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 Swift 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.swift
Replay: real traced execution (multi-file project)
let arr = [3, 5, 2, 5, 3, 8, 2]
var count: [Int: Int] = [:]
for value in arr {
    count[value, default: 0] += 1
}
for value in arr {
    if count[value] == 1 {
        print(value)
        break
    }
}

  1. arr ← [3, 5, 2, 5, 3, 8, 2]

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]
    values this step[3, 5, 2, 5, 3, 8, 2]arr

  2. count ← {}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for value in arr {
    values this step{}count

  3. count ← {3: 1}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for value in arr {
    values this step{} → {3: 1}count3value

  4. count ← {3: 1, 5: 1}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for value in arr {
    values this step{3: 1} → {3: 1, 5: 1}count5value

  5. count ← {3: 1, 5: 1, 2: 1}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for 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}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for 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}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for 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}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for 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}

    1let arr = [3, 5, 2, 5, 3, 8, 2]2var count: [Int: Int] = [:]3for 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

    6for value in arr {7    if count[value] == 1 {8        print(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

    6for value in arr {7    if count[value] == 1 {8        print(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

    6for value in arr {7    if count[value] == 1 {8        print(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

    6for value in arr {7    if count[value] == 1 {8        print(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

    6for value in arr {7    if count[value] == 1 {8        print(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

    6for value in arr {7    if count[value] == 1 {8        print(value)
    values this step5i8value1count[value]yesfound8result{3: 2, 5: 2, 2: 2, 8: 1}count

  16. stdout ← 8

    7if count[value] == 1 {8    print(value)9    break
    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/Character iteration: let arr = [3, 5, 2, 5, 3, 8, 2].
  • var count: [Int: Int] = [:] is the mutable Swift dictionary used as the frequency table.
  • The first pass iterates for value in arr and updates counts with count[value, default: 0] += 1, so missing keys start from 0 before the increment.
  • The trace records logical dictionary states, ending the count pass at {3: 2, 5: 2, 2: 2, 8: 1}; it does not expose bucket order or collision behavior.
  • The second pass scans arr again in input order and checks count[value] == 1 through Swift dictionary subscript lookup.
  • Values 3, 5, 2, 5, and 3 are traced as found no with frequency 2; value 8 at index 5 has frequency 1 and becomes the result.
  • print(value) writes the first non-repeating integer directly, so the exact stdout is 8.