minor reset

pkg/statistics/builder.go

@@ -326,18 +326,23 @@ func BuildHistAndTopN(
 			continue
 		}
 		// case 2, meet a different value: counting for the "current" is complete
-		// case 2-1, now topn is empty: append the "current" count directly
+		// case 2-1, do not add a count of 1
+		if curCnt == 1 {
+			cur, curCnt = sampleBytes, 1
+			continue
+		}
+		// case 2-2, now topn is empty: append the "current" count directly if the value is greater than 1
 		if len(topNList) == 0 {
 			topNList = append(topNList, TopNMeta{Encoded: cur, Count: uint64(curCnt)})
 			cur, curCnt = sampleBytes, 1
 			continue
 		}
-		// case 2-2, now topn is full, and the "current" count is less than the least count in the topn: no need to insert the "current"
+		// case 2-3, now topn is full, and the "current" count is less than the least count in the topn: no need to insert the "current"
 		if len(topNList) >= numTopN && uint64(curCnt) <= topNList[len(topNList)-1].Count {
 			cur, curCnt = sampleBytes, 1
 			continue
 		}
-		// case 2-3, now topn is not full, or the "current" count is larger than the least count in the topn: need to find a slot to insert the "current"
+		// case 2-4, now topn is not full, or the "current" count is larger than the least count in the topn: need to find a slot to insert the "current"
 		j := len(topNList)
 		for ; j > 0; j-- {
 			if uint64(curCnt) < topNList[j-1].Count {
@@ -454,8 +459,7 @@ func BuildHistAndTopN(
 //
 //	We assume that the ones not in the top-n list's selectivity is 1/remained_ndv which is the internal implementation of EqualRowCount
 func pruneTopNItem(topns []TopNMeta, ndv, nullCount, sampleRows, totalRows int64) []TopNMeta {
-	// If the sampleRows holds all rows, or NDV of samples equals to actual NDV, we just return the TopN directly.
-	if sampleRows == totalRows || totalRows <= 1 || int64(len(topns)) >= ndv || len(topns) == 0 {
+	if totalRows <= 1 || len(topns) == 0 {
 		return topns
 	}
 	// Sum the occurrence except the least common one from the top-n list. To check whether the lest common one is worth