137B - Permutation - CodeForces Solution


greedy *1000

Please click on ads to support us..

Python Code:


n = int(input())
a = set(map(int, input().split()))

counter = 0
for i in range(n):
    if i + 1 not in a:
        counter += 1

print(counter)

C++ Code:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
double pi = 3.14159;
const int N = 105;
void solve() {
  int n;
  cin >> n;
  int a[n + 1], freq[5005] = {0};
  int ans = 0;
  for(int i = 1; i <= n; i++) {
    cin >> a[i];
    freq[a[i]]++;
    if(a[i] > n) {
      ans++;
    }
  }
  for(int i = 1; i <= n; i++) {
    ans += max(freq[i], 1) - 1;
  }
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  // int t;
  // cin>>t;

  // while(t--)
      solve();
}




Comments

Submit
0 Comments
More Questions

811A - Vladik and Courtesy
1006B - Polycarp's Practice
1422A - Fence
21D - Traveling Graph
1559B - Mocha and Red and Blue
1579C - Ticks
268B - Buttons
898A - Rounding
1372B - Omkar and Last Class of Math
1025D - Recovering BST
439A - Devu the Singer and Churu the Joker
1323A - Even Subset Sum Problem
1095A - Repeating Cipher
630F - Selection of Personnel
630K - Indivisibility
20B - Equation
600B - Queries about less or equal elements
1015A - Points in Segments
1593B - Make it Divisible by 25
680C - Bear and Prime 100
1300A - Non-zero
1475E - Advertising Agency
1345B - Card Constructions
1077B - Disturbed People
653A - Bear and Three Balls
794A - Bank Robbery
157A - Game Outcome
3B - Lorry
1392A - Omkar and Password
489A - SwapSort