From f3abe4ff5bedc75284e00cad7947d9e00522ae6f Mon Sep 17 00:00:00 2001
From: "M.V. Hutz" <git@maximhutz.me>
Date: Sun, 11 Jan 2026 15:42:42 -0500
Subject: [PATCH] feat: sunday

---
 pkg/container_with_most_water/main.go | 36 +++++++++++++++++++++++++++
 1 file changed, 36 insertions(+)
 create mode 100644 pkg/container_with_most_water/main.go

diff --git a/pkg/container_with_most_water/main.go b/pkg/container_with_most_water/main.go
new file mode 100644
index 0000000..6bc1f18
--- /dev/null
+++ b/pkg/container_with_most_water/main.go
@@ -0,0 +1,36 @@
+package container_with_most_water
+
+func MaxArea(height []int) int {
+	if len(height) < 2 {
+		return 0
+	}
+
+	start := 0
+	end := len(height) - 1
+
+	// Let:
+	// - A = the smaller side,
+	// - B = the larger side.
+	//
+	// Notice that:
+	// - Any container with A cannot have a height larger than `min(A, B)`.
+	// - Any container with A cannot have a width larger than `B - A`.
+	//
+	// Because of this, all solutions containing A cannot be larger than `area`.
+	area := (end - start) * min(height[start], height[end])
+
+	// Futhermore, we can completely ignore A.
+	//
+	// The maximum area will either be the current `area`, or some other two
+	// lines in the subset without A.
+	var subset []int
+	if height[start] < height[end] {
+		subset = height[start+1:]
+	} else {
+		subset = height[:end]
+	}
+
+	// We can now evaluate the problem space in the subset, and compute the
+	// maximum area.
+	return max(area, MaxArea(subset))
+}