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Morgan Stanley Solution
Technical Round
In this implementation, we use a dictionary self.data to store the key-value pairs. The put method adds a new key-value pair to the map, the get method returns the value associated with a given key, the remove method removes a key-value pair from the map, the keys method returns a list of all keys, the values method returns a list of all values, the items method returns a list of all key-value pairs as tuples, the is_empty method returns True if the map is empty and False otherwise, and the size method returns the number of key-value pairs in the map.
The time complexity of the put, get, and remove operations is O(1) on average, as they use the dictionary data structure, which has average constant-time performance for these operations. The time complexity of the keys, values, and items methods is O(n), where n is the number of key-value pairs in the map, as they need to return a list of all keys, values, or key-value pairs. The is_empty and size methods have a time complexity of O(1).
The idea behind this solution is to use two pointers, one starting from the beginning of the array (left) and one starting from the end (right). The left pointer moves right until it finds an odd number, and the right pointer moves left until it finds an even number. When both pointers have found an odd number and an even number, respectively, they are swapped. The loop continues until the pointers cross each other, meaning that all even numbers are now at the beginning of the array and all odd numbers are at the end.
The time complexity of this solution is O(n), where n is the number of elements in the array, as it visits each element at most twice, once with the left pointer and once with the right pointer.
This solution uses a dictionary (frequency) to keep track of the frequency of each number in the array. The dictionary maps each number to its frequency, so if the same number appears multiple times, its frequency is updated in the dictionary. After looping through all the numbers in the array, the function then finds the number with the highest frequency by looping through the items in the dictionary and keeping track of the number with the maximum frequency.
The time complexity of this solution is O(n), where n is the number of elements in the array, as it visits each element once to update its frequency in the dictionary, and then visits each item in the dictionary once to find the most frequent number. The space complexity is O(n), as the frequency dictionary takes up O(n) space.
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