What if you could find the first unique letter in a huge text instantly, without checking each letter again and again?
Why First Non Repeating Character Using Hash in DSA Typescript?
Imagine you have a long list of letters from a story, and you want to find the first letter that appears only once. Doing this by checking each letter one by one and counting how many times it appears can take a lot of time, especially if the story is very long.
Manually counting each letter's occurrences means going back and forth through the list many times. This is slow and easy to make mistakes, like missing some letters or counting wrong. It feels like searching for a needle in a haystack by looking at every straw again and again.
Using a hash (a special kind of map) lets us count how many times each letter appears in just one pass. Then, we quickly find the first letter with a count of one. This saves time and avoids confusion, making the search fast and simple.
for (let i = 0; i < str.length; i++) { let count = 0; for (let j = 0; j < str.length; j++) { if (str[i] === str[j]) count++; } if (count === 1) return str[i]; }
const counts = new Map<string, number>(); for (const char of str) { counts.set(char, (counts.get(char) || 0) + 1); } for (const char of str) { if (counts.get(char) === 1) return char; }
This method lets you quickly find unique items in any list, making your programs faster and smarter.
Think about finding the first unique customer in a list of shoppers to send a special offer. Using this method, you can do it instantly, even if the list is huge.
Manual counting is slow and error-prone.
Hash maps let us count occurrences efficiently.
We can find the first unique character in just two quick steps.
Practice
Solution
Step 1: Understand the role of a hash in frequency counting
A hash (dictionary/map) stores each character as a key and counts how many times it appears in the string.Step 2: Use frequency counts to find the first non-repeating character
By checking the counts in order, we can find the first character with count 1, which is the first non-repeating character.Final Answer:
To count the frequency of each character efficiently -> Option AQuick Check:
Hash counts characters = To count the frequency of each character efficiently [OK]
- Thinking hash sorts characters
- Trying to reverse string instead
- Assuming hash removes duplicates
Solution
Step 1: Identify correct TypeScript syntax for a map
TypeScript uses Map to create a hash map. Here, keys are strings and values are numbers.Step 2: Check each option for correctness
const charCount = new Map(); uses Map(), which is correct. Options A, B, and D use incorrect types or structures.Final Answer:
const charCount = new Map<string, number>(); -> Option CQuick Check:
Map<string, number> is correct syntax [OK]
- Using Array instead of Map
- Using Set which stores only keys
- Incorrect Object initialization
function firstNonRepeatingChar(s: string): string {
const count = new Map();
for (const ch of s) {
count.set(ch, (count.get(ch) ?? 0) + 1);
}
for (const ch of s) {
if (count.get(ch) === 1) return ch;
}
return '_';
}
console.log(firstNonRepeatingChar('swiss'));Solution
Step 1: Count character frequencies in 'swiss'
Characters: s(3), w(1), i(1). The map stores these counts.Step 2: Find first character with count 1 in order
Order: 's' (count 3), 'w' (count 1), 'i' (count 1). The first with count 1 is 'w'.Final Answer:
'w' -> Option DQuick Check:
First unique char in 'swiss' = 'w' [OK]
- Choosing first character without checking count
- Confusing 'i' and 'w'
- Returning '_' when unique char exists
function firstNonRepeatingChar(s: string): string {
const count = new Map();
for (let i = 0; i <= s.length; i++) {
const ch = s[i];
count.set(ch, (count.get(ch) ?? 0) + 1);
}
for (const ch of s) {
if (count.get(ch) === 1) return ch;
}
return '_';
}Solution
Step 1: Check the for loop boundary
The loop runs from i = 0 to i <= s.length, which accesses s[s.length], an undefined index causing error.Step 2: Correct loop condition
Change condition to i < s.length to avoid out-of-bounds access and correctly count characters.Final Answer:
Loop condition should be i < s.length, not i <= s.length -> Option BQuick Check:
Loop boundary off by one error [OK]
- Off-by-one error in loop condition
- Misusing Map initialization
- Wrong loop type for string iteration
'aabbccddeeffg', which modification to the hash-based approach will correctly return the first non-repeating character when some characters appear multiple times and some only once?Solution
Step 1: Count frequencies of all characters
Use a hash map to count how many times each character appears in the string.Step 2: Iterate string in original order to find first count 1
Check each character in order; the first with count 1 is the first non-repeating character.Final Answer:
After counting, iterate the string and return the first character with count 1 -> Option AQuick Check:
Count then check order for first unique char [OK]
- Sorting string loses original order
- Returning highest count instead of unique
- Ignoring character order in result
Ready to prove it?
Pass the test and this pattern moves unknown
