HackerEarth Palindromic changes problem solution

YASH PAL, 31 July 202417 February 2026

In this HackerEarth Palindromic changes problem solution You are given the cost of changing M pairs of lowercase characters x to y. You are also given a string S.

You can change a character in string S to another lowercase character by spending the cost you were given earlier. Determine the minimum cost that is required to make S a palindrome.

You can assume that it is always possible to make S a palindrome.

HackerEarth Palindromic changes problem solution

HackerEarth Palindromic changes problem solution.

#include <algorithm>
#include <bitset>
#include <cassert>
#include <chrono>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <limits>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <ratio>
#include <set>
#include <sstream>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <utility>
#include <vector>
#include <climits>
#define ll long long
#define ld long double
#define mp make_pair
#define pb push_back
#define in insert
#define vll vector<ll>
#define endl "n"
#define pll pair<ll,ll>
#define f first
#define s second
#define FOR(i,a,b) for(int i=(a),_b=(b); i<=_b; i++)
#define int ll
#define sz(x) (ll)x.size()
#define all(x) (x.begin(),x.end())
using namespace std;

 
const ll INF = 1e12;
const ll N =(250002); // TODO : change value as per problem
const ll MOD = 1e9+7;
int dis[26][26];
void solve(){
    string s;
    cin >> s;
    int m;
    cin >> m;
    for(int i =0;i<26;i++){
        for(int j =0;j<26;j++){
            dis[i][j] = INF;
        }
        dis[i][i] = 0;
    }
    for(int i =1;i<=m;i++){
        char x,y;
        int c;
        cin >> x >> y >> c;
        dis[x-'a'][y-'a'] = c;
        dis[y-'a'][x-'a'] = c;
    }
    for (int k = 0; k < 26; ++k) {
    for (int i = 0; i < 26; ++i) {
        for (int j = 0; j < 26; ++j) {
            if (dis[i][k] < INF && dis[k][j] < INF)
                dis[i][j] = min(dis[i][j], dis[i][k] + dis[k][j]); 
        }
    }

}

    int ans =0;
    int n = (int)s.size();
    for(int i =0;i<n/2;i++){
        int mi =INF;
        for(char c = 'a';c<='z';c++){
            mi = min(dis[c-'a'][s[i]-'a'] + dis[c-'a'][s[n-i-1]-'a'],mi);
        }
        ans += mi;
    }
    cout << ans << endl;

}
signed main(){
 
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
     // freopen(".in","r",stdin);freopen(".out","w",stdout);
    
     ll tt=1;   
     // cin >> tt;
    while(tt--){    
        solve();
    }    
}

Second solution

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

const int z = 26;
int d[z][z];
int main(){
    string s;
    cin >> s;
    memset(d, 63, sizeof d);
    for(int i = 0; i < z; i++)
        d[i][i] = 0;
    int t;
    cin >> t;
    while(t--){
        char v, u;
        int w;
        cin >> v >> u >> w;
        v -= 'a', u -= 'a';
        d[u][v] = d[v][u] = min(d[v][u], w);
    }
    for(int k = 0; k < z; k++)
        for(int i = 0; i < z; i++)
            for(int j = 0; j < z; j++)
                d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
    ll ans = 0;
    for(int i = 0; i < s.size() / 2; i++){
        int cur = INT_MAX;
        for(int c = 0; c < z; c++)
            cur = min(cur, d[s[i] - 'a'][c] + d[s[s.size() - i - 1] - 'a'][c]);
        assert(cur < INT_MAX);
        ans += cur;
    }
    cout << ans << 'n';
}

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