ts-5: add sliding windows solutions

Author: aikhelis

typescript/Sliding Windows/longest_substring_with_unique_chars.ts

@@ -0,0 +1,20 @@
+function longestSubstringWithUniqueChars(s: string): number {
+    let maxLen = 0;
+    let hashSet: Set<string> = new Set();
+    let left = 0, right = 0;
+    while(right < s.length) {
+        // If we encounter a duplicate character in the window, shrink 
+        // the window until it's no longer a duplicate.
+        while(hashSet.has(s[right])){
+            hashSet.delete(s[left]);
+            left++;
+        }
+        // Once there are no more duplicates in the window, update
+        // 'maxLen' if the current window is larger.
+        maxLen = Math.max(maxLen, right - left + 1);
+        hashSet.add(s[right]);
+        // Expand the window.
+        right++;
+    }
+    return maxLen;
+}

typescript/Sliding Windows/longest_substring_with_unique_chars_optimized.ts

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+function longestSubstringWithUniqueCharsOptimized(s: string): number {
+    let maxLen = 0;
+    let prevIndexes: { [key: string]: number } = {};
+    let left = 0, right = 0;
+    while(right < s.length) {
+        // If a previous index of the current character is present
+        // in the current window, it's a duplicate character in the
+        // window. 
+        if (prevIndexes[s[right]] !== undefined 
+            && prevIndexes[s[right]] >= left) {
+            // Shrink the window to exclude the previous occurrence
+            // of this character.
+            left = prevIndexes[s[right]] + 1;
+        }
+        // Update 'maxLen' if the current window is larger.
+        maxLen = Math.max(maxLen, right - left + 1);
+        prevIndexes[s[right]] = right;
+        // Expand the window.
+        right++;
+    }
+    return maxLen;
+}

typescript/Sliding Windows/longest_uniform_substring_after_replacements.ts

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+function longestUniformSubstringAfterReplacements(s: string, k: number): number {
+    const freqs: { [key: string]: number } = {};
+    let highestFreq = 0, maxLen = 0;
+    let left = 0, right = 0;
+    while (right < s.length) {
+        // Update the frequency of the character at the right pointer 
+        // and the highest frequency for the current window.
+        freqs[s[right]] = (freqs[s[right]] || 0) + 1;
+        highestFreq = Math.max(highestFreq, freqs[s[right]]);
+        // Calculate replacements needed for the current window.
+        const numCharsToReplace = (right - left + 1) - highestFreq;
+        // Slide the window if the number of replacements needed exceeds 
+        // 'k'. The right pointer always gets advanced, so we just need 
+        // to advance 'left'.
+        if (numCharsToReplace > k) {
+            // Remove the character at the left pointer from the hash map 
+            // before advancing the left pointer.
+            freqs[s[left]]--;
+            left++;
+        }
+        // Since the length of the current window increases or stays the 
+        // same, assign  the length of the current window to 'maxLen'.
+        maxLen = right - left + 1;
+        // Expand the window.
+        right++;
+    }
+    return maxLen;
+}

typescript/Sliding Windows/substring_anagrams.ts

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+function substringAnagrams(s: string, t: string): number {
+    const lenS = s.length, lenT = t.length;
+    if (lenT > lenS) 
+        return 0;
+    let count = 0;
+    const expectedFreqs = new Array(26).fill(0);
+    const windowFreqs = new Array(26).fill(0);
+    // Populate 'expected_freqs' with the characters in string 't'.
+    for (const c of t)
+        expectedFreqs[c.charCodeAt(0) - 'a'.charCodeAt(0)]++;
+    let left = 0, right = 0;
+    while (right < lenS) {
+        // Add the character at the right pointer to 'window_freqs' 
+        // before sliding the window.
+        windowFreqs[s[right].charCodeAt(0) - 'a'.charCodeAt(0)]++;
+        // If the window has reached the expected fixed length, we 
+        // advance the left pointer as well as the right pointer to 
+        // slide the window.
+        if (right - left + 1 === lenT) {
+            if (windowFreqs.every((val, i) => val === expectedFreqs[i]))
+                count++;
+            // Remove the character at the left pointer from 
+            // 'window_freqs' before advancing the left pointer.
+            windowFreqs[s[left].charCodeAt(0) - 'a'.charCodeAt(0)]--;
+            left++;
+        }
+        right++;
+    }
+    return count;
+}