Merge 01c924567a into ca3d16dcce

README.md

@@ -129,6 +129,7 @@ a set of rules that precisely define a sequence of operations.
   * `B` [Jump Search](src/algorithms/search/jump-search) (or Block Search) - search in sorted array
   * `B` [Binary Search](src/algorithms/search/binary-search) - search in sorted array
   * `B` [Interpolation Search](src/algorithms/search/interpolation-search) - search in uniformly distributed sorted array
+  * `B` [Twin Pointers](src/algorithms/search/twin-pointers)
 * **Sorting**
   * `B` [Bubble Sort](src/algorithms/sorting/bubble-sort)
   * `B` [Selection Sort](src/algorithms/sorting/selection-sort)

package.json

@@ -43,6 +43,7 @@
     "eslint-plugin-import": "2.27.5",
     "eslint-plugin-jest": "27.2.1",
     "eslint-plugin-jsx-a11y": "6.7.1",
+    "eslint-plugin-react": "^7.33.2",
     "husky": "8.0.3",
     "jest": "29.4.1",
     "pngjs": "^7.0.0"

src/algorithms/search/twin-pointers/README.md

@@ -0,0 +1,25 @@
+# Twin Pointers
+
+The twin pointers method, also known as the two pointers method, is a searching algorithm 
+that can be used on both sorted and unsorted numerical arrays/lists, depending on the intent of the function. 
+At its simplest form the twin pointer method employes two "pointers" that either move at different 
+speeds/from different starting positions in order to draw comparisons between values in order to 
+find some specified target. In the case that the array/list being searched through is sorted, 
+a common usage of the twin pointers is to have one at the starting and one at the ending position; 
+in this manner, moving the left pointer to the right can be assumed to increase its value while moving 
+the right pointer to the left can be assumed to do vice versa. In the case of an unsorted arrays/list, 
+the usage methods are generally much more varied based on what the characteristics of the intended 
+target of the search are.
+
+Note that any array can be sorted to easily use the twin pointer method by using the Array.sort method.
+However, the Array.sort method inherently has a time complexity of O(n log n), which can be undesirable
+in many cases when the desired time complexity of your solution is simply O(n).
+
+## Complexity
+
+**Time Complexity**: `O(n)` - since we only need to look over every element of our array a single time when comparing, time complexity is O(n).
+
+## References
+
+- [GeeksForGeeks](https://www.geeksforgeeks.org/two-pointers-technique/)
+- [YouTube](https://youtu.be/VEPCm3BCtik?si=rH9O1My7Ym_83FrR)

src/algorithms/search/twin-pointers/__test__/twinPointer.test.js

@@ -0,0 +1,30 @@
+import { twinPointerSorted, twinPointerUnsorted } from '../twinPointers';
+
+describe('twinPointerSorted', () => {
+  it('should search for a specific combination sum', () => {
+    expect(twinPointerSorted([], 1)).toBe(-1);
+    expect(twinPointerSorted([0, 1, 2], 3)).toStrictEqual([1, 2]);
+    expect(twinPointerSorted([0, 1, 2], 1)).toStrictEqual([0, 1]);
+    expect(twinPointerSorted([1, 2, 5, 7, 9], 4)).toBe(-1);
+    expect(twinPointerSorted([1, 2, 5, 7, 9], 14)).toStrictEqual([2, 4]);
+    expect(twinPointerSorted([3, 5, 7, 9], 1)).toBe(-1);
+    expect(twinPointerSorted([4, 6, 10, 15, 16, 18, 20], 10)).toStrictEqual([0, 1]);
+    expect(twinPointerSorted([4, 6, 10, 15, 16, 18, 20], 38)).toStrictEqual([5, 6]);
+    expect(twinPointerSorted([0, 100, 300, 500, 700, 1000, 2000, 5000], 50)).toBe(-1);
+    expect(twinPointerSorted([0, 100, 300, 500, 700, 1000, 2000, 5000], 100)).toStrictEqual([0, 1]);
+    expect(twinPointerSorted([0, 100, 300, 500, 700, 1000, 2000, 5000], 1000)).toStrictEqual([0, 5]);
+    expect(twinPointerSorted([0, 100, 300, 500, 700, 1000, 2000, 5000], 5000)).toStrictEqual([0, 7]);
+  });
+});
+
+describe('twinPointerUnsorted', () => {
+  it('should search for the highest possible area', () => {
+    expect(twinPointerUnsorted([])).toBe(0);
+    expect(twinPointerUnsorted([2])).toBe(2);
+    expect(twinPointerUnsorted([0, 1, 2])).toBe(1);
+    expect(twinPointerUnsorted([1, 2, 5, 7, 9])).toBe(10);
+    expect(twinPointerUnsorted([3, 5, 7, 9])).toBe(10);
+    expect(twinPointerUnsorted([4, 6, 10, 15, 16, 18, 20])).toBe(45);
+    expect(twinPointerUnsorted([0, 100, 300, 500, 700, 1000, 2000, 5000])).toBe(2100);
+  });
+});

