The answer
No. If the inner loop bounds are constant (e.g., it always runs 10 times regardless of N), the overall complexity remains O(N).
Understanding the Time Complexity of O(n^2)
Do all nested loops result in O(N^2)?
More FAQs in Understanding the Time Complexity of O(n^2)
It is Quadratic Time. It means the execution time of an algorithm grows proportionally to the square of the input size.
Because the number of operations explodes exponentially. For an input of 100,000, an O(N^2) algorithm requires 10 billion operations, which takes too long for standard CPUs.
You can often reduce O(N^2) to O(N) by utilizing Hash Maps to look up data instantly, or to O(N log N) by sorting the data first.
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