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 R 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.R
Replay: real traced execution (multi-file project)
arr <- c(3, 5, 2, 5, 3, 8, 2)
count <- table(arr)
for (value in arr) {
  if (count[as.character(value)] == 1) {
    cat(value, "\n", sep = "")
    break
  }
}

  1. arr ← [3, 5, 2, 5, 3, 8, 2]

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    values this step[3, 5, 2, 5, 3, 8, 2]arr

  2. count ← {}

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)3for (value in arr) {
    values this step{}count

  3. count ← {3: 1}

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)3for (value in arr) {
    values this step{} → {3: 1}count3value

  4. count ← {3: 1, 5: 1}

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)3for (value in arr) {
    values this step{3: 1} → {3: 1, 5: 1}count5value

  5. count ← {3: 1, 5: 1, 2: 1}

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)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 <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)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 <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)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 <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)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 <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)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

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    values this step0i3value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    values this step1i5value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    values this step2i2value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    values this step3i5value2count[value]nofound{3: 2, 5: 2, 2: 2, 8: 1}count

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

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    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

    1arr <- c(3, 5, 2, 5, 3, 8, 2)2count <- table(arr)
    values this step5i8value1count[value]yesfound8result{3: 2, 5: 2, 2: 2, 8: 1}count

  16. stdout ← 8

    4if (count[as.character(value)] == 1) {5  cat(value, "\n", sep = "")6  break
    values this step8stdout8result

Complexity

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

Implementation notes

  • arr <- c(3, 5, 2, 5, 3, 8, 2) is the pinned R vector for this replay.
  • count <- table(arr) builds R's named frequency table. The trace shows it filling from {} through {3: 1}, {3: 1, 5: 1}, and the final {3: 2, 5: 2, 2: 2, 8: 1}.
  • The second pass uses for (value in arr), so it checks values in the original vector order instead of walking the table names.
  • table() stores its names as labels, so lookup uses count[as.character(value)]: numeric 3 is read from the table name "3".
  • Values 3, 5, and 2 are skipped because their counts are 2.
  • When the pass reaches 8, its count is 1; cat(value, "\n", sep = "") prints 8, and break stops the loop.

Replay steps

counts:  {} -> {3:1} -> {3:1,5:1} -> ... -> {3:2,5:2,2:2,8:1}
scan:    3(count 2), 5(count 2), 2(count 2), 5(count 2), 3(count 2)
result:  8(count 1) prints 8