Adding extra clause to the array pair sum solution.

Author: blakeembrey

array-pair-sum/Readme.md

@@ -2,6 +2,8 @@
 
 Given an integer array, output all pairs that sum up to a specific value k. Consider the fact that the same number can add up to `k` with its duplicates in the array.
 
+> For example the array is [1, 1, 2, 3, 4] and the desired sum is 4. Should we output the pair (1, 3) twice or just once? Also do we output the reverse of a pair, meaning both (3, 1) and (1, 3)? Let’s keep the output as short as possible and print each pair only once. So, we will output only one copy of (1, 3). Also note that we shouldn’t output (2, 2) because it’s not a pair of two distinct elements.
+
 ## Example
 
 ```
@@ -11,4 +13,4 @@ f(8, [3, 4, 5, 4, 4]) // [ [3, 5], [4, 4], [4, 4], [4, 4] ]
 
 ## Source
 
-[http://www.ardendertat.com/2012/01/09/programming-interview-questions/](http://www.ardendertat.com/2012/01/09/programming-interview-questions/)
+[http://www.ardendertat.com/2011/09/17/programming-interview-questions-1-array-pair-sum/](http://www.ardendertat.com/2011/09/17/programming-interview-questions-1-array-pair-sum/)

array-pair-sum/array-pair-sum.js

@@ -7,11 +7,14 @@ var arrayPairSum = function (k, array) {
   // time complexity of O(n) - the naive solution consists of two for loops
   // which results in a complexity of O(n^2)
   array.forEach(function (number) {
-    for (var i = 0; i < hash[k - number]; i++) {
+    // Make sure the value in unused and it's a unique pair
+    if (hash[k - number] === false && k - number !== number) {
       pairs.push([number, k - number]);
+      hash[k - number] = true; // Set it to "used"
     }
 
-    hash[number] = hash[number] + 1 || 1;
+    // If the hash value is not true, set the hash to "unused"
+    !hash[k - number] && (hash[number] = false);
   });
 
   return pairs;