Career Programming

How to solve coding interview questions: Looking deeper into finding the first NON-recurring character in a string

In the previous installment in the How to solve coding interview questions series, I showed you my Python solution for the challenge “Find the first NON-recurring character in a string,” which is a follow-up to the challenge “Find the first recurring character in a string.”

A couple of readers posted their solution in the comments, and I decided to take a closer look at them. I also decided to write my own JavaScript solution, which led me to refine my Python solution.

“CK”’s JavaScript solution

The first reader-submitted solution comes from “CK,” who submitted a JavaScript solution that uses sets:

// JavaScript

function findFirstNonRepeatingChar(chars) {
    let uniqs = new Set();
    let repeats = new Set();
    for (let ch of chars) {
        if (! uniqs.has(ch)) {
        } else {
    for (let ch of uniqs) {
        if (! repeats.has(ch)) {
            return ch;
    return null;

The one thing that all programming languages that implement a set data structure is that every element in a set must be unique — you can’t put more than one of a specific element inside a set. If you try to add an element to a set that already contains that element, the set will simply ignore the addition.

In mathematics, the order of elements inside a set doesn’t matter. When it comes to programming languages, it varies — JavaScript preserves insertion order (that is, the order of items in a set is the same as the order in which they were added to the set) while Python doesn’t.

CK’s solution uses two sets: uniqs and repeats. It steps through the input string character by character and does the following:

  • If uniqs does not contain the current character, it means that we haven’t seen it before. Add the current character to uniqs, the set of characters that might be unique.
  • If uniqs contains the current character, it means that we’ve seen it before. Add the current character to repeats, the set of repeated characters.

Once the function has completed the above, it steps through uniqs character by character, looking for the first character in uniqs that isn’t in repeats.

For a string of length n, the function performs the first for loop n times, and the second for loop some fraction of n times. So its computational complexity is O(n).

Thanks for the submission, CK!

Dan Budiac’s TypeScript solution

Dan Budiac provided the second submission, which he implemented in TypeScript:

// TypeScript

function firstNonRecurringCharacter(val: string): string | null {

    // First, we split up the string into an
    // array of individual characters.
    const all = val.split('');

    // Next, we de-dup the array so we’re not
    // doing redundant work.
    const unique = [ Set(all)];

    // Lastly, we return the first character in
    // the array which occurs only once.
    return unique.find((a) => {
        const occurrences = all.filter((b) => a === b);
        return occurrences.length === 1;
    }) ?? null;


I really need to take up TypeScript, by the way.

Anyhow, Dan’s approach is similar to CK’s, except that it uses one set instead of two:

  1. It creates an array, all, by splitting the input string into an array made up of its individual characters.
  2. It then creates a set, unique, using the contents of all. Remember that sets ignore the addition of elements that they already contain, meaning that unique will contain only individual instances of the characters from all.
  3. It then goes character by character through unique, checking the number of times each character appears in all.

For a string of length n:

  • Splitting the input string into the array all is O(n).
  • Creating the set unique from all is O(n).
  • The last part of the function features a find() method — O(n) — containing a filter() method — also O(n). That makes this operation O(n2). So the whole function is O(n2).

Thanks for writing in, Dan!

My JavaScript solution that doesn’t use sets

In the meantime, I’d been working on a JavaScript solution. I decided to play it safe and use JavaScript’s Map object, which holds key-value pairs and remembers the order in which they were inserted — in short, it’s like Python’s dictionaries, but with get() and set() syntax.

// JavaScript

function firstNonRecurringCharacter(text) {
    // Maps preserve insertion order.
    let characterRecord = new Map()

    // This is pretty much the same as the first
    // for loop in the Python version.
    for (const character of text) {
        if (characterRecord.has(character)) {
            newCount = characterRecord.get(character) + 1
            characterRecord.set(character, newCount)
        } else {
            characterRecord.set(character, 1)

    // Just go through characterRecord item by item
    // and return the first key-value pair
    // for which the value of count is 1.
    for (const [character, count] of characterRecord.entries()) {
        if (count == 1) {
            return character

    return null

The final part of this function takes an approach that’s slightly different from the one in my original Python solution. It simply goes through characterRecord one key-value pair at a time and returns the key for the first pair it encounters whose value is 1.

It remains an O(n) solution.

My revised Python solution

If I revise my original Python solution to use the algorithm from my JavaScript solution, it becomes this:

# Python

def first_non_recurring_character(text):
    character_record = {}
    # Go through the text one character at a time
    # keeping track of the number of times
    # each character appears.
    # In Python 3.7 and later, the order of items
    # in a dictionary is the order in which they were added.
    for character in text:
        if character in character_record:
            character_record[character] += 1
            character_record[character] = 1
    # The first non-recurring character, if there is one,
    # is the first item in the list of non-recurring characters.
    for character in character_record:
        if character_record[character] == 1:
            return character

    return None

This is also an O(n) solution.

Previously in this series

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