refactor: clean up formatting and improve JSDoc comments

Author: RomanSanaye

Sprint-1/JavaScript/calculateSumAndProduct/calculateSumAndProduct.js

@@ -9,9 +9,13 @@
  *   "product": 30 // 2 * 3 * 5
  * }
  *
- * Time Complexity: O(n) - one loop through the list (the original used two loops, 2n steps, which is also O(n)).
- * Space Complexity: O(1) - only two variables (sum, product), regardless of input size
- * Optimal Time Complexity: O(n) - every number must be read at least once, so we can't do better than linear time
+ * Time Complexity: O(n) - one loop through the list (the original used two
+ * loops, 2n steps, which is also O(n)).
+ *
+ * Space Complexity: O(1) - only two variables (sum, product), regardless of
+ * input size
+ *
+ * Optimal Time Complexity: O(n) - every number must be read at least once, so * we can't do better than linear time
  *
  * Refactor: the complexity can't be reduced, but merging the two loops into
  * one makes the code cleaner and does half the steps.
@@ -22,11 +26,12 @@
 export function calculateSumAndProduct(numbers) {
   let sum = 0;
   let product = 1;
+
   for (const num of numbers) {
     sum += num;
     product *= num;
   }
- 
+
   return {
     sum: sum,
     product: product,

Sprint-1/JavaScript/findCommonItems/findCommonItems.js

@@ -1,35 +1,41 @@
 /**
- * Finds common items between two arrays.
+* Finds common items between two arrays.
  
-  * Inefficiency: the original used secondArray.includes() inside firstArray.filter(), which is a loop inside a loop.
+* Inefficiency: the original used secondArray.includes() inside firstArray
+filter(), which is a loop inside a loop.
   
-  * Originally:
-  * Time Complexity: O(n × m) - filter runs n times and each includes call can scan up to m items (O(n²) if the arrays are the same size).
+* Originally:
+* Time Complexity: O(n × m) - filter runs n times and each includes call can
+* scan up to m items (O(n²) if the arrays are the same size).
   
-  * Space Complexity: O(n) - the filtered array and the Set hold up to n items.
+* Space Complexity: O(n) - the filtered array and the Set hold up to n items.
   
-  * Optimal Time Complexity: O(n + m) - every item in both arrays must be read at least once
+* Optimal Time Complexity: O(n + m) - every item in both arrays must be read at
+least once.
   
-  * After refactor (the complexity was reduced from quadratic to linear time by
-    using a Set for O(1) average lookups instead of includes):
+* After refactor (the complexity was reduced from quadratic to linear time by
+using a Set for O(1) average lookups instead of includes):
   
-  * Time Complexity: O(n + m) - building secondSet is O(m); the loop over firstArray is O(n) with O(1) average  has() checks.
+* Time Complexity: O(n + m) - building secondSet is O(m); the loop over
+firstArray is O(n) with O(1) average  has() checks.
+
+* Space Complexity: O(n + m) - secondSet holds up to m items; commonElements
+and the final Set hold up to n    items.
 
-* Space Complexity: O(n + m) - secondSet holds up to m items; commonElements and the final Set hold up to n    items.
 * Optimal Time Complexity: O(n + m) - every item in both arrays must be read at least once, so we can't do better than linear
- * 
- * @param {Array} firstArray - First array to compare
- * @param {Array} secondArray - Second array to compare
- * @returns {Array} Array containing unique common items
- */
+* 
+* @param {Array} firstArray - First array to compare
+* @param {Array} secondArray - Second array to compare
+* @returns {Array} - Array containing unique common items
+*/
 export const findCommonItems = (firstArray, secondArray) => {
-const secondSet = new Set(secondArray);
-let commonElements = [];
+  const secondSet = new Set(secondArray);
+  let commonElements = [];
 
-for(let ele of firstArray){
-  if(secondSet.has(ele)){
-    commonElements.push(ele);
+  for (let ele of firstArray) {
+    if (secondSet.has(ele)) {
+      commonElements.push(ele);
+    }
   }
-}
-return [...new Set(commonElements)];
+  return [...new Set(commonElements)];
 };

Sprint-1/JavaScript/hasPairWithSum/hasPairWithSum.js

@@ -6,22 +6,22 @@
  *
  * Originally:
  * Time Complexity: O(n²) - for each number, the inner loop scans the
- *   remaining numbers, so about n * (n - 1) / 2 pair checks.
- * 
+ * remaining numbers, so about n * (n - 1) / 2 pair checks.
+ *
  * Space Complexity: O(1) - only the loop counters i and j are used.
- * 
+ *
  * Optimal Time Complexity: O(n) - every number must be read at least once,
- *   and a Set gives O(1) average lookups, so one pass is enough
+ * and a Set gives O(1) average lookups, so one pass is enough
  *
  * After Refactor (reduced from quadratic to linear time by remembering the
  * numbers already seen in a Set instead of re-scanning the list):
- * 
+ *
  * Time Complexity: O(n) - one loop, with O(1) average has() and add().
- * 
+ *
  * Space Complexity: O(n) - visitedNumbers can hold up to n numbers.
- * 
+ *
  * Optimal Time Complexity: O(n) - every number must be read at least once,
- *   so we can't do better than linear.
+ * so we can't do better than linear.
  *
  * @param {Array<number>} numbers - Array of numbers to search through
  * @param {number} target - Target sum to find
@@ -40,4 +40,4 @@ export function hasPairWithSum(numbers, target) {
   }
 
   return false;
-}
+}

Sprint-1/JavaScript/removeDuplicates/removeDuplicates.mjs

@@ -1,28 +1,29 @@
 /**
- * Remove duplicate values from a sequence, preserving the order of the first occurrence of each value.
- *
-/**
+* Remove duplicate values from a sequence, preserving the order of the first
+* occurrence of each value.
+*
 * Originally (nested loops):
 * Time Complexity: O(n²) - the outer loop runs n times and the inner loop
-*   scans uniqueItems, which can grow up to n items (about n(n - 1) / 2
-*   checks in the worst case, when all items are unique)
-* Space Complexity: O(n) - uniqueItems can hold up to n items
+* scans uniqueItems, which can grow up to n items (about n(n - 1) / 2
+* checks in the worst case, when all items are unique).
+* 
+* Space Complexity: O(n) - uniqueItems can hold up to n items.
+* 
 * Optimal Time Complexity: O(n) - every item must be read at least once, and
-*   a Set gives O(1) average lookups
+* a Set gives O(1) average lookups.
 
 * After refactor ([...new Set(inputSequence)]):
 * Time Complexity: O(n) - building the Set visits each item once with O(1)
-*   average inserts; spreading it into an array is also O(n)
-* Space Complexity: O(n) - the Set and the result array each hold up to n items
+* average inserts; spreading it into an array is also O(n).
+* 
+* Space Complexity: O(n) - the Set and the result array each hold up to n items.
+* 
 * Optimal Time Complexity: O(n) - every item must be read at least once, so we
-*   can't do better than linear
-* /
- * 
- *
- * @param {Array} inputSequence - Sequence to remove duplicates from
- * @returns {Array} New sequence with duplicates removed
- */
+* can't do better than linear.
+*
+* @param {Array} inputSequence - Sequence to remove duplicates from
+* @returns {Array} New sequence with duplicates removed
+*/
 export function removeDuplicates(inputSequence) {
-  
-  return [... new Set(inputSequence)];
+  return [...new Set(inputSequence)];
 }