Python Question / Quiz; What is the output of the following Python code, and why? Comment your answers below!

@PythonPr Answer: (C) 0 0 The range(1) function generates a sequence of numbers starting from 0 up to (but not including) 1, which means it only contains the number 0. The outer loop for i in range(1) runs once, with i assigned the value 0. Inside the outer loop, the inner

@PythonPr Answer: C) 0 0 range(1) starts at 0 and stops before 1, so the only value is 0. That means the loops run once with i = 0 and j = 0.

@PythonPr Answer: C) 0 0 ✅ Why: range(1) runs only once with value 0. Outer loop: i = 0 Inner loop: j = 0 print(i, j) prints 0 0 once.

@PythonPr Answer: C. 0 0 Solution: Just need to be clear on `range` usage with a single argument. range(a) is equivalent to range(0, a) ie, starting from 0, stopping at a (a won't be in the range). So, range(1) is range(0, 1) starting from 0, stopping before 1. So just 0 +

@PythonPr Option C. 0 0 Fundamental syntax. But in our HFT architecture, we avoid nested loops for execution. Vectorized calculations with NumPy are the only way to beat the latency. ⚡️

@PythonPr Option C

@PythonPr Answer 0, 0 Because the range of iteration starts from 0 till less than 1 for both nested loops so it will print 0, 0.

@PythonPr the answer is C

@PythonPr The output is 0,0: in inner and outer range(1), starts at zero and stops before 1.

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@PythonPr Option -C Reason- because indexing is starting from 0 and it excludes the last given value

@PythonPr C. With range(1) the only thing that can print is zero.

@PythonPr C. 0, 0

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@PythonPr C . 0 0 Range passed is 1 which means it will print number from 0 to 1 but excluding 1 .

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@PythonPr Answer 0, 0

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@PythonPr B)1 1 Good Health.

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