List Format

Languages: JS / TS

Difficulty: Medium

Time: ~20 mins

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The Task

Write a function called listFormat that takes a list of names (strings) and joins them into a single, readable sentence.

Basic joining rules:

• No items → return an empty string ''
• One item → return it as-is
• Two items → join with " and "
• Three or more → separate with commas, and use " and " before the last one

listFormat([])                          // ''
listFormat(['Bob'])                     // 'Bob'
listFormat(['Bob', 'Alice'])            // 'Bob and Alice'
listFormat(['Bob', 'Ben', 'Tim', 'Jane', 'John'])
                                        // 'Bob, Ben, Tim, Jane and John'

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Options (second parameter)

The function accepts an optional settings object as the last argument:

sorted — Sort the names alphabetically before formatting.

length — Only show the first N names. Summarize the rest as "and X other(s)". Ignore this option if the value is 0, negative, or otherwise invalid.

unique — Remove duplicate names before formatting.

// length: only show first 3, summarize the rest
listFormat(['Bob', 'Ben', 'Tim', 'Jane', 'John'], { length: 3 })
// 'Bob, Ben, Tim and 2 others'

// length: when only 1 is hidden, use "other" (not "others")
listFormat(['Bob', 'Ben', 'Tim', 'Jane', 'John'], { length: 4 })
// 'Bob, Ben, Tim, Jane and 1 other'

// sorted + length: sort first, then trim
listFormat(['Bob', 'Ben', 'Tim', 'Jane', 'John'], { length: 3, sorted: true })
// 'Ben, Bob, Jane and 2 others'

// unique + length: remove duplicates first, then trim
listFormat(['Bob', 'Ben', 'Tim', 'Jane', 'John', 'Bob'], { length: 3, unique: true })
// 'Bob, Ben, Tim and 2 others'

// Empty strings should be ignored
listFormat(['Bob', 'Ben', '', '', 'John'])
// 'Bob, Ben and John'

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Things to Watch Out For

• Empty strings in the list should be silently dropped.
• length values that are 0, negative, or non-numeric should be ignored (treat the list as if no length was given).
• Order of operations when multiple options are combined: apply unique → sorted → length, in that order.
• “other” vs “others”: use "1 other" for exactly one hidden item, "X others" for two or more.

How to Solve It

The solution has two main parts:

1. Clean up and process the input list — strip empty strings, sort, deduplicate.
2. Format the result — join everything into a readable sentence.

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Part 1: Processing the List

Apply the options in this order:

• Filter empty/falsy strings out first (always)
• Sort alphabetically if sorted: true
• Deduplicate if unique: true

These are all straightforward array operations.

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Part 2: Formatting the Sentence

The key insight is that length splits the array into two parts:

• First part — the names to show, joined by ", "
• Second part — either the last name, or "X other(s)" if length is active

['Bob', 'Ben', 'Tim', 'Jane', 'John']
 └── first part: 'Bob, Ben, Tim' ──┘  └── second part: '2 others'

Use " and " to connect the two parts.

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The Code

const SEPARATOR = ', ';
const OTHERS_SEPARATOR = ' and ';
const OTHERS_LABEL = 'other';

export default function listFormat(itemsParam, options = {}) {
  // Step 1: Remove empty/falsy strings
  let items = itemsParam.filter((item) => !!item);

  // Early exits for trivial cases
  if (items.length === 0) return '';
  if (items.length === 1) return items[0];

  // Step 2: Sort alphabetically if requested
  if (options.sorted) {
    items.sort();
  }

  // Step 3: Remove duplicates if requested
  if (options.unique) {
    items = Array.from(new Set(items));
  }

  // Step 4a: If `length` is valid and cuts the list short...
  if (options.length > 0 && options.length < items.length) {
    const shown    = items.slice(0, options.length).join(SEPARATOR);
    const count    = items.length - options.length;
    const label    = count + ' ' + OTHERS_LABEL + (count > 1 ? 's' : '');
    return shown + OTHERS_SEPARATOR + label;
  }

  // Step 4b: No `length` limit — show everything
  const allButLast = items.slice(0, -1).join(SEPARATOR);
  const last       = items[items.length - 1];
  return allButLast + OTHERS_SEPARATOR + last;
}

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Key Details to Get Right

Edge case	How it’s handled
Empty strings in input	Filtered out with .filter(Boolean)
length: 0 or negative	Treated as if length wasn’t given
Exactly 1 item hidden	"1 other" (no “s”)
2+ items hidden	"2 others", "3 others", etc.
Single item in list	Returned directly, no joining needed

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Worth Knowing

This solution hardcodes English separators ("," and "and"). The real Intl.ListFormat API handles this properly — it adapts separators based on the user’s locale, so "and" might become "y" in Spanish, for example. For any real production use, you’d want to either use Intl.ListFormat directly, or accept the separators as configurable options rather than hardcoding them.