Equal MEX CodeChef Solution

Problem -Equal MEX CodeChef Solution

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Equal MEX CodeChef Solution in C++17

#include <bits/stdc++.h>
using namespace std;

int main() {
	// your code goes here
	int t;
	cin>>t;
	while(t--) {
	    int n;
	    cin>>n;
	    int tmp;
	    vector<int>occ(n+1,0);
	    for(int i=0; i<2*n; i++) {
	        cin>>tmp;
	        occ[tmp]++;
	    }
	    
	    int smallest = n+1;
	    for(int i=0; i<=n; i++) {
	        if(occ[i] == 0) {
	            smallest = i;
	            break;
	        }
	    }
        
        if(smallest == n+1) {
            cout<<"NO\n";
            continue;
        }
        
        bool ans = true;
        for(int i=0; i<smallest; i++) {
            if(occ[i]<2) {
                ans = false;
                break;
            }
        }
        
        if(ans) cout<<"YES\n";
        else cout<<"NO\n";
	    
	}
	return 0;
}

Equal MEX CodeChef Solution in C++14

#include <bits/stdc++.h>
using namespace std;

int main() {
	int T;
	cin>>T;
	while(T--)
	{
	    int N;
	    cin>>N;
	    int A[2*N];
	    map<int,int>mp;
	    for(int i = 0; i < 2*N; i++)
	    {
	        cin>>A[i];
	        mp[A[i]]++;
	    }
        int i=0;
        bool flag = 1;
        for(auto it : mp)
	    {
	        if(it.first==i)
            {
                if(it.second<2)
                {
                    flag = 0;
                    break;
                }
            }
            i++;
	    }
        (flag) ? cout<<"YES" : cout<<"NO";
        cout<<endl;
	}
	return 0;
}

Equal MEX CodeChef Solution in PYTH 3

# cook your dish here
from collections import defaultdict
for _ in range(int(input())):
    N = (2*int(input()))
    A = sorted(list(map(int, input().split()))); cnt = i = 0
    while i < N:
        cn = 0
        while i < N and A[i] == cnt:
            cn += 1
            i += 1
        if cn == 1: print("NO"); break
        if cn == 0: print("YES"); break
        cnt += 1
    else:
        print("YES")
    

Equal MEX CodeChef Solution in C

#include <stdio.h>

int main()
{
long long int T,i,N,j,C,k;
scanf("%lld",&T);
for(i=0;i<T;i++){
    scanf("%lld",&N);
    long long int A[N+1];
    for(j=0;j<=N;j++){
        A[j]=0;
    }
    for(j=0;j<2*N;j++){
        scanf("%lld",&C);
        A[C]=A[C]+1;
    }
    for(k=0,j=0;j<=N;j++){
        if(A[j]==0){
            break;
        }
        else if(A[j]==1){
            k=1;
            break;
        }
    }
    (k==0)?(printf("YES\n")):(printf("NO\n"));
}
return 0;
}

Equal MEX CodeChef Solution in JAVA

/* package codechef; // don't place package name! */

import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
class Codechef
{
    public static int solve(){
        int res=0;
        
        return res;
    }
	public static void main (String[] args) throws java.lang.Exception
	{
		PrintWriter out = new PrintWriter(System.out);
		Scanner sc = new Scanner(System.in);
		int testCases = sc.nextInt();
		while(testCases-->0){
		    int n = sc.nextInt();
		    TreeMap<Integer,Integer> map = new TreeMap<>();
		    for(int i=0;i<2*n;i++){
		        int x = sc.nextInt();
		        map.put(x,map.getOrDefault(x,0)+1);
		    }
		    boolean ans = true;
		    int st = 0;
		    for(int key : map.keySet()){
		        if(key != st++) break;
		        int x = map.get(key);
		        if(x == 1) {
		            ans = false;
		            break;
		        }
		    }
		    if(ans) out.println("YES"); 
		    else out.println("NO");
		    
		    
		}
		out.close();
	}
}

Equal MEX CodeChef Solution in PYPY 3

testcases = int(input())
for eachcase in range(testcases):
    length = int(input())
    array = list(map(int, input().split()))
    
    hashmap = {}
    for nums in array:
        hashmap[nums] = 1 + hashmap.get(nums, 0)
    
    currmin = min(array)
    if currmin == 0:
        while True:
            count = hashmap.get(currmin, 0)
            if count == 0:
                print("YES")
                break
            if count < 2:
                print("NO")
                break
            currmin += 1
    else:
        print("YES")

Equal MEX CodeChef Solution in PYTH

t = int(raw_input())
for i in range(t):
	N = int(raw_input())
	st = raw_input().split()
	A = []
	for x in st:
		A.append(int(x))
	# endfor x
	A.sort()
	A.append(10**6)
	found = True
	cnt = 1
	n = 0
	p = 0
	while found and (cnt > 0):
		cnt = 0
		while A[p] == n:
			cnt += 1
			p += 1
		# endwhile
		if cnt == 1:
			found = False
		# endif
		n += 1
	# endwhile
	if found:
		print 'YES'
	else:
		print 'NO'
	# endif
# endfor i

