Returning an array containing last n even numbers from input array in JavaScript

Problem

We are required to write a JavaScript function that takes in an array of numbers as the first argument and a number as the second argument.

Our function should pick and return an array of last n even numbers present in the input array.

Example

Following is the code ?

const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const num = 3;

const pickEvens = (arr = [], num = 1) => {
    const res = [];
    for(let index = arr.length - 1; index >= 0; index -= 1){
        if (res.length === num){
            break;
        };
        const number = arr[index];
        if (number % 2 === 0){
            res.unshift(number);
        };
    };
    return res;
};

console.log(pickEvens(arr, num));

[4, 6, 8]

How It Works

The function uses a reverse loop to traverse the array from the last element. When it finds an even number (number % 2 === 0), it adds it to the beginning of the result array using unshift(). This maintains the original order of the last n even numbers.

Alternative Approach Using filter() and slice()

Here's a more concise approach using built-in array methods:

const arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12];
const num = 4;

const pickEvensAlternative = (arr = [], num = 1) => {
    return arr.filter(number => number % 2 === 0).slice(-num);
};

console.log(pickEvensAlternative(arr, num));

[6, 8, 10, 12]

Comparison

Edge Cases

// When array has fewer even numbers than requested
console.log(pickEvens([1, 3, 5, 7, 8], 3)); // [8]

// When num is 0
console.log(pickEvens([2, 4, 6], 0)); // []

// Empty array
console.log(pickEvens([], 2)); // []

[8]
[]
[]

Conclusion

The reverse loop approach is more efficient for large arrays as it stops early when the required count is reached. Use the filter-slice method for better readability in smaller datasets.