Most Beautiful Item For Each Query - Leetcode Solution

The problem "Most Beautiful Item For Each Query" on LeetCode asks us to find the most beautiful item for each query given a list of items and a list of "beauty" values for each item.

We can solve this problem using a hash table to keep track of the most beautiful item for each query. We can loop through each item in the list, and for each item, we can check if the beauty value is greater than the beauty value of the current most beautiful item for that query.

If it is, we update the hash table with the new most beautiful item for that query. After we loop through all the items, we return the values in the hash table.

Here is the detailed solution in Python:

def mostBeautiful(self, nums: List[int], queries: List[List[int]]) -> List[int]:
    # Create a hash table to store the most beautiful item for each query
    most_beautiful = {}
    
    # Loop through each query
    for q in queries:
        # Set the initial most beautiful item for the query to -1
        most = -1
        
        # Loop through each item in the list
        for i, n in enumerate(nums):
            # Check if the item is within the range of the query
            if i >= q[0] and i <= q[1]:
                # Check if the beauty value is greater than the current most beautiful item
                if n > most:
                    most = n
        
        # Update the hash table with the most beautiful item for the query
        most_beautiful[tuple(q)] = most
    
    # Return the values in the hash table
    return [most_beautiful[tuple(q)] for q in queries]

The time complexity of this solution is O(nq), where n is the number of items in the list and q is the number of queries. This is because we loop through each item for each query. The space complexity is O(q) because we store the most beautiful item for each query in the hash table.

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