HackerEarth Rhezo and character frequency problem solution

YASH PAL,
In this HackerEarth Rhezo and character frequency problem solution Rhezo got a string S, a character C, and an integer P as a birthday gift from his best friend JK.
He got curious about the gift and started analyzing it. He found out that the maximum frequency of character C over all the P length substrings of S is equal to some integer Z.
He doesn’t know to program, so asks you to find the largest position, where, if we insert some character after that position, the maximum frequency of C over all the P length substrings becomes Z + 1. If there is no such position, output 1.
If the answer is the position before the first character, output 0.
HackerEarth Rhezo and character frequency problem solution

HackerEarth Rhezo and character frequency 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;
}
const int MAXN = 1e3+3;
char str[MAXN];
int fre[MAXN];
int main() {
  // freopen("TASK.in","r",stdin);freopen("TASK.out","w",stdout);
  precompute();
  scanf("%s",str+1);
  int len=strlen(str+1);
  assert(len>=1 and len<=1000);
  char ch;
  cin>>ch;
  int p;
  cin>>p;
  assert(p>=1 and p<=len);
  if(p==1) {
    bool f=false;
    for(int i=1;i<=len;i++) if(str[i]==ch) f=true;
    if(f) cout<<-1<<endl;
    else cout<<len<<endl;
    return 0;
  }
  if(str[1]==ch) fre[1]=1;
  else fre[1]=0;
  for(int i=2;i<=len;i++) {
    fre[i]=fre[i-1];
    if(str[i]==ch) fre[i]++;
  }
  int maxim=-1;
  for(int i=p;i<=len;i++) maxim=max(maxim,fre[i]-fre[i-p]);
  int ans=-1;
  for(int i=p-1;i<=len;i++) {
    int Z = fre[i]-fre[i-(p-1)];
    Z++;
    if(Z==maxim+1) ans=i;
  }
  cout<<ans<<endl;
  return 0;
}

Second solution

#include <bits/stdc++.h>

using namespace std;

int main()
{
    string s;
    char c;
    int n, p, cnt = 0, z = 0, idx = -1;
    
    cin >> s;
    n = (int)s.size();
    assert(n >= 1 && n <= 1000);
    
    for ( int i = 0; i < n; i++ ) assert(s[i] >= 'a' && s[i] <= 'z');
    
    cin >> c;
    assert(c >= 'a' && c <= 'z');
    
    cin >> p;
    assert(p >= 1 && p <= n);
    
    for ( int i = 0; i < p; i++ ) cnt += (s[i] == c);
    z = max(z, cnt);
    
    for ( int i = p; i < n; i++ ) {
        cnt -= (s[i - p] == c);
        cnt += (s[i] == c);
        z = max(z, cnt);
    }
    
    if ( z == p ) {
        puts("-1");
        return 0;
    }
    
    p--;
    
    cnt = 0;

    for ( int i = 0; i < p; i++ ) cnt += (s[i] == c);
    
    if ( cnt == z ) idx = p;
    
    for ( int i = p; i < n; i++ ) {
        cnt -= (s[i - p] == c);
        cnt += (s[i] == c);
        if ( cnt == z ) idx = i + 1;
    }
    
    cout << idx << endl;
    
    return 0;
}
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