Determine the frequency counts for all statement in the following algorithm Segment
Determine the frequency counts for all statement in the following algorithm Segment.
I=1;
While(I<=n)
{
X=X+I;
I=I+1;
}1 answer
Total frequency count = \(3n + 2\), so the time complexity is \(O(n)\).
| Statement | Frequency count |
|---|---|
I = 1; | 1 |
while (I <= n) | \(n + 1\) |
X = X + I; | \(n\) |
I = I + 1; | \(n\) |
| Total | \(3n + 2\) |
Why:
- The initialisation runs once.
- The loop body runs for \(I = 1, 2, \dots, n\), i.e. \(n\) times, so each statement inside runs \(n\) times.
- The condition is tested once more than the body runs: \(n\) times it is true and the \((n+1)\)th time (when \(I = n + 1\)) it is false and the loop exits.
\[1 + (n + 1) + n + n = 3n + 2 = O(n)\]
Doubtly AI
· 23 Oct 2026
AI-verified · Oct 2026
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