Problem setter: Kamonasish Roy (Bullet)
Problem tag : Bit mask DP
Tutorial (Prepared by Abu Sufian Rafi (ASRafi) : To solve this problem, we use a Bit-masking Dynamic Programming (DP) approach.
The DP state is defined as dp[i][last][mask], where i is the current index in the string, last represents the last vowel group used, and mask is a bitmask that tracks which vowel groups have been used.
The transitions involve either extending the current group by transforming S[i] to match last, or starting a new group by transforming S[i] to a different vowel type not yet used (tracked via mask).
Time Complexity: O(N * 5 * 2^5).
Note : You can solve it using the permutation method, which is also accepted. Time complexity: O( 5! * 5 * N).

**Permutation method Code **:

#include<bits/stdc++.h>
using namespace std;
const long long M=2e5+10,MOD=1000000007;
typedef long long ll;
int cost[26][26];
vector<string>all;
void precal(){
	string vowel="aeiou";
	for(int i=0;i<(int)vowel.size();i++){
		int l=i;
		int cnt=0;
		while(cnt<5){
			cost[vowel[i]-'a'][vowel[l]-'a']=cnt;
			l++;
			cnt++;
			if(l==5)l=0;
			
		}
		
	}
	all.push_back(vowel);
	while(next_permutation(vowel.begin(),vowel.end())){
		all.push_back(vowel);
	}
	
}
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t=1;
    precal();
    cin>>t;
    while(t--){
    	int n;
    	cin>>n;
    	string s;
    	cin>>s;
    	int mx=1e9;
    	for(auto it:all){
    		string p=it;
    		vector<vector<int>>dp(n+1,vector<int>(5,0));
    		for(int i=0;i<5;i++){
    			for(int j=1;j<=n;j++)
    			{
    				int pro=cost[s[j-1]-'a'][p[i]-'a'];
    				if(i==0){
    					dp[j][i]=dp[j-1][i]+pro;
    				}
    				else{
    					dp[j][i]=dp[j-1][i]+pro;
    					dp[j][i]=min(dp[j][i],dp[j][i-1]);
    				}
    			}
    			mx=min(mx,dp[n][i]);
    		}
    	}
    	cout<<mx<<"\n";
    	
    	
   
     

    
    
}

    
    return 0;
 
}

Using Bit mask ( _Ash__ ):

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

int get_index(char ch) {
    string vowels = "aeiou";
    for(int i = 0; i < vowels.size(); i++) {
        if(vowels[i] == ch) {
            return i;
        }
    }
    assert(false);
}

// i -> j
int get_cost(int i, int j) {
    if(j >= i) return j - i;
    return j - i + 5;
}

int solve(string s) {
    int n = s.size();
    vector<int> v(n);
    for(int i = 0; i < n; i++) {
        v[i] = get_index(s[i]);
    }
    vector<vector<vector<int>>> dp(n, vector<vector<int>>(1<<5, vector<int>(5, 1e9)));
    for(int i = 0; i < n; i++) {
        for(int mask = 0; mask < (1<<5); mask++) {
            for(int j = 0; j < 5; j++) {
                if((mask >> j) & 1) {
                    continue;
                }
                // 
                int prev_best = (i > 0 ? dp[i - 1][mask][j] : 0);
                for(int k = 0; k < 5; k++) {
                    if((mask >> k) & 1) {
                        prev_best = min(prev_best, (i > 0 ? dp[i - 1][mask ^ (1<<k)][k] : 0));
                    }
                }
                dp[i][mask][j] = prev_best + get_cost(v[i], j);
            }
        }
    }

    int ans = 1e9;
    for(int mask = 0; mask < (1<<5); mask++) {
        for(int j = 0; j < 5; j++) {
            ans = min(ans, dp[n - 1][mask][j]);
        }
    }
    return ans;
}


int main() {
    int tc;
    cin >> tc;
    while(tc--) {
        int n;
        cin >> n;
        string s;
        cin >> s;
        cout << solve(s) << endl;
    }
}