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Lesson 1.3

Hunt edge cases

7 min · 7 steps

  1. Step 1 of 7 · Explanation

    An edge case is an input at the border of the rules. Check this list every time: empty, one item, duplicates or ties, negatives, max size.

    Example problem: "Return the second largest distinct number, or -1 if there isn't one." For [4, 9, 9, 2], the answer is 4, not 9. The second 9 is a duplicate, so it can't count as the second largest.

  2. Step 2 of 7 · Pick the approach

    Constraints: 1 ≤ nums.length ≤ 10⁵ and -10⁹ ≤ nums[i] ≤ 10⁹.
    Which input can never appear in a test?

    1. []
    2. [-3]
    3. [7, 7, 7]
    4. An array of 100

  3. Step 3 of 7 · Predict the result

    This code returns the second largest distinct number, or -1 if there isn't one. What does it return for [5, 5]?

    function solution(nums) {  let first = -Infinity  let second = -Infinity  for (let i = 0; i < nums.length; i++) {    const n = nums[i]    if (n > first) {      second = first      first = n    } else if (n > second && n < first) {      second = n    }  }  return second === -Infinity ? -1 : second}
    1. -1
    2. 5
    3. 0

  4. Step 4 of 7 · Find the bug

    This code should return the second largest distinct number, or -1 if there isn't one. It passes [4, 9, 2] but fails the hidden test [9, 9] (expected -1). Tap the wrong line.

    function solution(nums) {  let first = -Infinity  let second = -Infinity  for (let i = 0; i < nums.length; i++) {    const n = nums[i]    if (n > first) {      second = first      first = n    } else if (n > second) {      second = n    }  }  return second === -Infinity ? -1 : second}

  5. Step 5 of 7 · Predict the result

    This code returns the second largest distinct number, or -1 if there isn't one. What does it return for [-3, -7]?

    function solution(nums) {  let first = -Infinity  let second = -Infinity  for (let i = 0; i < nums.length; i++) {    const n = nums[i]    if (n > first) {      second = first      first = n    } else if (n > second && n < first) {      second = n    }  }  return second === -Infinity ? -1 : second}
    1. -7
    2. -1
    3. -3

  6. Step 6 of 7 · Write the code

    solution(nums)

    Return the smallest number in nums. If nums is empty, return null.
    Constraints: 0 ≤ nums.length ≤ 100, -1000 ≤ nums[i] ≤ 1000.

    function solution(nums) {  // [3, 1, 2] -> 1  // [] -> null }

  7. Step 7 of 7 · Recap

    • Edge case checklist: empty, one item, duplicates, negatives, max size.
    • The constraints tell you which ones can happen and which can't.
    • A starting value of -Infinity or nums[0] is safe for negatives. A starting value of 0 isn't.

    Pattern card

    Edge case checklist
    When to use it: After you understand the problem, before coding. Write one small test case for each item that can happen.
    // []        kosong / empty// [x]       satu item / one item// [x, x]    duplikat / duplicates// [-x]      negatif / negatives// n = max   ukuran maksimal / max size