Hash Tables

First Non-Repeating Value

Find the first input value whose final frequency is one.

Algorithm

Canonical input [3, 5, 2, 5, 3, 8, 2] prints 8. The replay uses the same input in every language, so this Scala DSA implementation can be compared directly with the rest of the DSA track.

two-pass lookup The first pass builds a frequency table. The second pass keeps the original order and stops at the first value with frequency one.

Basic Implementation

basic.scala
Replay: real traced execution (multi-file project)
import scala.collection.mutable.HashMap
object Main {
  def main(args: Array[String]): Unit = {
    val arr = List(3, 5, 2, 5, 3, 8, 2)
    val count = HashMap.empty[Int, Int]
    for (value <- arr) {
      count(value) = count.getOrElse(value, 0) + 1
    }
    for (value <- arr) {
      if (count(value) == 1) {
        println(value)
        return
      }
    }
  }
}

  1. arr ← [3, 5, 2, 5, 3, 8, 2]

    1import scala.collection.mutable.HashMap2object Main {
    values this step[3, 5, 2, 5, 3, 8, 2]arr

  2. count ← {}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{}count

  3. count ← {3: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{} → {3: 1}count3value

  4. count ← {3: 1, 5: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{3: 1} → {3: 1, 5: 1}count5value

  5. count ← {3: 1, 5: 1, 2: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{3: 1, 5: 1} → {3: 1, 5: 1, 2: 1}count2value

  6. count ← {3: 1, 5: 2, 2: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{3: 1, 5: 1, 2: 1} → {3: 1, 5: 2, 2: 1}count5value

  7. count ← {3: 2, 5: 2, 2: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{3: 1, 5: 2, 2: 1} → {3: 2, 5: 2, 2: 1}count3value

  8. count ← {3: 2, 5: 2, 2: 1, 8: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{3: 2, 5: 2, 2: 1} → {3: 2, 5: 2, 2: 1, 8: 1}count8value

  9. count ← {3: 2, 5: 2, 2: 2, 8: 1}

    4val arr = List(3, 5, 2, 5, 3, 8, 2)5val count = HashMap.empty[Int, Int]6for (value <- arr) {
    values this step{3: 2, 5: 2, 2: 1, 8: 1} → {3: 2, 5: 2, 2: 2, 8: 1}count2value

  10. i ← 0, value ← 3, count[value] ← 2, found ← no

    1import scala.collection.mutable.HashMap2object Main {
    values this step0i3value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

  11. i ← 1, value ← 5, count[value] ← 2, found ← no

    1import scala.collection.mutable.HashMap2object Main {
    values this step1i5value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

  12. i ← 2, value ← 2, count[value] ← 2, found ← no

    1import scala.collection.mutable.HashMap2object Main {
    values this step2i2value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

  13. i ← 3, value ← 5, count[value] ← 2, found ← no

    1import scala.collection.mutable.HashMap2object Main {
    values this step3i5value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

  14. i ← 4, value ← 3, count[value] ← 2, found ← no

    1import scala.collection.mutable.HashMap2object Main {
    values this step4i3value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

  15. i ← 5, value ← 8, count[value] ← 1, found ← yes, result ← 8

    1import scala.collection.mutable.HashMap2object Main {
    values this step5i8value1count[value]yesfound8result{3: 2, 5: 2, 2: 2, 8: 1}count

  16. stdout ← 8

    10if (count(value) == 1) {11  println(value)12  return
    values this step8stdout8result

Complexity

  • Time: O(n) average
  • Space: O(k) for k distinct values

Implementation notes

  • This checked source uses integer input, not string or character iteration: val arr = List(3, 5, 2, 5, 3, 8, 2).
  • val count = HashMap.empty[Int, Int] binds a mutable Scala HashMap; the binding is val, but the table contents are updated in place.
  • The first pass runs for (value <- arr) in list order and writes count(value) = count.getOrElse(value, 0) + 1.
  • getOrElse(value, 0) supplies the default count for a missing key before the increment.
  • The trace records logical table states only; it does not expose bucket order, resizing, or collision behavior.
  • After counting, the table is {3: 2, 5: 2, 2: 2, 8: 1}.
  • The second pass scans the original list again. Values 3, 5, 2, 5, and 3 have frequency 2; value 8 has frequency 1.
  • On the first unique value, the code calls println(value) and return, so the exact output is 8. If no value matched, this source would fall through without printing a sentinel.