add new solutions

solutions/0/q4/solution.go

@@ -0,0 +1,59 @@
+package q4
+
+import "math"
+
+func findMedianSortedArrays(nums1 []int, nums2 []int) float64 {
+	l1, l2 := len(nums1), len(nums2)
+	even := (l1+l2)%2 == 0
+
+	if l1+l2 == 1 {
+		if l1 == 1 {
+			return float64(nums1[0])
+		}
+		return float64(nums2[0])
+	}
+
+	// Find how many numbers `nums1` has that are to the left of the median
+	nLeft := (len(nums1) + len(nums2)) / 2
+	l, r := max(0, nLeft-l2), min(nLeft, len(nums1))+1
+
+Loop:
+	for l+1 < r {
+		m := (l + r) / 2
+		p1 := m - 1
+		p2 := nLeft - m - 1
+
+		switch {
+		case p2 >= 0 && p1+1 < l1 && nums2[p2] > nums1[p1+1]:
+			l = m + 1
+		case p1 >= 0 && p2+1 < l2 && nums1[p1] > nums2[p2+1]:
+			r = m
+		default:
+			l = m
+			break Loop
+		}
+	}
+
+	lNum, rNum := math.MinInt, math.MaxInt
+	p1 := l - 1
+	p2 := nLeft - l - 1
+	if p1 >= 0 {
+		lNum = max(lNum, nums1[p1])
+	}
+	if p2 >= 0 {
+		lNum = max(lNum, nums2[p2])
+	}
+	if p1+1 < l1 {
+		rNum = min(rNum, nums1[p1+1])
+	}
+	if p2+1 < l2 {
+		rNum = min(rNum, nums2[p2+1])
+	}
+
+	if even {
+		return float64(lNum+rNum) / 2
+	}
+	return float64(rNum)
+}
+
+var _ = findMedianSortedArrays