## Most asked interview question – Check if the array is a dominant array or not Problem Statement

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 T — the number of test cases.
Then the test cases follow.
The first line of each test case contains an integer N — the size of the array A .
The second line of each test case contains N space-separated integers denoting the array A.

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 Sample 1:
1 ≤ T ≤ 500
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 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
N

## Source Code – Below source code is using JavaScript Language

```function isDominantElementPresent(arr) {

var count = {};
arr.forEach(ele=>{
if(count[ele]){
count[ele] +=1;

}else{
count[ele] = 1;
}
});

var maxFrequency = 0;
for (var ele in count) {
maxFrequency = Math.max(maxFrequency, count[ele]);
}
var isDominant = Object.values(count).filter(freq => freq === maxFrequency).length === 1;

return isDominant;
}

function main() {
const t = parseInt(prompt("Enter the number of test cases:"));

for (let i = 0; i < t; i++) {
const n = parseInt(prompt("Enter the size of the array:"));
const arr = prompt("Enter the array elements separated by space:")
.split(" ");

if (isDominantElementPresent(arr)) {
console.log("YES");
} else {
console.log("NO");
}
}
}

main();```
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