# JavaScript: Rotate the elements left of a given array of integers of length 3

## JavaScript Basic: Exercise-70 with Solution

Write a JavaScript program to rotate the elements left of a given array of integers of length 3.

Pictorial Presentation:

Sample Solution:

HTML Code:

``````<!DOCTYPE html>
<html>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width">
<title>JavaScript program to rotate the elements left of a given array of integers of length 3.</title>
<body>

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

JavaScript Code:

``````function rotate_elements_left(array)
{
return [array[1], array[2], array[0]];
}
console.log(rotate_elements_left([3, 4, 5]));
console.log(rotate_elements_left([0, -1, 2]));
console.log(rotate_elements_left([7, 6, 5]));
```
```

Sample Output:

```[4,5,3]
[-1,2,0]
[6,5,7]
```

Flowchart:

ES6 Version:

``````function rotate_elements_left(array)
{
return [array[1], array[2], array[0]];
}
console.log(rotate_elements_left([3, 4, 5]));
console.log(rotate_elements_left([0, -1, 2]));
console.log(rotate_elements_left([7, 6, 5]));
``````

Live Demo:

See the Pen JavaScript - Rotate the elements left of a given array of integers of length 3 - basic-ex-70 by w3resource (@w3resource) on CodePen.

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## JavaScript: Tips of the Day

Returns the sum of the powers of all the numbers from start to end (both inclusive)

Example:

```const sumPower = (end, power = 2, start = 1) =>
Array(end + 1 - start)
.fill(0)
.map((x, i) => (i + start) ** power)
.reduce((a, b) => a + b, 0);
console.log(sumPower(10)); // 385
console.log(sumPower(10, 3)); // 3025
console.log(sumPower(10, 3, 5)); // 2925
```

Output:

```385
3025
2925
```