HackerEarth Multiple of 3 problem solution YASH PAL, 31 July 2024 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. #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) coding problems