Equal MEX CodeChef Solution in C#

using System;

class Program
{
    void Solve()
    {
        int n=int.Parse(Console.ReadLine());
        string[] nr=Console.ReadLine().Split(' ');
        int[] v = new int[2*n];
        for(int i=0;i<n*2;i++)
        {
            v[i]=int.Parse(nr[i]);
        }
        Array.Sort(v);
        int mex=0;
        int freq=0;
        for(int i=0;i<n*2;i++)
        {
            if(v[i]==mex)
            {
                freq++;
                if(freq==2)
                {
                    freq=0;
                    mex++;
                }
            }
        }
        if(freq==0)Console.WriteLine("YES");
        else Console.WriteLine("NO");
    }
    static void Main()
    {
        int t=int.Parse(Console.ReadLine());
        while(t>0)
        {
            t--;
            new Program().Solve();
        }
    }
}

Equal MEX CodeChef Solution in NODEJS

"use strict";

process.stdin.resume();
process.stdin.setEncoding("utf-8");

let inputString = "";
let currentLine = 0;

process.stdin.on("data", (inputStdin) => {
  inputString += inputStdin;
});

process.stdin.on("end", (_) => {
  inputString = inputString
    .trim()
    .split("\n")
    .map((string) => {
      return string.trim();
    });

  main();
});

function readline() {
  return inputString[currentLine++];
}
// Make a Snippet for the code above this and then write your logic in main();

function main() {
  let T = parseInt(readline());
  
  while(T--){
    let n = parseInt(readline(), 10);
    let arr = readline().split(" ").map(s => parseInt(s, 10));
    const has = {}
    for(let i=0; i < arr.length; i++){
        if (has[arr[i]]) {
            has[arr[i]]+=1
        } else {
            has[arr[i]] = 1
        }
    }
    
    for(let i=0; i<=n; i++){
        if (!has[i]) {
            console.log("YES");
            break;
        } else {
            if(has[i] == 1) {
                console.log("NO");
                break
            }
        }
    }
    //console.log(has);
  }
  return 0;
}

Equal MEX CodeChef Solution in GO

package main

import (
	"bufio"
	"bytes"
	"fmt"
	"os"
	"sort"
)

func main() {

	reader := bufio.NewReader(os.Stdin)

	tc := readNum(reader)
	var buf bytes.Buffer
	for tc > 0 {
		tc--
		n := readNum(reader)
		arr := readNNums(reader, 2*n)
		res := solve(arr)
		if res {
			buf.WriteString("YES\n")
		} else {
			buf.WriteString("NO\n")
		}
	}
	fmt.Print(buf.String())
}

func readInt(bytes []byte, from int, val *int) int {
	i := from
	sign := 1
	if bytes[i] == '-' {
		sign = -1
		i++
	}
	tmp := 0
	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
		tmp = tmp*10 + int(bytes[i]-'0')
		i++
	}
	*val = tmp * sign
	return i
}

func readString(reader *bufio.Reader) string {
	s, _ := reader.ReadString('\n')
	for i := 0; i < len(s); i++ {
		if s[i] == '\n' {
			return s[:i]
		}
	}
	return s
}

func readNum(reader *bufio.Reader) (a int) {
	bs, _ := reader.ReadBytes('\n')
	readInt(bs, 0, &a)
	return
}

func readTwoNums(reader *bufio.Reader) (a int, b int) {
	res := readNNums(reader, 2)
	a, b = res[0], res[1]
	return
}

func readThreeNums(reader *bufio.Reader) (a int, b int, c int) {
	res := readNNums(reader, 3)
	a, b, c = res[0], res[1], res[2]
	return
}

func readNNums(reader *bufio.Reader, n int) []int {
	res := make([]int, n)
	x := 0
	bs, _ := reader.ReadBytes('\n')
	for i := 0; i < n; i++ {
		for x < len(bs) && (bs[x] < '0' || bs[x] > '9') && bs[x] != '-' {
			x++
		}
		x = readInt(bs, x, &res[i])
	}
	return res
}

func readUint64(bytes []byte, from int, val *uint64) int {
	i := from

	var tmp uint64
	for i < len(bytes) && bytes[i] >= '0' && bytes[i] <= '9' {
		tmp = tmp*10 + uint64(bytes[i]-'0')
		i++
	}
	*val = tmp

	return i
}

func solve(arr []int) bool {
	// n := len(arr) / 2
	sort.Ints(arr)
	// first no existing number or single number is the mex
	var mex int
	var i int
	for i < len(arr) {
		if arr[i] > mex {
			return true
		}
		j := i
		for i < len(arr) && arr[i] == arr[j] {
			i++
		}
		if i-j == 1 {
			return false
		}
		mex++
	}
	return true
}

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