 # JavaScript: Find the number of elements which presents in both of the given arrays

## JavaScript Basic: Exercise-141 with Solution

Write a JavaScript program to find the number of elements which presents in both of the given arrays.

Pictorial Presentation: Sample Solution:

HTML Code:

``````<!DOCTYPE html>
<html>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>Find the number of elements which presents in both of the given arrays.</title>
<body>

</body>
</html>
```
```

JavaScript Code:

``````function test_same_elements_both_arrays(arra1, arra2) {
var result = 0;
for(var i = 0; i < arra1.length; i++) {
for(var j = 0; j < arra2.length; j++){
if(arra1[i] === arra2[j])
{
result++;
}
}
}
return result;
}
console.log(test_same_elements_both_arrays([1,2,3,4], [1,2,3,4]));
console.log(test_same_elements_both_arrays([1,2,3,4], [1,2,3,5]));
console.log(test_same_elements_both_arrays([1,2,3,4], [11,22,33,44]));
``````

Sample Output:

```4
3
0
```

Flowchart: ES6 Version:

``````function test_same_elements_both_arrays(arra1, arra2) {
let result = 0;
for(let i = 0; i < arra1.length; i++) {
for(let j = 0; j < arra2.length; j++){
if(arra1[i] === arra2[j])
{
result++;
}
}
}
return result;
}
console.log(test_same_elements_both_arrays([1,2,3,4], [1,2,3,4]));
console.log(test_same_elements_both_arrays([1,2,3,4], [1,2,3,5]));
console.log(test_same_elements_both_arrays([1,2,3,4], [11,22,33,44]));
``````

Live Demo:

See the Pen javascript-basic-exercise-141 by w3resource (@w3resource) on CodePen.

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## JavaScript: Tips of the Day

Classes/function constructors

```class Person {
constructor() {
this.name = 'Owen';
}
}

Person = class AnotherPerson {
constructor() {
this.name = 'Eddie';
}
};

const member = new Person();
console.log(member.name);
```

We can set classes equal to other classes/function constructors. In this case, we set Person equal to AnotherPerson. The name on this constructor is Eddie, so the name property on the new Person instance member is "Eddie".

Ref: https://bit.ly/3jFRBje