src/algorithms/search/twin-pointers/twinPointers.js

@@ -0,0 +1,101 @@
+import Comparator from '../../../utils/comparator/Comparator';
+
+/**
+ * Some twin pointer implementations.
+ *
+ * @param {*[]} sortedArray
+ * @param {*} seekElement
+ * @param {function(a, b)} [comparatorCallback]
+ * @return {[number, number]}
+ */
+
+// Example of a twin pointer application in a sorted array where we are seeking the indices of two elements that sum to equal the target.
+export function twinPointerSorted(sortedArray, seekElement, comparatorCallback) {
+  const comparator = new Comparator(comparatorCallback);
+
+  // These variables will be our pointers; since the array is sorted, we can set them to the left and rightmost elements.
+  let left = 0;
+  let right = sortedArray.length - 1
+
+  // If our left and right pointers have met then we have iterated through the entire array.
+  while (left < right) {
+    
+    /** 
+     * If our sum is less than the target then we can increase said sum but by increasing the left value; 
+     * since the array is sorted, this will always result in array[left] becoming a larger number.
+    */
+    if (comparator.lessThan(sortedArray[left] + sortedArray[right], seekElement)) {
+        left++;
+
+    // Same concept as before, only now we decrease our sum because it's greater than the target.    
+    } else if (comparator.greaterThan(sortedArray[left] + sortedArray[right], seekElement)) {
+        right--;
+    
+    // Assuming we have found our target, return left and right since they represent the indices that our correct sum is located at.
+    } else {
+        return [left, right]
+    }
+  }
+
+  // Return -1 if we haven't found any combination of numbers that works.
+  return -1;
+}
+
+/* An example of a twin pointer method on an unsorted array. In this problem, we aim to get the heighest possible area from two numbers by using the
+    small of the two heights, assuming that each number n is a rectangle of 1 width and n height. (Problem and solution taken from Leetcode #11)
+*/
+
+/**
+ *
+ * @param {*[]} unsortedArray
+ * @param {function(a, b)} [comparatorCallback]
+ * @return {number}
+ */
+
+export function twinPointerUnsorted(unsortedArray, comparatorCallback) {
+    const comparator = new Comparator(comparatorCallback);
+  
+    // Edge cases; not relevant to the pointer method.
+    if (unsortedArray.length === 0) {
+        return 0
+    } else if (unsortedArray.length === 1) {
+        return unsortedArray[0]
+    }
+    
+    // Again, we set our two pointers to the left and rightmost elements of the array.
+    let left = 0;
+    let right = unsortedArray.length - 1;
+
+    // We initialize two area variables; one for our current area between our two pointers and one for the highest that we'll return.
+    let area = 0;
+    let mostArea = 0;
+
+    // Functionally equivalent to the while conditional we set in the first example.
+    while (left !== right) {
+
+        // In this situation, since we don't have a specific "target" in mind we instead compare the two values at our two pointers to each other.
+        if (comparator.lessThan(unsortedArray[left], unsortedArray[right])) {
+            
+            // Here we simply calculate our current area and whether we need to change our highest area by comparing it with the current.
+            area = (Math.min(unsortedArray[left], unsortedArray[right]) * (right - left));
+            mostArea = Math.max(area, mostArea);
+            
+            /**
+             * Again, we move the left pointer forward or the right pointer backwards. You may be thinking that this is basically the same as with the 
+             * sorted array; while that is correct from a pure code standpoint, conceptually the reasoning is different. In the first example, because the array
+             * is sorted we can move the left pointer forward with the knowledge that this will DEFINITELY either keep the value the same or increase it. 
+             * In this situation however, our array isn't sorted and thus moving the left pointer forward isn't guaranteed to increase the value; all we know
+             * is that it will change. However, because we are calculating area with the smallest height and our value (heght) at the left pointer is smaller than the right pointer,
+             * we know that the ONLY way to get a higher area is if there is a potentially higher value for the left pointer.
+             */
+            left++;
+        } else {
+            area = (Math.min(unsortedArray[left], unsortedArray[right]) * (right - left));
+            mostArea = Math.max(area, mostArea);
+            right--;
+        }
+    }
+
+    // Our greatest area should be correct since we re-state if the current area is greater.
+    return mostArea
+  }