Add the Fibonacci Search Algortihm

src/algorithms/search/fibonacci-search/README.md

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+# Fibonacci Search Algorithm 
+Let k be defined as an element in F, the array of Fibonacci numbers. n = Fm is the array size. If n is not a Fibonacci number, let Fm be the smallest number in F that is greater than n.
+
+The array of Fibonacci numbers is defined where Fk+2 = Fk+1 + Fk, when k ≥ 0, F1 = 1, and F0 = 0.
+
+To test whether an item is in the list of ordered numbers, follow these steps:
+```sh
+Set k = m.
+If k = 0, stop. There is no match; the item is not in the array.
+Compare the item against element in Fk−1.
+If the item matches, stop.
+If the item is less than entry Fk−1, discard the elements from positions Fk−1 + 1 to n. Set k = k − 1 and return to step 2.
+If the item is greater than entry Fk−1, discard the elements from positions 1 to Fk−1. Renumber the remaining elements from 1 to Fk−2, set k = k − 2, and return to step 2.
+Alternative implementation (from "Sorting and Searching" by Knuth[4]):
+
+Given a table of records R1, R2, ..., RN whose keys are in increasing order K1 < K2 < ... < KN, the algorithm searches for a given argument K. Assume N+1 = Fk+1
+
+Step 1. [Initialize] i ← Fk, p ← Fk-1, q ← Fk-2 (throughout the algorithm, p and q will be consecutive Fibonacci numbers)
+
+Step 2. [Compare] If K < Ki, go to Step 3; if K > Ki go to Step 4; and if K = Ki, the algorithm terminates successfully.
+
+Step 3. [Decrease i] If q=0, the algorithm terminates unsuccessfully. Otherwise set (i, p, q) ← (p, q, p - q) (which moves p and q one position back in the Fibonacci sequence); then return to Step 2
+
+Step 4. [Increase i] If p=1, the algorithm terminates unsuccessfully. Otherwise set (i,p,q) ← (i + q, p - q, 2q - p) (which moves p and q two positions back in the Fibonacci sequence); and return to Step 2
+```
+The two variants of the algorithm presented above always divide the current interval into a larger and a smaller subinterval. The original algorithm,[1] however, would divide the new interval into a smaller and a larger subinterval in Step 4. This has the advantage that the new i is closer to the old i and is more suitable for accelerating searching on magnetic tape.
+

src/algorithms/search/fibonacci-search/__test__/fibonacciSearch.test.js

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+import fibonacciSearch from '../fibonacciSearch';
+
+describe('fibonacciSearch', () => {
+  it('should search number in sorted array', () => {
+    expect(fibonacciSearch([], 1)).toBe(-1);
+    expect(fibonacciSearch([1], 1)).toBe(0);
+    expect(fibonacciSearch([1, 2], 1)).toBe(0);
+    expect(fibonacciSearch([1, 2], 2)).toBe(1);
+    expect(fibonacciSearch([1, 5, 10, 12], 1)).toBe(0);
+    expect(fibonacciSearch([1, 5, 10, 12, 14, 17, 22, 100], 17)).toBe(5);
+    expect(fibonacciSearch([1, 5, 10, 12, 14, 17, 22, 100], 1)).toBe(0);
+    expect(fibonacciSearch([1, 5, 10, 12, 14, 17, 22, 100], 100)).toBe(7);
+    expect(fibonacciSearch([1, 5, 10, 12, 14, 17, 22, 100], 0)).toBe(-1);
+  });
+});

src/algorithms/search/fibonacci-search/fibonacciSearch.js

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+/** Author Slim Gharbi
+ * Fibonacci search implementation.
+ *
+ * @param {*[]} integers
+ * @param {*} elementToSearch
+ * @return {number}
+ */
+export default function fibonacciSearch(integers, elementToSearch) {
+  let i;
+
+  /* Initialize fibonacci numbers */
+  let fibonacciMinus2 = 0;// (m-2)'th Fibonacci No.
+  let fibonacciMinus1 = 1;// (m-1)'th Fibonacci No.
+  let fibonacciNumber = fibonacciMinus2 + fibonacciMinus1; // m'th Fibonacci
+  /* fibonacciNumber is going to store the smallest
+ Fibonacci Number greater than or equal to the length of the array */
+  while (fibonacciNumber < integers.length) {
+    fibonacciMinus2 = fibonacciMinus1;
+    fibonacciMinus1 = fibonacciNumber;
+    fibonacciNumber = fibonacciMinus2 + fibonacciMinus1;
+  }
+  // Marks the eliminated range from front
+  let offset = -1;
+  /* while there are elements to be inspected.
+ Note that we compare integers[fibonacciMinus2] with elementToSearch.
+ When fibonacciNumber becomes 1, fibonacciMinus2 becomes 0 */
+  while (fibonacciNumber > 1) {
+  // Check if fibonacciMinus2 is a valid location
+    i = Math.min((offset + fibonacciMinus2), (integers.length - 1));
+    /* If elementToSearch is greater than the value at
+  index fibonacciMinus2, cut the subarray array
+  from offset to i */
+    if (integers[i] < elementToSearch) {
+      fibonacciNumber = fibonacciMinus1;
+      fibonacciMinus1 = fibonacciMinus2;
+      fibonacciMinus2 = fibonacciNumber - fibonacciMinus1;
+      offset = i;
+    } else if (integers[i] > elementToSearch) {
+      fibonacciNumber = fibonacciMinus2;
+      fibonacciMinus1 -= fibonacciMinus2;
+      fibonacciMinus2 = fibonacciNumber - fibonacciMinus1;
+      /* If elementToSearch is greater than the value at index
+      fibonacciMinus2, cut the subarray after i+1 */
+    } else return i;
+    // element found. return index
+  }
+  // comparing the last element with elementToSearch
+  if (fibonacciMinus1 === 1 && integers[offset + 1] === elementToSearch) return offset + 1;
+  // element not found. return -1
+  return -1;
+}