Merge pull request #54 from aikhelis/typescript-solutions-two-pointers

Adding TypeScript with Two Pointers chapter

Author: Destiny-02

typescript/Two Pointers/is_palindrome_valid.ts

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+function is_palindrome_valid(s: string): boolean {
+    let left = 0, right = s.length - 1
+    while (left < right) {
+        // Skip non-alphanumeric characters from the left. 
+        while (left < right && !isalnum(s[left]))
+            left++
+        // Skip non-alphanumeric characters from the right.
+        while (left < right && !isalnum(s[right]))
+            right--
+        // If the characters at the left and right pointers don't
+        // match, the string is not a palindrome.
+        if (s[left] !== s[right])
+            return false
+        left++
+        right--
+    }
+    return true
+}
+
+const isalnum = (c: string): boolean => /^[a-z0-9]+$/i.test(c);

typescript/Two Pointers/largest_container.ts

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+function largest_container(heights: number[]): number {
+    let max_water = 0
+    let left = 0, right = heights.length - 1
+    while (left < right) {
+        // Calculate the water contained between the current pair of 
+        // lines.
+        let water = Math.min(heights[left], heights[right]) * (right - left)
+        max_water = Math.max(max_water, water)
+        // Move the pointers inward, always moving the pointer at the 
+        // shorter line. If both lines have the same height, move both 
+        // pointers inward.
+        if (heights[left] < heights[right]) 
+            left++
+        else if (heights[left] > heights[right])
+            right--     
+        else 
+            left++, right--
+    }
+    return max_water
+}

typescript/Two Pointers/largest_container_brute_force.ts

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+function largest_container_brute_force(heights: number[]): number {
+    let n = heights.length
+    let max_water = 0
+    // Find the maximum amount of water stored between all pairs of
+    // lines.
+    for (let i = 0; i < n - 1; i++) {
+        for (let j = i + 1; j < n; j++) {
+            let water = Math.min(heights[i], heights[j]) * (j - i)
+            max_water = Math.max(max_water, water)
+        }
+    }
+    return max_water
+}

typescript/Two Pointers/next_lexicographical_sequence.ts

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+function next_lexicographical_sequence(s: string): string {
+    let letters = [...s];
+    // Locate the pivot, which is the first character from the right that breaks 
+    // non-increasing order. Start searching from the second-to-last position.
+    let pivot = letters.length - 2;
+    while (pivot >= 0 && letters[pivot] >= letters[pivot + 1])
+        pivot--;
+    // If pivot is not found, the string is already in its largest permutation. In
+    // this case, reverse the string to obtain the smallest permutation.
+    if (pivot === -1) 
+        return letters.reverse().join('');
+    // Find the rightmost successor to the pivot.
+    let rightmost_successor = letters.length - 1;
+    while (letters[rightmost_successor] <= letters[pivot])
+        rightmost_successor--;
+    // Swap the rightmost successor with the pivot to increase the lexicographical
+    // order of the suffix.
+    [letters[pivot], letters[rightmost_successor]] = [letters[rightmost_successor], letters[pivot]];
+    // Reverse the suffix after the pivot to minimize its permutation.
+    const suffix = letters.slice(pivot + 1).reverse();
+    letters.splice(pivot + 1, suffix.length, ...suffix);
+    return letters.join('');
+} 

typescript/Two Pointers/pair_sum_sorted.ts

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+function pair_sum_sorted(nums: number[], target: number): number[] {
+    let left = 0, right = nums.length - 1
+    while (left < right) {
+        let sum = nums[left] + nums[right]
+        // If the sum is smaller, increment the left pointer, aiming
+        // to increase the sum toward the target value.
+        if (sum < target)
+            left++
+        // If the sum is larger, decrement the right pointer, aiming
+        // to decrease the sum toward the target value.
+        else if (sum > target)
+            right--
+        // If the target pair is found, return its indexes.
+        else
+            return [left, right]
+    }
+    return []
+}

