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Update Interview Questions.md
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‎Coding Interview Prep/Interview Questions.md

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@@ -207,3 +207,56 @@ const throttleFunc = (func, interval) => {
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* Implement the Promise.all function
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* Flatten a nested JavaScript array without using Array.prototype.flat()
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* Implement a custom Sort() function
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* To find the **longest substring without repeating characters**, you can use a **sliding window** approach with two pointers (left and right) to track the current window, and a hash map to store the index of the characters.
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Here is the solution in JavaScript:
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```javascript
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function lengthOfLongestSubstring(s) {
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let seen = new Map(); // Hash map to store the last index of each character
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let maxLength = 0;
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let left = 0; // Left pointer for the sliding window
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for (let right = 0; right < s.length; right++) {
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const currentChar = s[right];
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// If the character is already in the map and its index is greater than or equal to left, update left
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if (seen.has(currentChar) && seen.get(currentChar) >= left) {
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left = seen.get(currentChar) + 1;
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}
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// Update or add the current character's index in the map
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seen.set(currentChar, right);
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// Calculate the maximum length of substring so far
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maxLength = Math.max(maxLength, right - left + 1);
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}
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return maxLength;
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}
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```
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### Example Usage:
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```javascript
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console.log(lengthOfLongestSubstring("abcabcbb")); // Output: 3 ("abc")
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console.log(lengthOfLongestSubstring("bbbbb")); // Output: 1 ("b")
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console.log(lengthOfLongestSubstring("pwwkew")); // Output: 3 ("wke")
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console.log(lengthOfLongestSubstring("")); // Output: 0
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```
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### Explanation:
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1. **Sliding Window**:
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- Use two pointers (`left` and `right`) to maintain the current window of characters.
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- As you move `right` through the string, check if the current character is already in the map and its last index is within the window (i.e., greater than or equal to `left`).
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2. **Updating Window**:
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- If a repeating character is found, move the `left` pointer just after the previous occurrence of that character.
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- Store or update the index of each character in the hash map (`seen`).
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3. **Calculate Length**:
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- For each new character, update the maximum length of the substring if the current window is longer.
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### Time Complexity:
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- **O(n)** where `n` is the length of the string. Each character is processed at most twice (once when the right pointer moves and once when the left pointer moves).

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