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DARKcoder37 · web-flow · commit 847200fe3563 · 2023-04-11T22:50:33.000+05:30
diff --git a/Problem Solution/1) 2 sum (easy)/2 sum cpp.txt b/Problem Solution/1) 2 sum (easy)/2 sum cpp.txt
@@ -0,0 +1,44 @@
+1) Brute Force [O(n^2)]
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        vector<int> ret;
+        int size = nums.size();
+        for(int i=0 ; i < size - 1 ;i++)
+        {
+            for(int j = i+1 ; j<size ; j++)
+            {
+                if(nums[i]+nums[j] == target)
+                {
+                    ret.push_back(i);
+                    ret.push_back(j);
+                    return ret;
+                }   
+            }
+        }
+        return ret;
+    }
+};
+
+2) Hash Table [O(n)]
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        vector<int> ret;
+        int size = nums.size();
+        int diff;
+        unordered_map<int,int>m;
+        for (int i=0 ; i<size ; i++)
+        {
+            diff = target - nums[i];
+            if(m.find(diff) != m.end() && m.find(diff) -> second != i )
+            {
+                ret.push_back(i);
+                ret.push_back(m.find(diff) -> second);
+                return ret;
+            }
+            m[nums[i]] = i;
+        }
+        return ret;
+    }
+};
diff --git a/Problem Solution/1) 2 sum (easy)/2 sum python.txt b/Problem Solution/1) 2 sum (easy)/2 sum python.txt
@@ -0,0 +1,10 @@
+1) Brute force
+class Solution(object):
+    def twoSum(self, nums, target):
+        #Using Two pointers P1 and P2 to represent every possible pairs of the array.
+        for p1 in range(len(nums)):           
+             for p2 in range(p1+1, len(nums)): 
+                #If one of those pairs add together equal to the target return the answer else return None.
+                 if nums[p1] + nums[p2] == target:    
+                    return [p1, p2]              
+        return 'None' 
