HackerEarth Digit Problem solution

In this HackerEarth Digit problem solution we have Given two integers X and K, find the largest number that can be formed by changing digits at atmost K places in the number X.

HackerEarth Digit Problem solution.

#include<bits/stdc++.h>

using namespace std;

#define ff first
#define ss second
#define pb push_back
#define mp make_pair
#define all(x) x.begin(),x.end()
#define sz(x) ((int)x.size())
#define eps 1e-9

const int MOD = 1e9+7;

typedef long long ll;
typedef pair<int,int> pii;

ll POWER[65];
ll power(ll a, ll b) {ll ret=1;while(b) {if(b&1) ret*=a;a*=a;if(ret>=MOD) ret%=MOD;if(a>=MOD) a%=MOD;b>>=1;}return ret;}
ll inv(ll x) {return power(x,MOD-2);}

void precompute() {
  POWER[0]=1;
  for(int i=1;i<63;i++) POWER[i]=POWER[i-1]<<1LL;
}
int main() {
  precompute();
  ll n,k;
  cin>>n>>k;
  assert(n>=1LL and n<=1000000000000000000LL);
  assert(k>=0 and k<=9);
  vector<int> v;
  while(n) {
    v.pb(n%10);
    n/=10;
  }
  reverse(all(v));
  for(int i=0;i<sz(v) and k--;i++) {
    if(v[i]==9) {
      k++;
      continue;
    }
    else v[i]=9;
  }
  for(auto it:v) cout<<it;
  cout<<endl;
  return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;

int main()
{
    string s;
    int k;
    
    cin >> s >> k;
    
    assert(k >= 0 && k <= 9);
    assert(s.size() >= 1 && s.size() <= 19);
    
    if ( s.size() == 19 ) assert(s == "1000000000000000000");
    
    for ( int i = 0; i < s.size(); i++ ) assert(s[i] >= '0' && s[i] <= '9');
    
    for ( int i = 0; i < s.size() && k > 0; i++ ) {
        if ( s[i] == '9' ) continue;
        s[i] = '9', k--;
    }
    
    cout << s << endl;
    
    return 0;
}

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