HackerEarth Multiple of 3 problem solution

YASH PAL, 31 July 202415 February 2026

In this HackerEarth Multiple of 3 problem solution, You are given an integer N and your task is to make N a multiple of 3. In order to make N multiple of 3, you can insert at most one digit in N.

Your task is to find the minimum possible N which is a multiple of 3 after inserting at most one digit.

HackerEarth Multiple of 3 problem solution

HackerEarth Multiple of 3 problem solution.

#include<bits/stdc++.h>
using namespace std;
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007
#define endl "n"
#define test ll t; cin>>t; while(t--)
typedef long long int ll;
void solve(){
    string s; cin>>s;
    int n=(int)s.length();
    int sum=0;
    for(int i=0;i<n;i++){
        sum+=(s[i]-'0');
    }
    sum%=3;
    if(sum==0){
        cout<<s<<endl; return;
    }
    bool ok=false;
    int dig=3-sum;
    for(int i=0;i<n;i++){
        if(!ok && s[i]-'0'>dig){
            cout<<dig; ok=true;
        }
        cout<<s[i];
    }
    if(!ok) cout<<dig;
    cout<<endl;
}
int main() {
    FIO;
    test
    {
       solve();
    }
    return 0;
}

Second solution

t = int(input())
while t > 0:
    t -= 1
    n = int(input())
    d = n % 3
    if d == 0:
        print(n)
        continue
    d = str(3 - d)
    n = str(n)
    mn = int(d + n)
    for i in range(len(n)):
        mn = min(mn, int(n[:i + 1] + d + n[i + 1:]))
    print(mn)

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