Add 3sum solution

Author: Jeehay28

3sum/Jeehay28.js

@@ -0,0 +1,86 @@
+/**
+ * @param {number[]} nums
+ * @return {number[][]}
+ */
+
+
+// Time Complexity: O(n^2)
+// Space Complexity: O(n^2)
+
+var threeSum = function (nums) {
+
+    const sorted = [...nums].sort((a, b) => a - b);
+
+    let result = [];
+
+    // Loop through the array and pick each number as the first number for the triplet
+    for (let i = 0; i < sorted.length - 2; i++) {
+
+        // skip duplicate values for sorted[middle]
+        if(i > 0 && sorted[i - 1] === sorted[i]) {
+            continue;
+        }
+
+        let left = i + 1; // Left pointer starts right after the current middle
+        let right = sorted.length - 1; // Right pointer starts at the last element
+
+        while (left < right) {
+            const sum = sorted[i] + sorted[left] + sorted[right];
+
+            if (sum === 0) {
+                result.push([sorted[left], sorted[i], sorted[right]]);
+
+                // skip duplicates for sorted[left] and sorted[right]
+                while(left < right && sorted[left] === sorted[left + 1]){
+                    left += 1; // Move left pointer to the right to skip duplicate values
+                } 
+
+                while(left < right && sorted[right] === sorted[right - 1]) {
+                    right -= 1; // Move right pointer to the left to skip duplicate values
+                }
+
+                left += 1; 
+                right -= 1;
+
+            } else if (sum > 0) {
+                right -= 1;
+
+            } else {
+                left += 1
+            }
+        }
+    }
+
+     return result;
+  
+};
+
+
+// var threeSum = function (nums) {
+
+// i != j, i != k, and j != k can be interpreted as i < j < k
+
+    // three nested loops
+    // time complexity of O(n³)
+    // Time Limit Exceeded
+
+    // let result = [];
+
+    // for (let i = 0; i < nums.length - 2; i++) {
+    //     for (let j = i + 1; j < nums.length - 1; j++) {
+    //         for (let k = j + 1; k < nums.length; k++) {
+    //             if (nums[i] + nums[j] + nums[k] === 0) {
+    //                 const str = [nums[i], nums[j], nums[k]].sort((a, b) => a - b).join(",")
+    //                 result.push(str)
+    //             }
+    //         }
+    //     }
+
+    // }
+
+    // result = [... new Set(result)].map(str => str.split(",").map(str => +str))
+
+    // return result;
+    
+// }
+