Prevent unnecessary heap sort when buckets needs to be ordered by key in NumericTermsAggregation (#17252) (#17290)

(cherry picked from commit 3f793b6)

Signed-off-by: Rishabh Maurya <rishabhmaurya05@gmail.com>
Signed-off-by: github-actions[bot] <github-actions[bot]@users.noreply.github.com>
Signed-off-by: Ankit Jain <akjain@amazon.com>
Co-authored-by: github-actions[bot] <github-actions[bot]@users.noreply.github.com>
Co-authored-by: Ankit Jain <akjain@amazon.com>

Author: 3 people

CHANGELOG.md

@@ -5,6 +5,7 @@ The format is based on [Keep a Changelog](https://keepachangelog.com/en/1.0.0/),
 
 ## [Unreleased 2.x]
 ### Added
+- Improve performace of NumericTermAggregation by avoiding unnecessary sorting([#17252](https://github.com/opensearch-project/OpenSearch/pull/17252))
 
 ### Dependencies
 - Bump `dnsjava:dnsjava` from 3.6.2 to 3.6.3 ([#17231](https://github.com/opensearch-project/OpenSearch/pull/17231))

server/src/main/java/org/opensearch/search/aggregations/bucket/terms/NumericTermsAggregator.java

@@ -64,6 +64,7 @@
 import java.io.IOException;
 import java.math.BigInteger;
 import java.util.Arrays;
+import java.util.Iterator;
 import java.util.List;
 import java.util.Map;
 import java.util.function.BiConsumer;
@@ -202,9 +203,19 @@ private InternalAggregation[] buildAggregations(long[] owningBucketOrds) throws
                 // Get the top buckets
                 B[] bucketsForOrd = buildBuckets(ordered.size());
                 topBucketsPerOrd[ordIdx] = bucketsForOrd;
-                for (int b = ordered.size() - 1; b >= 0; --b) {
-                    topBucketsPerOrd[ordIdx][b] = ordered.pop();
-                    otherDocCounts[ordIdx] -= topBucketsPerOrd[ordIdx][b].getDocCount();
+                if (isKeyOrder(order)) {
+                    for (int b = ordered.size() - 1; b >= 0; --b) {
+                        topBucketsPerOrd[ordIdx][b] = ordered.pop();
+                        otherDocCounts[ordIdx] -= topBucketsPerOrd[ordIdx][b].getDocCount();
+                    }
+                } else {
+                    // sorted buckets not needed as they will be sorted by key in buildResult() which is different from
+                    // order in priority queue ordered
+                    Iterator<B> itr = ordered.iterator();
+                    for (int b = ordered.size() - 1; b >= 0; --b) {
+                        topBucketsPerOrd[ordIdx][b] = itr.next();
+                        otherDocCounts[ordIdx] -= topBucketsPerOrd[ordIdx][b].getDocCount();
+                    }
                 }
             }