Create median of two sorted arrays

LeetCodeQuestions/median of two sorted arrays

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+class Solution {
+    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
+      /* brute force , tc - O(m + n) , sc - O(n)  
+      int m = nums1.length;
+        int n= nums2.length;
+        int res[]  = new int[m + n];
+
+        int k = 0;
+        for(int i = 0; i < m; i++ ){
+            res[k++] = nums1[i];
+        }
+
+        for( int  i = 0;i < n; i++){
+            res[k++] = nums2[i];
+        }
+
+        Arrays.sort(res);
+
+
+
+        int mid = res.length/2;
+
+        if(res.length % 2 == 0){
+            return (double) (res[mid -1 ] + res[mid])/2;
+        }
+        return res[mid]; 
+
+
+
+
+
+          linear search (better approach) , tc - O(n) , sc - O(1) 
+        int n1 = nums1.length;
+        int n2 = nums2.length;
+        int count = 0;
+        int i = 0,j =0;
+        int mid = (n1 + n2) /2;
+        int ele1 = -1;
+        int ele2 = -1;
+
+        while(i < n1 && j < n2 ){
+            if(nums1[i] < nums2[j]){
+                if( count == mid -1)ele1 = nums1[i];
+               else if ( count == mid)ele2 = nums1[i];
+                i++;
+                count++;
+            }
+            else{
+                if( count == mid -1)ele1 = nums2[j];
+                else if( count == mid)ele2 = nums2[j];
+                j++;
+                count++;
+            }
+
+
+
+        }
+        while( i < n1){
+          if( count == mid-1)ele1= nums1[i];
+        else if( count == mid)ele2 = nums1[i];
+          i++;
+          count++;
+        }
+
+         while( j < n2){
+          if( count == mid-1)ele1= nums2[j];
+          else if( count == mid)ele2 = nums2[j];
+          j++;
+          count++;
+        }
+
+        if( (n1 + n2) % 2 == 0 ){
+            return 1D * (ele1 + ele2)/2;
+        }
+        return 1D * ele2;
+        */
+
+        // optimal approach binary search  ,tc -  O(log n) , sc - O(1)
+
+          int n1 = nums1.length, n2 = nums2.length;
+
+        // Ensure nums1 is the smaller array for simplicity
+        if (n1 > n2)
+            return findMedianSortedArrays(nums2, nums1);
+
+        int n = n1 + n2;
+        int left = (n1 + n2 + 1) / 2; // Calculate the left partition size
+        int low = 0, high = n1;
+
+        while (low <= high) {
+            int mid1 = (low + high) >> 1; // Calculate mid index for nums1
+            int mid2 = left - mid1; // Calculate mid index for nums2
+
+            int l1 = Integer.MIN_VALUE, l2 = Integer.MIN_VALUE, r1 = Integer.MAX_VALUE, r2 = Integer.MAX_VALUE;
+
+            // Determine values of l1, l2, r1, and r2
+            if (mid1 < n1)
+                r1 = nums1[mid1];
+            if (mid2 < n2)
+                r2 = nums2[mid2];
+            if (mid1 - 1 >= 0)
+                l1 = nums1[mid1 - 1];
+            if (mid2 - 1 >= 0)
+                l2 = nums2[mid2 - 1];
+
+            if (l1 <= r2 && l2 <= r1) {
+                // The partition is correct, we found the median
+                if (n % 2 == 1)
+                    return Math.max(l1, l2);
+                else
+                    return ((double)(Math.max(l1, l2) + Math.min(r1, r2))) / 2.0;
+            }
+            else if (l1 > r2) {
+                // Move towards the left side of nums1
+                high = mid1 - 1;
+            }
+            else {
+                // Move towards the right side of nums1
+                low = mid1 + 1;
+            }
+        }
+
+        return 0; // If the code reaches here, the input arrays were not sorted.
+
+
+        }
+
+
+    }