typescript/Two Pointers/pair_sum_sorted_brute_force.ts

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+function pair_sum_sorted_brute_force(nums: number[], target: number): number[] {
+    const n = nums.length;
+    for (let i = 0; i < n; i++) {
+        for (let j = i + 1; j < n; j++) {
+            if (nums[i] + nums[j] === target) 
+                return [i, j]
+        }
+    }
+    return []
+}

typescript/Two Pointers/shift_zeros_to_the_end.ts

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+function shift_zeros_to_the_end(nums: number[]): void {
+    // The 'left' pointer is used to position non-zero elements.
+    let left = 0
+    // Iterate through the array using a 'right' pointer to locate non-zero 
+    // elements.
+    for (let right = 0; right < nums.length; right++) {
+        if (nums[right] !== 0) {
+            [nums[left], nums[right]] = [nums[right], nums[left]]
+            // Increment 'left' since it now points to a position already occupied 
+            // by a non-zero element.
+            left++
+        }
+    }
+}

typescript/Two Pointers/shift_zeros_to_the_end_naive.ts

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+function shift_zeros_to_the_end_naive(nums: number[]): void {
+    let temp = Array(nums.length).fill(0);
+    let i = 0
+    // Add all non-zero elements to the left of 'temp'.
+    for (let number of nums) {
+        if (number !== 0) {
+            temp[i] = number
+            i++
+        }
+    }
+    // Set 'nums' to 'temp'.
+    for (let j = 0; j < nums.length; j++) {
+        nums[j] = temp[j]
+    }
+}

typescript/Two Pointers/triplet_sum.ts

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+function triplet_sum(nums: number[]): number[][] {
+    const triplets = []
+    nums.sort()
+    for (let i = 0; i < nums.length; i++) {
+        // Optimization: triplets consisting of only positive numbers 
+        // will never sum to 0.
+        if (nums[i] > 0 ) 
+            break
+        // To avoid duplicate triplets, skip 'a' if it's the same as 
+        // the previous number.
+        if (i > 0 && nums[i] === nums[i - 1]) 
+            continue
+        // Find all pairs that sum to a target of '-a' (-nums[i]).
+        const pairs = pair_sum_sorted_all_pairs(nums, i + 1, -nums[i])
+        for (const pair of pairs) {
+            triplets.push([nums[i]].concat(pair))
+        }
+    }
+    return triplets 
+}
+
+function pair_sum_sorted_all_pairs(nums: number[], start: number, target: number): number[]{
+    const pairs = []
+    let left = start, right = nums.length - 1
+    while (left < right) {
+        const sum = nums[left] + nums[right]
+        if (sum === target) {
+            pairs.push([nums[left], nums[right]])
+            left++
+            // To avoid duplicate '[b, c]' pairs, skip 'b' if it's the 
+            // same as the previous number.
+            while (left < right && nums[left] === nums[left - 1]) 
+                left++
+        } else if (sum < target) {
+            left++
+        } else {
+            right--
+        }
+    }
+    return pairs
+}

typescript/Two Pointers/triplet_sum_brute_force.ts

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+function triplet_sum_brute_force(nums: number[]): number[][] {
+    let n = nums.length
+    // Use a hash set to ensure we don't add duplicate triplets.
+    let triplets = new Set()
+    // Iterate through the indexes of all triplets.
+    for (let i = 0; i < n - 2; i++) {
+        for (let j = i + 1; j < n - 1; j++) {
+            for (let k = j + 1; k < n; k++) {
+                if (nums[i] + nums[j] + nums[k] === 0) {
+                    // Sort the triplet before including it in the hash set.
+                    // [javascript] convert triplet array into a JSON string to use as a unqiue Set value.
+                    let triplet = JSON.stringify([nums[i], nums[j], nums[k]].sort())
+                    triplets.add(triplet)
+                }
+            }
+        }
+    }
+    // [javascript] convert the Set back into an array of triplets.
+    return Array.from(triplets).map(JSON.parse)
+}