HackerEarth An Easy Problem solution

In this HackerEarth An Easy problem solution Mr X loves his clock. His clock displays time in the HH : MM Format using the 24 hour system, the first 2 characters in the string are used to display the current hour, with possible leading zeros. Also, the last 2 characters are used to display the current minutes, also with possible leading zeroes.

Now, Mr X came up with a new thing, where he calls a particular time displayed by the clock good if the sum of all digits written on the clock during that time is divisible by a given number x.

You have been given the current time displayed by the clock and an integer x. You need to find the minimum number of minutes Mr X needs to wait, to see a good time being displayed by the clock. If the clock is already displaying a good time, Mr X does not need to wait at all.

If the clock will never ever display a good time, then print -1 instead as the answer.

HackerEarth An Easy Problem solution.

#include<bits/stdc++.h>

using namespace std;

typedef complex<double> base;
typedef long double ld;
typedef long long ll;

#define pb push_back
#define pii pair<int,int>
#define pll pair< ll , ll >
#define vi vector<int>
#define vvi vector< vi >

const int maxn=(int)(1e5+5);
const ll mod=(ll)(1e9+7);

bool check(int h,int m,int x)
{
    string s1=to_string(h),s2=to_string(m);

    int ret=0;

    for(char ch:s1)
    {
        ret+=(ch-'0');
    }

    for(char ch:s2)
    {
        ret+=(ch-'0');
    }

    return (ret%x==0);
}

int main()
{
    string s;cin>>s;

    int x;cin>>x;

    int h=((s[0]-'0')*10)+((s[1]-'0'));

    int m=((s[3]-'0')*10)+(s[4]-'0');

    int ctr=0,ans=-1;

    while(ctr<=1440)
    {
        if(check(h,m,x))
        {
            ans=ctr;break;
        }

        m++;

        if(m==60)
        {
            h=(h+1)%24;m=0;
        }

        ctr++;
    }

    cout<<ans<<endl;

    return 0;
}

Second solution

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

const int maxt = 24 * 60;
int x;
int main(){
    auto getTime = [](){
        int h, m, ret = 0;
        scanf("%d:%d", &h, &m);
        return m + h * 60;
    };
    int l = getTime();
    scanf("%d", &x);
    for(int i = l; i < maxt; i++){
        string s;
        int y = i;
        for(int j = 0; j < 2; j++, y /= 60)
            s.push_back('0' + y % 60 % 10), s.push_back('0' + y % 60 / 10);
        if((accumulate(s.begin(), s.end(), 0) - '0' * 4) % x == 0)
            return cout << i - l << 'n', 0;
    }
    cout << "-1n";
}

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