Merge pull request #1971 from hongseoupyun/main

[hongseoupyun] WEEK 01  Solutions

Author: dalestudy[bot]

two-sum/hongseoupyun.py

@@ -0,0 +1,21 @@
+# Create a empty hash table(seen) and store numbers as a key and their indices as a value
+# Iterate through the given array(nums) 
+# For each numbers in given array, calculate complement (target - num)
+# If complement is in hash table(seen), return array of index of the currnet number and index of complement in hash table(seen)
+# If complement is not in has table(seen), store the current number in hash table(seen) and continue checking the numbers
+# The algorithm uses a single loop that goes through each element in nums exactly one time.
+# If the array contains n elements, the loop runs n times → O(n).
+# We use a hash table (seen) to store numbers we've already visited.
+# seen could contain up to n − 1 entries, so its size grows linearly with input size → O(n).
+
+
+
+class Solution:
+    def twoSum(self, nums: List[int], target: int) -> List[int]:
+        seen = {}
+        for i, num in enumerate(nums):
+            complement = target - num
+            if complement in seen:      
+                return [seen[complement], i]
+            seen[num] = i               
+        return []

two-sum/hongseoupyun.ts

@@ -0,0 +1,27 @@
+/*
+
+
+// Create a empty hash table(seen) and store numbers as a key and their indices as a value
+// Iterate through the given array(nums) 
+// For each numbers in given array, calculate complement (target - num)
+// If complement is in hash table(seen), return array of index of the currnet number and index of complement in hash table(seen)
+// If complement is not in has table(seen), store the current number in hash table(seen) and continue checking the numbers
+// The algorithm uses a single loop that goes through each element in nums exactly one time.
+// If the array contains n elements, the loop runs n times → O(n).
+   We use a hash table (seen) to store numbers we've already visited.
+  seen could contain up to n − 1 entries, so its size grows linearly with input size → O(n).
+
+
+*/
+
+function twoSum2(nums: number[], target: number): number[] {
+    const seen:{[key:number] : number} = {};
+    for (let i = 0; i<nums.length; i++){
+      let complement = target - nums[i]
+      if (complement in seen) {
+        return [seen[complement],i]
+      }
+      seen[nums[i]] = i
+    }
+    return []
+};