Even index sum in JavaScript

We need to write a JavaScript function that calculates the sum of all integers at even indices (0, 2, 4, etc.) in an array, then multiplies this sum by the last element of the array.

Problem Statement

Given an array of integers, find the sum of elements at even indices and multiply by the last element.

Input: [4, 1, 6, 8, 3, 9]
Even indices: 0, 2, 4 ? elements: 4, 6, 3
Sum: 4 + 6 + 3 = 13
Last element: 9
Result: 13 × 9 = 117

Solution

const arr = [4, 1, 6, 8, 3, 9];

const evenLast = (arr = []) => {
    if (arr.length === 0) {
        return 0;
    } else {
        const evenIndexElements = arr.filter((_, index) => index % 2 === 0);
        const sum = evenIndexElements.reduce((a, b) => a + b);
        const lastElement = arr[arr.length - 1];
        const result = sum * lastElement;
        return result;
    }
};

console.log(evenLast(arr));

117

Step-by-Step Breakdown

const arr = [4, 1, 6, 8, 3, 9];

// Step 1: Filter elements at even indices
const evenIndexElements = arr.filter((_, index) => index % 2 === 0);
console.log("Even index elements:", evenIndexElements);

// Step 2: Sum the filtered elements
const sum = evenIndexElements.reduce((a, b) => a + b);
console.log("Sum:", sum);

// Step 3: Get last element
const lastElement = arr[arr.length - 1];
console.log("Last element:", lastElement);

// Step 4: Multiply sum by last element
const result = sum * lastElement;
console.log("Final result:", result);

Even index elements: [4, 6, 3]
Sum: 13
Last element: 9
Final result: 117

Alternative Approach

We can solve this more efficiently using a single loop:

const evenLastOptimized = (arr = []) => {
    if (arr.length === 0) return 0;
    
    let sum = 0;
    for (let i = 0; i < arr.length; i += 2) {
        sum += arr[i];
    }
    
    return sum * arr[arr.length - 1];
};

const arr = [4, 1, 6, 8, 3, 9];
console.log(evenLastOptimized(arr));

117

Edge Cases

// Empty array
console.log(evenLast([]));

// Single element
console.log(evenLast([5]));

// Two elements
console.log(evenLast([2, 7]));

0
25
14

Conclusion

This solution filters elements at even indices, sums them, and multiplies by the last element. The optimized approach using a single loop is more efficient for large arrays.