Find duplicate element in a progression of first n terms JavaScript

When you have an array of first n natural numbers with one duplicate element, finding the duplicate efficiently is a common programming challenge. We'll explore two mathematical approaches that solve this in linear time.

Method 1: Using Array.prototype.reduce()

This approach uses a mathematical trick with the reduce function to find the duplicate in a single pass:

const arr = [2, 3, 1, 2];
const duplicate = a => a.reduce((acc, val, ind) => val + acc - (ind + 1)) + a.length - 1;
console.log(duplicate(arr));

2

How the Reduce Method Works

Let's trace through the algorithm step by step with array [2, 3, 1, 2]:

Since no initial value is provided to reduce(), it starts from index 1 with the first element as the accumulator:

First Pass:

acc = 2, val = 3, ind = 1
return value = 2 + 3 - (1 + 1) = 3

Second Pass:

acc = 3, val = 1, ind = 2
return value = 3 + 1 - (2 + 1) = 1

Third Pass:

acc = 1, val = 2, ind = 3
return value = 1 + 2 - (3 + 1) = -1

Finally: -1 + (4 - 1) = 2, which is our duplicate element.

Method 2: Using Array.prototype.forEach()

This method calculates the sum of all elements and subtracts the sum of first (n-1) natural numbers:

const arr = [1, 4, 8, 5, 6, 7, 9, 2, 3, 7];
const duplicate = a => {
    let sum = 0;
    const { length: n } = a;
    a.forEach(num => sum += num);
    return sum - ((n * (n - 1)) / 2);
}
console.log(duplicate(arr));

7

Mathematical Explanation

The forEach method works because:

• Sum of first n natural numbers = n × (n + 1) / 2
• Sum of first (n-1) natural numbers = (n-1) × n / 2
• Array sum - Sum of (n-1) numbers = duplicate element

Comparison

Conclusion

Both methods achieve linear time complexity using mathematical formulas. The forEach approach is more readable, while the reduce method is more compact but harder to understand.