Add week 6 solutions : longestSubstringWithoutRepeatingCharacters, longestRepeatingCharacterReplacement

Author: yolophg

longest-repeating-character-replacement/yolophg.js

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+// Time Complexity: O(n)
+// Space Complexity: O(1)
+
+var characterReplacement = function(s, k) {
+    let maxLen = 0; 
+    // initialize an array to store the count of each character.
+    let charCounts = new Array(26).fill(0); 
+    let start = 0;
+
+    // iterate the string with the end pointer.
+    for (let end = 0; end < s.length; end++) {
+        // calculate the index of the character.
+        let charIndex = s.charCodeAt(end) - 65; 
+        // update maxCount with the maximum count.
+        maxLen = Math.max(maxLen, ++charCounts[charIndex]); 
+
+        // move the start pointer and decrement the count at that position.
+        if (end - start + 1 - maxLen > k) {
+            charCounts[s.charCodeAt(start) - 65]--;
+            start++;
+        }
+    }
+
+    return s.length - start;
+};

longest-substring-without-repeating-characters/yolophg.js

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+// Time Complexity: O(n)
+// Space Complexity: O(n)
+
+var lengthOfLongestSubstring = function(s) {
+    let charSet = new Set();
+    
+    let left = 0;
+    let right = 0;
+    
+    let maxLength = 0;
+    
+    // iterate the string.
+    while (right < s.length) {
+        if (!charSet.has(s[right])) {
+            // if the character isn't in the set, add it.
+            charSet.add(s[right]);
+            // move the right pointer to the right.
+            right++;
+            // update the maximum length if the current window is larger.
+            maxLength = Math.max(maxLength, right - left);
+        } else {
+            // if the character is in the set, remove the leftmost character.
+            charSet.delete(s[left]);
+            // move the left pointer to the right.
+            left++;
+        }
+    }
+    
+    return maxLength;
+};