Reuse start position in binarySearchKey (#49641)

* Reuse start position in binarySearchKey

* Kick out early if end < position

Author: amcasey

src/services/utilities.ts

@@ -1147,13 +1147,20 @@ namespace ts {
                 // position and whose end is greater than the position.
 
 
+                // There are more sophisticated end tests later, but this one is very fast
+                // and allows us to skip a bunch of work
+                const end = children[middle].getEnd();
+                if (end < position) {
+                    return Comparison.LessThan;
+                }
+
                 const start = allowPositionInLeadingTrivia ? children[middle].getFullStart() : children[middle].getStart(sourceFile, /*includeJsDoc*/ true);
                 if (start > position) {
                     return Comparison.GreaterThan;
                 }
 
                 // first element whose start position is before the input and whose end position is after or equal to the input
-                if (nodeContainsPosition(children[middle])) {
+                if (nodeContainsPosition(children[middle], start, end)) {
                     if (children[middle - 1]) {
                         // we want the _first_ element that contains the position, so left-recur if the prior node also contains the position
                         if (nodeContainsPosition(children[middle - 1])) {
@@ -1181,13 +1188,16 @@ namespace ts {
             return current;
         }
 
-        function nodeContainsPosition(node: Node) {
-            const start = allowPositionInLeadingTrivia ? node.getFullStart() : node.getStart(sourceFile, /*includeJsDoc*/ true);
+        function nodeContainsPosition(node: Node, start?: number, end?: number) {
+            end ??= node.getEnd();
+            if (end < position) {
+                return false;
+            }
+            start ??= allowPositionInLeadingTrivia ? node.getFullStart() : node.getStart(sourceFile, /*includeJsDoc*/ true);
             if (start > position) {
                 // If this child begins after position, then all subsequent children will as well.
                 return false;
             }
-            const end = node.getEnd();
             if (position < end || (position === end && (node.kind === SyntaxKind.EndOfFileToken || includeEndPosition))) {
                 return true;
             }