answer:Here is a Python function that uses a dictionary to store the indices of the numbers it has seen so far, along with their complements: ```python def find_zero_sum_pair(nums): num_dict = {} for i, num in enumerate(nums): complement = -num if complement in num_dict: return [num_dict[complement], i] num_dict[num] = i return None # Example usage: numbers = [1, 2, 3, -2, 5] result = find_zero_sum_pair(numbers) if result: print(Indices of zero sum pair:, result) else: print(No zero sum pair found.) ``` This function has a time complexity of O(n), where n is the number of elements in the input list, because it makes a single pass through the list. The space complexity is also O(n) because in the worst case, it might need to store every element in the dictionary.

answer:Here is a Python function that checks whether a number is prime: ```python def is_prime(n): if n < 2: return False for i in range(2, int(n ** 0.5) + 1): if n % i == 0: return False return True # Example usage: number = 23 if is_prime(number): print(number, is a prime number.) else: print(number, is not a prime number.) ``` This function works by checking whether the input number `n` has any divisors other than 1 and itself. If it finds any such divisor, it immediately returns `False`. If it checks all possible divisors without finding any, it returns `True`. The reason it only checks up to the square root of `n` is that a larger factor of the number would be a multiple of a smaller factor that has already been checked. This optimization greatly reduces the function's running time for large inputs. Note that this function has a time complexity of O(sqrt(n)), where n is the input number.

answer:To check for multiple factors of a given integer, you can iterate over all numbers from 1 to the square root of the integer and check if the integer is divisible by the current number. If it is, then both the current number and the result of the division are factors. Here is a Python function that finds all factors of a given integer: ```python def find_factors(n): factors = set() for i in range(1, int(n ** 0.5) + 1): if n % i == 0: factors.add(i) factors.add(n // i) return sorted(list(factors)) # Example usage: number = 24 factors = find_factors(number) print(Factors of, number, are:, factors) ``` This function returns a sorted list of all factors of the input number. The `set` data structure is used to automatically eliminate any duplicate factors. Note that this function has a time complexity of O(sqrt(n)), where n is the input number. Alternatively, you can also use the following function to find all factors of a number, including their multiplicities (i.e., how many times each factor appears in the prime factorization of the number): ```python def find_factors_with_multiplicity(n): factors = {} i = 2 while i * i <= n: if n % i: i += 1 else: n //= i factors[i] = factors.get(i, 0) + 1 if n > 1: factors[n] = factors.get(n, 0) + 1 return factors # Example usage: number = 24 factors = find_factors_with_multiplicity(number) print(Factors of, number, with multiplicity are:, factors) ``` This function returns a dictionary where the keys are the prime factors and the values are their multiplicities. The time complexity is still O(sqrt(n)).

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