added js solution to _3

.gitignore

@@ -7,4 +7,5 @@ out/
 *.vscode/
 src/main/java/com/fishercoder/solutions/_99999RandomQuestions.java
 src/main/java/com/fishercoder/solutions/_Contest.java
-.project
+.project
+bin

javascript/_17.js

@@ -0,0 +1,43 @@
+function letterCombinations(digits) {
+  // If the input is an empty string, return an empty array.
+  if (digits.length === 0) {
+      return [];
+  }
+
+  // Mapping of digits to letters as per the telephone keypad using a javascript dictionary.
+  const digitToChar = {
+      '2': ['a', 'b', 'c'],
+      '3': ['d', 'e', 'f'],
+      '4': ['g', 'h', 'i'],
+      '5': ['j', 'k', 'l'],
+      '6': ['m', 'n', 'o'],
+      '7': ['p', 'q', 'r', 's'],
+      '8': ['t', 'u', 'v'],
+      '9': ['w', 'x', 'y', 'z']
+  };
+
+  // Resultant array to store all possible combinations
+  const result = [];
+
+  // Backtracking function to generate combinations
+  function backtrack(index, currentCombination) {
+      // if the current combination has the same length as the input digits.
+      if (index === digits.length) {
+          result.push(currentCombination);
+          return;
+      }
+
+      // Get the letters that the current digit maps to.
+      let letters = digitToChar[digits[index]];
+
+      // Loop through the letters and call backtrack recursively for the next digit.
+      for (let letter of letters) {
+          backtrack(index + 1, currentCombination + letter);
+      }
+  }
+
+  // Start backtracking from the first digit (index 0) with an empty string as the initial combination.
+  backtrack(0, '');
+
+  return result;
+};

javascript/_3.js

@@ -0,0 +1,23 @@
+function lengthOfLongestSubstring(s) {
+  // Using the "sliding window" data structure.
+  // Create a javascript set to store unique characters.
+  let charSet = new Set();
+  let left = 0; // Left pointer of the sliding window.
+  let maxLength = 0;
+
+  // This moves the right pointer of the sliding window.
+  for (let right = 0; right < s.length; right++) {
+      // If the character at the right pointer is already in the set, move the left pointer.
+      while (charSet.has(s[right])) {
+          charSet.delete(s[left]);
+          left++;
+      }
+      // Add the current character at the right pointer to the set.
+      charSet.add(s[right]);
+
+      // Update the maximum length of substring without repeating characters.
+      maxLength = Math.max(maxLength, right - left + 1);
+  }
+
+  return maxLength;
+}