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 Kotlin 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.kt
Replay: real traced execution (multi-file project)
fun main() {
    val arr = listOf(3, 5, 2, 5, 3, 8, 2)
    val count = mutableMapOf<Int, Int>()
    for (value in arr) {
        count[value] = count.getOrDefault(value, 0) + 1
    }
    for (value in arr) {
        if (count[value] == 1) {
            println(value)
            break
        }
    }
}

  1. arr ← [3, 5, 2, 5, 3, 8, 2]

    1fun main() {2    val arr = listOf(3, 5, 2, 5, 3, 8, 2)
    values this step[3, 5, 2, 5, 3, 8, 2]arr

  2. count ← {}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in arr) {
    values this step{}count

  3. count ← {3: 1}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in arr) {
    values this step{} → {3: 1}count3value

  4. count ← {3: 1, 5: 1}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in arr) {
    values this step{3: 1} → {3: 1, 5: 1}count5value

  5. count ← {3: 1, 5: 1, 2: 1}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in arr) {
    values this step{3: 1, 5: 1} → {3: 1, 5: 1, 2: 1}count2value

  6. count ← {3: 1, 5: 2, 2: 1}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in 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}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in 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}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in 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}

    2val arr = listOf(3, 5, 2, 5, 3, 8, 2)3val count = mutableMapOf<Int, Int>()4for (value in 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

    7for (value in arr) {8    if (count[value] == 1) {9        println(value)
    values this step0i3value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    7for (value in arr) {8    if (count[value] == 1) {9        println(value)
    values this step1i5value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    7for (value in arr) {8    if (count[value] == 1) {9        println(value)
    values this step2i2value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    7for (value in arr) {8    if (count[value] == 1) {9        println(value)
    values this step3i5value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    7for (value in arr) {8    if (count[value] == 1) {9        println(value)
    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

    7for (value in arr) {8    if (count[value] == 1) {9        println(value)
    values this step5i8value1count[value]yesfound8result{3: 2, 5: 2, 2: 2, 8: 1}count

  16. stdout ← 8

    8if (count[value] == 1) {9    println(value)10    break
    values this step8stdout8result

Complexity

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

Implementation notes

  • Kotlin stores the input as val arr = listOf(3, 5, 2, 5, 3, 8, 2), a read-only List<Int> reference; the list is scanned twice and not mutated.
  • The frequency table is val count = mutableMapOf<Int, Int>(). The binding is stable, but the map contents mutate during the first pass.
  • Count updates use count[value] = count.getOrDefault(value, 0) + 1, so a missing Int key reads as zero before the first write.
  • The second pass uses the original arr order, not map iteration order, so the first non-repeating result does not depend on hash bucket ordering.
  • count[value] == 1 compares the nullable map lookup result with 1; for these values every key was inserted by the first pass before the lookup.
  • The trace records count states from {} through {3: 2, 5: 2, 2: 2, 8: 1}, then scans values 3, 5, 2, 5, 3 as repeated before finding 8.
  • On the first frequency-one value, println(value) prints 8 and break stops the scan.