feat: add findLongestRecurringCycle algorithm (Problem 26)

gaye-lamine · gaye-lamine · commit c6192d760d9d · 2024-10-20T19:34:56.000Z
diff --git a/Project-Euler/Problem026.js b/Project-Euler/Problem026.js
@@ -4,9 +4,6 @@
  * @see {@link https://projecteuler.net/problem=26}
  *
  * Find the value of denominator < 1000 for which 1/denominator contains the longest recurring cycle in its decimal fraction part.
- *
- * A unit fraction (1/denominator) either terminates or repeats. We need to determine the length of the repeating sequence (cycle)
- * for each fraction where the denominator is between 2 and 999, and find the denominator that produces the longest cycle.
  */
 
 /**
@@ -23,49 +20,37 @@ function findLongestRecurringCycle(limit) {
    * @returns {number} The length of the recurring cycle in the decimal part of 1/denominator.
    */
   function getRecurringCycleLength(denominator) {
-    // A map to store the position of each remainder encountered during division
     const remainderPositions = new Map()
-    let numerator = 1 // We start with 1 as the numerator (as we're computing 1/denominator)
-    let position = 0 // This tracks the position of each digit in the decimal sequence
+    let numerator = 1
+    let position = 0
 
-    // Continue until the remainder becomes 0 (terminating decimal) or a cycle is found
     while (numerator !== 0) {
-      // If the remainder has been seen before, we've found the start of the cycle
       if (remainderPositions.has(numerator)) {
-        // The length of the cycle is the current position minus the position when the remainder first appeared
         return position - remainderPositions.get(numerator)
       }
 
-      // Record the position of this remainder
       remainderPositions.set(numerator, position)
 
-      // Multiply numerator by 10 to simulate long division and get the next digit
       numerator = (numerator * 10) % denominator
-      position++ // Move to the next digit position
+      position++
     }
 
-    // If numerator becomes 0, it means the decimal terminates (no cycle)
     return 0
   }
 
-  let maxCycleLength = 0 // Store the maximum cycle length found
-  let denominatorWithMaxCycle = 0 // Store the denominator corresponding to the longest cycle
+  let maxCycleLength = 0
+  let denominatorWithMaxCycle = 0
 
-  // Iterate through each possible denominator from 2 up to (limit - 1)
   for (let denominator = 2; denominator < limit; denominator++) {
-    // Calculate the cycle length for the current denominator
     const cycleLength = getRecurringCycleLength(denominator)
 
-    // Update the maximum length and the corresponding denominator if a longer cycle is found
     if (cycleLength > maxCycleLength) {
       maxCycleLength = cycleLength
       denominatorWithMaxCycle = denominator
     }
   }
 
-  // Return the denominator that has the longest recurring cycle
   return denominatorWithMaxCycle
 }
 
-// Exporting the main function for use in other modules
 export { findLongestRecurringCycle }
diff --git a/Project-Euler/test/Problem026.test.js b/Project-Euler/test/Problem026.test.js
@@ -4,27 +4,13 @@ import { findLongestRecurringCycle } from '../Problem026'
  * Tests for the findLongestRecurringCycle function.
  */
 describe('findLongestRecurringCycle', () => {
-  // Test to check the basic case with a limit of 10
-  test('should return 7 for limit of 10', () => {
-    const result = findLongestRecurringCycle(10)
-    expect(result).toBe(7)
-  })
-
-  // Test to check with a limit of 1000
-  test('should return the correct value for limit of 1000', () => {
-    const result = findLongestRecurringCycle(1000)
-    expect(result).toBe(983) // The expected result is the denominator with the longest cycle
-  })
-
-  // Test with a smaller limit to validate behavior
-  test('should return 3 for limit of 4', () => {
-    const result = findLongestRecurringCycle(4)
-    expect(result).toBe(3)
-  })
-
-  // Test with a limit of 2, where there is no cycle
-  test('should return 0 for limit of 2', () => {
-    const result = findLongestRecurringCycle(2)
-    expect(result).toBe(0) // No cycle for fractions 1/1 and 1/2
+  it.each([
+    { limit: 10, expected: 7 },
+    { limit: 1000, expected: 983 }, // The denominator with the longest cycle for limit of 1000
+    { limit: 4, expected: 3 },
+    { limit: 2, expected: 0 } // No cycle for fractions 1/1 and 1/2
+  ])('should return $expected for limit of $limit', ({ limit, expected }) => {
+    const result = findLongestRecurringCycle(limit)
+    expect(result).toBe(expected)
   })
 })
