Algorithm
Mishka has got n empty boxes. For every i (1 ≤ i ≤ n), i-th box is a cube with side length ai.
Mishka can put a box i into another box j if the following conditions are met:
- i-th box is not put into another box;
- j-th box doesn't contain any other boxes;
- box i is smaller than box j (ai < aj).
Mishka can put boxes into each other an arbitrary number of times. He wants to minimize the number of visible boxes. A box is called visible iff it is not put into some another box.
Help Mishka to determine the minimum possible number of visible boxes!
The first line contains one integer n (1 ≤ n ≤ 5000) — the number of boxes Mishka has got.
The second line contains n integers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the side length of i-th box.
Print the minimum possible number of visible boxes.
3
1 2 3
1
4
4 2 4 3
2
In the first example it is possible to put box 1 into box 2, and 2 into 3.
In the second example Mishka can put box 2 into box 3, and box 4 into box 1.
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define endl "\n"
#define debug(n) cout<<(n)<<endl;
const ll INF = 2e18 + 99;
int main(){
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll n;
cin>>n;
unordered_map<ll, ll> mp;
ll x;
ll ans = -1;
while(n--){
cin>>x;
mp[x]++;
ans = max(ans, mp[x]);
}
cout<<ans<<endl;
}Input
1 2 3
Output
Demonstration
Codeforcess Solution 903-C C. Boxes Packing ,C++, Java, Js and Python ,903-C,Codeforcess Solution