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certainly! the problem "sort array by increasing frequency" (leetcode 1636) requires us to sort an array based on the frequency of its elements. if two elements have the same frequency, they should be sorted by their value in ascending order.
problem statement
given an array of integers, you need to sort the array based on the frequency of its elements. if two elements have the same frequency, the smaller number should come first.
steps to solve the problem
1. **count frequencies**: use a dictionary to count the occurrences of each element in the array.
2. **sort the elements**: use the frequency counts to sort the elements. we can use python's built-in `sorted()` function, providing a custom sorting key.
3. **construct the result**: create the output array by repeating each element based on its frequency.
code example
here's a sample implementation:
explanation of the code
1. **importing counter**: we import the `counter` class from the `collections` module, which helps us to count the frequency of each element in the list easily.
2. **counting frequencies**: we create a `counter` object called `frequency`, which counts how many times each number appears in the list `nums`.
3. **sorting**:
we use the `sorted()` function to sort the elements. the `key` parameter is a lambda function that sorts primarily by frequency (`frequency[x]`) and secondarily by the element value itself (`x`).
this ensures that if two numbers have the same frequency, the smaller number will come first.
4. **returning the result**: finally, the sorted list `sorted_nums` is returned.
complexity analysis
**time complexity**: the time complexity is \(o(n \log n)\), where \(n\) is the number of elements in the input array. the sorting step is the most time-consuming operation.
**space complexity**: the space complexity is \(o(n)\) due to the storage of frequency counts.
conclusion
this problem is a good exercise in using dictionaries for counti ...
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