Leetcode day1

leetcode/lc4/TwoPointerMed.cpp

@@ -0,0 +1,40 @@
+class Solution {
+public:
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        int a = nums1.size();
+        int b = nums2.size();
+        int i=0,j=0,m1=0,m2=0;
+
+        for(int count=0;count<=(a+b)/2;count++){
+            m2=m1;
+            if(i!=a && j!=b){
+                if(nums1[i]<nums2[j]){
+                    m1= nums1[i++];
+                }
+                else{
+                     m1 = nums2[j++];
+                     }
+            }
+            else if(i<a){
+                m1 = nums1[i++];
+            }
+            else  {
+                m1 = nums2[j++];
+            }
+           
+
+                
+            
+
+            
+        }
+         if((a+b)%2==1){
+                return (double)m1;
+            }
+            else{
+                return (m1+m2)/2.0;
+            }
+      
+        }
+    
+};

leetcode/lc4/bruteMedian.cpp

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+class Solution{
+    public : 
+        double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2){
+            int n = nums1.size();
+            int m = nums2.size();
+            vector<int> merger;
+            for(auto nums: nums1)
+                    merger.push_back(nums);
+            for(auto nums : nums2)
+                    merger.push_back(nums);
+            sort(merger.begin(), meger.end());
+
+            int size = meger.size();
+            if(size%2==1)
+                return (double) merger[size/2];
+            else
+                return (double) (merger[size/2]+ merger[size/2 - 1])/2.0;
+
+            
+        }
+};

leetcode/lc4/notes.md

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+The brute force approach is quite intuitive but doesnt meet the specified Run time complexity of O(log(m+n)) instead takes O((m+n)log(m+n)) taken by sorting
+
+TwoPointer method takes O(n+m) where we only merge till half and then extract the median which is the middle element
+
+Yet to learn binarySearch approach