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What is the time complexity of the following code snippet?
int count = 0;
for (int i = 1; i <= n; i *= 2) {
for (int j = 0; j < n; j++) {
count++;
}
}
cout << count;
int count = 0;
for (int i = 1; i <= n; i *= 2) {
for (int j = 0; j < n; j++) {
count++;
}
}
cout << count;
- a)O(n)
- b)O(log n)
- c)O(n log n)
- d)O(n^2)
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
What is the time complexity of the following code snippet?int count = ...
The outer loop runs log2(n) times, and the inner loop runs n times. Hence, the time complexity is O(n log n).
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Community Answer
What is the time complexity of the following code snippet?int count = ...
Understanding the Code Snippet
The provided code snippet consists of nested loops, and analyzing their behavior will help determine the time complexity.
Outer Loop Analysis
- The outer loop iterates with `i` starting from 1 and doubling each time (`i *= 2`).
- This means `i` takes values: 1, 2, 4, 8, ..., up to `n`.
- The number of iterations of the outer loop can be calculated as follows:
- Let `k` be the number of iterations; then, `2^k ≤ n`.
- Taking logarithm base 2 of both sides, we find `k ≤ log_2(n)`.
- Thus, the outer loop runs `O(log n)` times.
Inner Loop Analysis
- The inner loop runs from `j = 0` to `j < n`,="" which="" means="" it="" iterates="" `n`="" times="" for="" each="" iteration="" of="" the="" outer="" />
Combining the Two Loops
- For each of the `O(log n)` iterations of the outer loop, the inner loop executes `n` iterations.
- Therefore, the total number of iterations across both loops is:
- Total iterations = (Number of outer loop iterations) * (Number of inner loop iterations)
- This results in: `O(log n) * O(n) = O(n log n)`.
Final Conclusion
The correct time complexity of the given code snippet is:
- **O(n log n)**
Thus, the answer is option 'C'.
The provided code snippet consists of nested loops, and analyzing their behavior will help determine the time complexity.
Outer Loop Analysis
- The outer loop iterates with `i` starting from 1 and doubling each time (`i *= 2`).
- This means `i` takes values: 1, 2, 4, 8, ..., up to `n`.
- The number of iterations of the outer loop can be calculated as follows:
- Let `k` be the number of iterations; then, `2^k ≤ n`.
- Taking logarithm base 2 of both sides, we find `k ≤ log_2(n)`.
- Thus, the outer loop runs `O(log n)` times.
Inner Loop Analysis
- The inner loop runs from `j = 0` to `j < n`,="" which="" means="" it="" iterates="" `n`="" times="" for="" each="" iteration="" of="" the="" outer="" />
Combining the Two Loops
- For each of the `O(log n)` iterations of the outer loop, the inner loop executes `n` iterations.
- Therefore, the total number of iterations across both loops is:
- Total iterations = (Number of outer loop iterations) * (Number of inner loop iterations)
- This results in: `O(log n) * O(n) = O(n log n)`.
Final Conclusion
The correct time complexity of the given code snippet is:
- **O(n log n)**
Thus, the answer is option 'C'.
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