Two-sum solution

leetcode/two-sum/TwoPass.cpp

@@ -0,0 +1,16 @@
+class Solution{
+public:
+    vector<int> twoSum(vector<int> &nums, int target) {
+        unordered_map<int, int> hash;
+        for (int i = 0; i < nums.size(); i++) {
+            hash[nums[i]] = i;
+        }
+        for (int i = 0; i < nums.size(); i++) {
+            int comp = target - nums[i];
+            if (hash.find(comp) != hash.end() && hash[comp] != i) {
+                return {i, hash[comp]};
+            }
+        }
+        return {};
+    }
+};

leetcode/two-sum/brute.cpp

@@ -0,0 +1,13 @@
+class Solution {
+public:
+    vector<int> twoSum(vector<int>& nums, int target) {
+        for(int i=0;i<nums.size();i++)
+        {
+            for(int j=i+1;j<nums.size();j++)
+                if(nums[i]+nums[j] == target)
+                    return {i,j};
+            }
+            return {};
+    }
+        
+};

leetcode/two-sum/notes.md

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+One pass:
+one pass starts from index 0 and stores it in a map,
+and for next number in array,
+it computes the new complement and checks if it’s present in map,
+if yes then prints
+else stores the original number in hash and continues
+
+
+
+
+Two pass:
+Two-pass hash table solution first builds a map of all values to their indices,
+then makes a second scan through the array to compute each element’s complement
+and check if it exists in the map.
+

leetcode/two-sum/onePass.cpp

@@ -0,0 +1,13 @@
+public:
+	vector<int> twoSum(vector<int>& nums, int target){
+		unordered_map<int,int>hash;
+		for(int i=0;i<nums.size();++i){
+			int comp = target - nums[i];
+			if(hash.find(complement) != hash.end(){
+					return {hash[complement],i};
+			hash[nums[i]] = i;
+			
+		       }
+		}
+	}
+