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 Python 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.py
Replay: real traced execution (multi-file project)
arr = [3, 5, 2, 5, 3, 8, 2]
count = {}
for value in arr:
    count[value] = count.get(value, 0) + 1
for value in arr:
    if count[value] == 1:
        print(value)
        break

  1. arr ← [3, 5, 2, 5, 3, 8, 2]

    1arr = [3, 5, 2, 5, 3, 8, 2]2count = {}
    values this step[3, 5, 2, 5, 3, 8, 2]arr

  2. count ← {}

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

  3. count ← {3: 1}

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

  4. count ← {3: 1, 5: 1}

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

  5. count ← {3: 1, 5: 1, 2: 1}

    1arr = [3, 5, 2, 5, 3, 8, 2]2count = {}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}

    1arr = [3, 5, 2, 5, 3, 8, 2]2count = {}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}

    1arr = [3, 5, 2, 5, 3, 8, 2]2count = {}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}

    1arr = [3, 5, 2, 5, 3, 8, 2]2count = {}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}

    1arr = [3, 5, 2, 5, 3, 8, 2]2count = {}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

    5for value in arr:6    if count[value] == 1:7        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

    5for value in arr:6    if count[value] == 1:7        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

    5for value in arr:6    if count[value] == 1:7        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

    5for value in arr:6    if count[value] == 1:7        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

    5for value in arr:6    if count[value] == 1:7        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

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

  16. stdout ← 8

    6if count[value] == 1:7    print(value)8    break
    values this step8stdout8result

Complexity

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

Implementation notes

  • Python uses a plain dict named count, with immutable int values as keys and integer frequencies as values. count.get(value, 0) + 1 performs a hash lookup with a default, then the assignment inserts or updates that entry.
  • The implementation is intentionally two pass: the first loop builds the full count table, and the second loop walks arr again so original list order, not dict insertion order, decides the first non-repeating value.
  • Hash collisions and table resizing are not visible for this small fixture; Python manages count while it is referenced and can reclaim it once unreferenced. The replay shows each count update, then scans until count[8] == 1 and prints 8.