Optimize Color.Equals performance (O(n²) → O(n))

color.go

@@ -526,27 +526,24 @@ func (c *Color) Equals(c2 *Color) bool {
 	if c == nil || c2 == nil {
 		return false
 	}
+
 	if len(c.params) != len(c2.params) {
 		return false
 	}
 
+	counts := make(map[Attribute]int, len(c.params))
 	for _, attr := range c.params {
-		if !c2.attrExists(attr) {
-			return false
-		}
+		counts[attr]++
 	}
 
-	return true
-}
-
-func (c *Color) attrExists(a Attribute) bool {
-	for _, attr := range c.params {
-		if attr == a {
-			return true
+	for _, attr := range c2.params {
+		if counts[attr] == 0 {
+			return false
 		}
+		counts[attr]--
 	}
 
-	return false
+	return true
 }
 
 func boolPtr(v bool) *bool {

color_test.go

@@ -105,6 +105,20 @@ func TestColorEquals(t *testing.T) {
 	}
 }
 
+func TestColorEquals_DuplicateAttributes(t *testing.T) {
+	ordered := New(FgRed, Bold).Add(FgRed)
+	reordered := New(Bold, FgRed).Add(FgRed)
+	differentCounts := New(FgRed, Bold).Add(Bold)
+
+	if !ordered.Equals(reordered) {
+		t.Error("Colors with the same attributes in different orders are not equal")
+	}
+
+	if ordered.Equals(differentCounts) {
+		t.Error("Colors with different duplicate attribute counts are equal")
+	}
+}
+
 func TestNoColor(t *testing.T) {
 	rb := new(bytes.Buffer)
 	Output = rb