IndexOf

Problem Link

Problem

Implement the type version of Array.indexOf, IndexOf<T, U> takes an Array T, any U and returns the index of the first U in Array T.

type Res = IndexOf<[1, 2, 3], 2>          // 1
type Res1 = IndexOf<[2, 6, 3, 8, 4, 1, 7, 3, 9], 3>  // 2
type Res2 = IndexOf<[0, 0, 0], 2>         // -1

Solution

Approach: Linear Scan with Index Counter

Walk through the tuple, comparing each element to U using IsEqual.

type IsEqual<A, B> =
  (<T>() => T extends A ? 1 : 2) extends
  (<T>() => T extends B ? 1 : 2)
    ? true : false

type IndexOf<T extends unknown[], U, Count extends unknown[] = []> =
  T extends [infer Head, ...infer Tail]
    ? IsEqual<Head, U> extends true
      ? Count['length']
      : IndexOf<Tail, U, [...Count, unknown]>
    : -1

How it works:

1. Extract Head from T and compare to U using the strict IsEqual helper.
2. If equal, return Count['length'] as the current index.
3. Otherwise, recurse on Tail with Count incremented.
4. If T is empty, return -1 (not found).

Why use IsEqual instead of extends? Simple extends doesn’t distinguish any, never, or literal vs. union types correctly. The IsEqual trick using conditional type identity is fully strict.

Key Takeaways

• IsEqual via the <T>() => T extends A ? 1 : 2 trick is the standard way to do strict equality in TypeScript types.
• Index tracking via tuple accumulator Count is reusable across many array-scanning problems.