Hello coders, today we are going to solve Dominant Element Codechef Solutions whose Problem Code is DOMINANT2.
Dominant Element Codechef Solution
Problem
You are given an array A of length N. An element X is said to be dominant if the frequency of X in A is strictly greater than the frequency of any other element in the A.
For example, if A = [2, 1, 4, 4, 4] then 4 is a dominant element since its frequency is higher than the frequency of any other element in A.
Find if there exists any dominant element in A.
Input Format
- The first line of input contains a single integer TT — the number of test cases. Then the test cases follow.
- The first line of each test case contains an integer NN — the size of the array AA.
- The second line of each test case contains NN space-separated integers A_1, A_2, \ldots, A_NA1,A2,…,AN denoting the array AA.
Output Format
For each test case, output YES if there exists any dominant element in A. Otherwise, output NO.
You may print each character of YES and NO in uppercase or lowercase (for example, yes, yEs, Yes will be considered identical).
Constraints
- 1 ≤ T ≤ 500
- 1 ≤ N ≤ 1000
- 1 ≤ Ai ≤ N
Sample 1:
Input
4 5 2 2 2 2 2 4 1 2 3 4 4 3 3 2 1 6 1 1 2 2 3 4
Output
YES NO YES NO
Explanation:
Test case 1 : 2 is the dominant element.
Test case 2: There does not exist any dominant element.
Test case 3: 3 is the dominant element.
Dominant Element Codechef Solution in CPP
#include <bits/stdc++.h>
using namespace std;
int main() {
// your code goes here
int t;
cin>>t;
while(t--) {
int n;
cin>>n;
int arr[n];
for(int i=0; i<n; i++){
cin>>arr[i];
}
vector<int> v;
unordered_map<int, int> m;
for(int i=0; i<n; i++){
m[arr[i]]++;
}
for(auto it: m) v.push_back(it.second);
sort(v.begin(), v.end());
if(v[v.size()-1] > v[v.size()-2]) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}
return 0;
}Dominant Element Codechef Solution in Python
# cook your dish here
z=int(input())
for i in range(z):
n=int(input())
l=list(map(int,input().split()))
h={}
for i in range(len(l)):
if l[i] not in h:
h[l[i]]=1
else:
h[l[i]]+=1
l2=[]
for keys,values in h.items():
l2.append(values)
a=max(l2)
if l2.count(a)>1:
print("NO")
else:
print("YES")Dominant Element 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 void main (String[] args) throws java.lang.Exception{
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
while(T-->0){
int n = sc.nextInt();
int[] input = new int[n];
for(int i=0;i<n;i++){
input[i] = sc.nextInt();
}
Map<Integer,Integer> mp = new HashMap<>();
for(int x: input){
mp.put(x, mp.getOrDefault(x,0)+1);
}
ArrayList<int[]> l = new ArrayList<>();
for(int k : mp.keySet()){
int v = mp.get(k);
l.add(new int[]{k,v});
}
Collections.sort(l, (o1,o2)->o2[1]-o1[1]);
if(l.size()==1 || l.get(0)[1] > l.get(1)[1]){
System.out.println("YES");
}else{
System.out.println("NO");
}
}
}
}Disclaimer: The above Problem (Dominant Element ) is generated by CodeChef but the solution is provided by Chase2learn.This tutorial is only for Educational and Learning purpose.
