_base::sync_with_stdio(0);
cin.tie(0);
int ans = 0;
string a , b;
cin >> a >> b;
for(int i = 0; i < a.length(); i++) {
f[a[i] - 'a']++;
}
for(int i = 0; i < b.length(); i++) {
f[b[i] - 'a']++;
}
bool flag = 0;
for(int i = 0; i < 26; i++) {
ans = ans + f[i] - (f[i] % 2);
if(f[i] % 2) {
flag = 1;
}
}
cout << ans + flag;
}

Second solution

#include <bits/stdc++.h>

using namespace std;

vector<int> f(26, 0);

int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
string s, t;
cin >> s >> t;
int n = s.size();
for(int i = 0; i < n; i ++)
f[s[i] - 'a'] ++;
for(int i = 0; i < n; i ++)
f[t[i] - 'a'] ++;
bool odd = 0;
int total_len = 0;
for(int i = 0; i < 26; i ++) {
if(f[i] & 1) {
odd = 1;
f[i] --;
}
total_len += f[i];
}
if(odd)
total_len ++;
cout << total_len;
return 0;
}

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