Big O

What is the Big O notation for the following algorithm?

ctr = 0
while (ctr

Question

Big O

What is the Big O notation for the following algorithm?

ctr = 0
while (ctr < N) {
   myAge = ctr
   val = ctr + 1
   while (val < N) {
      if (ageOfPeople[val] < ageOfPeople[myAge]) 
         myAge = val
      ++val
   }  
   tempList = ageOfPeople[ctr]
   ageOfPeople [ctr] = ageOfPeople [myAge]
   ageOfPeople [myAge] = tempList
   ++ctr
}

BlackTom AI · Accepted Answer

Selection-sort behavior: for each position ctr the inner loop scans the remaining elements to find the minimum, so the algorithm does a quadratic number of comparisons.

1) Relevant knowledge points and concepts
- Big O measures asymptotic growth for large N, focusing on the highest-order term.
- Nested loops typically multiply their iteration counts.
- Constant-time operations (assignments, swaps, increments) do not change the highest-order term when dominated by a quadratic term.

2) Step-by-step reasoning
- Outer loop: ctr starts at 0 and increments until ctr < N, so it executes N times (ctr = 0..N-1).
- Inner loop: for a given ctr, val runs from ctr+1 up to N-1, so it executes (N - 1 - ctr) iterations.
- The dominant work is the comparison if (ageOfPeople[val] < ageOfPeople[myAge]) inside the inner loop.

3) Calculations with intermediate steps
- Total number of inner-loop comparisons = sum_{ctr=0}^{N-1} (N - 1 - ctr)
- Let k = N - 1 - ctr; as ctr runs 0..N-1, k runs  N-1 down to 0, so sum = 0 + 1 + 2 + ... + (N-1) = (N-1)N/2
- Expand: (N^2 - N)/2, which is Θ(N^2) comparisons.
- Outer-loop also does one swap and a few constant-time operations per iteration: O(N) extra work, but O(N) is dominated by O(N^2).

4) Why the correct answer is correct
- The (N^2 - N)/2 term dominates growth for large N, so the algorithm's time complexity is O(N^2). Constant and linear terms are negligible asymptotically.

5) Why the other multiple-choice options are incorrect
- O(1): incorrect because the runtime increases with N; the algorithm is not constant-time.
- O(N): incorrect because work grows faster than linear (sums to about N^2/2).
- O(log N): incorrect because there is no logarithmic reduction; nested scanning yields polynomial growth, not logarithmic.

6) Conclusion
- The time complexity is O(N^2). This matches the provided correct answer: O(N^2).

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