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Consider an array which stores a maximum of 100 elements. For case 1, the user inserts 1 element into an array. For case 2, the user inserts 100 elements into an array. What is the time complexity for both the cases?
- a)O(1), O(n)
- b)O(n), O(n)
- c)O(n), O(1)
- d)O(1), O(1)
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Consider an array which stores a maximum of 100 elements. For case 1, ...
To store 1 element, 100 memory spaces are required.
To store 2 elements, 100 memory spaces are required.
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To store n element, 100 memory spaces are required.
Most Upvoted Answer
Consider an array which stores a maximum of 100 elements. For case 1, ...
Time Complexity for Case 1:
- In Case 1, the user inserts only 1 element into the array.
- The time complexity for inserting a single element into an array is constant, denoted by O(1).
- This is because regardless of the size of the array, inserting a single element requires a fixed number of operations, such as finding the position to insert and updating the array.
- Therefore, the time complexity for Case 1 is O(1).
Time Complexity for Case 2:
- In Case 2, the user inserts 100 elements into the array.
- The time complexity for inserting elements into an array depends on the number of elements being inserted.
- As the number of elements being inserted is constant (100), the time complexity can be considered as O(n), where n is the number of elements being inserted.
- This is because for each element being inserted, the algorithm needs to find the position and update the array, which takes a constant amount of time.
- Since the number of elements being inserted is constant, the time complexity does not depend on the size of the array and remains O(1).
- Therefore, the time complexity for Case 2 is also O(1).
Conclusion:
- In both Case 1 and Case 2, the time complexity is O(1).
- This means that regardless of the number of elements being inserted, the time taken to insert the elements into the array remains constant.
- The time complexity is not dependent on the size of the array, but rather on the number of elements being inserted.
- Therefore, the correct answer is option 'D' - O(1), O(1).
- In Case 1, the user inserts only 1 element into the array.
- The time complexity for inserting a single element into an array is constant, denoted by O(1).
- This is because regardless of the size of the array, inserting a single element requires a fixed number of operations, such as finding the position to insert and updating the array.
- Therefore, the time complexity for Case 1 is O(1).
Time Complexity for Case 2:
- In Case 2, the user inserts 100 elements into the array.
- The time complexity for inserting elements into an array depends on the number of elements being inserted.
- As the number of elements being inserted is constant (100), the time complexity can be considered as O(n), where n is the number of elements being inserted.
- This is because for each element being inserted, the algorithm needs to find the position and update the array, which takes a constant amount of time.
- Since the number of elements being inserted is constant, the time complexity does not depend on the size of the array and remains O(1).
- Therefore, the time complexity for Case 2 is also O(1).
Conclusion:
- In both Case 1 and Case 2, the time complexity is O(1).
- This means that regardless of the number of elements being inserted, the time taken to insert the elements into the array remains constant.
- The time complexity is not dependent on the size of the array, but rather on the number of elements being inserted.
- Therefore, the correct answer is option 'D' - O(1), O(1).
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Consider an array which stores a maximum of 100 elements. For case 1, the user inserts 1 element into an array. For case 2, the user inserts 100 elements into an array. What is the time complexity for both the cases?a)O(1), O(n)b)O(n), O(n)c)O(n), O(1)d)O(1), O(1)Correct answer is option 'D'. Can you explain this answer?
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