Merge pull request #138 from rmn5124/master

4. Median of Two Sorted Arrays

Author: pranjay-poddar

LeetCode/Hard/4. Median of Two Sorted Arrays.cpp

@@ -0,0 +1,78 @@
+// Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.
+
+// The overall run time complexity should be O(log (m+n)).
+
+// Example 1:
+
+// Input: nums1 = [1,3], nums2 = [2]
+// Output: 2.00000
+// Explanation: merged array = [1,2,3] and median is 2.
+// Example 2:
+
+// Input: nums1 = [1,2], nums2 = [3,4]
+// Output: 2.50000
+// Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
+
+ 
+
+
+//Approach 1
+
+class Solution {
+public:
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        for(int i=0;i<nums2.size();i++){
+            nums1.push_back(nums2[i]);
+        }
+        sort(nums1.begin(),nums1.end());
+        int n=nums1.size();
+        double res;
+        if(n&1)
+            res=nums1[n/2];
+        else if(!(n&1))
+            res=(double)(nums1[n/2]+nums1[(n/2)-1])/2;
+    
+        return res;
+            
+    }
+};
+
+
+// Approach 2
+
+class Solution {
+public:
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        if(nums2.size() < nums1.size()) return findMedianSortedArrays(nums2, nums1);
+        int n1 = nums1.size();
+        int n2 = nums2.size(); 
+        int low = 0, high = n1;
+        
+        while(low <= high) {
+            int cut1 = (low+high) >> 1;
+            int cut2 = (n1 + n2 + 1) / 2 - cut1; 
+            
+        
+            int left1 = cut1 == 0 ? INT_MIN : nums1[cut1-1];
+            int left2 = cut2 == 0 ? INT_MIN : nums2[cut2-1]; 
+            
+            int right1 = cut1 == n1 ? INT_MAX : nums1[cut1];
+            int right2 = cut2 == n2 ? INT_MAX : nums2[cut2]; 
+            
+            
+            if(left1 <= right2 && left2 <= right1) {
+                if( (n1 + n2) % 2 == 0 ) 
+                    return (max(left1, left2) + min(right1, right2)) / 2.0; 
+                else 
+                    return max(left1, left2); 
+            }
+            else if(left1 > right2) {
+                high = cut1 - 1; 
+            }
+            else {
+                low = cut1 + 1; 
+            }
+        }
+        return 0.0; 
+    }
+};