Problem Description

The problem at hand is a classic algorithmic challenge where we are given an array of integers, which we will refer to as nums. The goal is to find, for each element in the array, how many numbers to its right are smaller than it. To clarify, for every index i in the array, we want to calculate the count of elements that are located at any index greater than i and have a value less than nums[i]. The output should be an array of integers where each element counts[i] represents the number of smaller elements to the right of nums[i].

For example, if the input array nums is [5, 2, 6, 1], the output array should be [2, 1, 1, 0]. For the first element 5, there are two elements (2 and 1) to the right that are smaller. For the second element 2, there is only one element (1) to the right that is smaller, and so on.

Intuition

The straightforward brute force solution would be to go through each element in the array and count the number of smaller elements to its right with nested loops. However, this method would result in a time complexity of O(n^2), which is not efficient for large arrays.

To optimize this, we can use a data structure called a Binary Indexed Tree (BIT), also known as a Fenwick Tree, or alternatively, a Segment Tree. The Binary Indexed Tree is a data structure that allows us to update values and calculate prefix sums in a logarithmic time complexity, which is much faster than the brute force approach.

The intuition behind using BIT in this solution is as follows:

1. First, we need to discretize the values in the array to make sure they can fit into our Binary Indexed Tree without taking too much space. Discretization involves mapping each value to its corresponding rank in the sorted list of unique values.

2. We process the elements of the original array in reverse order. This means we start with the rightmost element and move to the left. This allows us to use the tree to keep track of the number of elements that have already been seen and are less than the current element.

3. For each element, we find its index in the discretized array, which represents its value in the BIT.

4. Before we add the current element to the tree, we query the tree to find out how many elements less than the current one have already been processed (those to the right in the original array). This query gives us the count of smaller numbers to the right.

5. Update the Binary Indexed Tree to indicate that we have now seen an instance of the current element.

6. We store the query result in the result list, which we then reverse at the end since we processed the elements from right to left.

Using the BIT, both the query and update operations are performed in O(log n) time, which significantly improves the efficiency for large datasets.

Learn more about Segment Tree, Binary Search, Divide and Conquer and Merge Sort patterns.

Solution Approach

The solution relies on two key components: the discretization of array values and a Binary Indexed Tree (BIT). Let's walk through the implementation step by step, highlighting how the algorithms, data structures, and patterns are used:

1. Discretization of Values: Before we can work with the BIT, we need to make sure array indices won't be too large. This is achieved by discretizing the array values. In Python, this can be done by sorting the unique values and then creating a mapping from the original values to their indices in the sorted array, starting with 1 (since BIT works with 1-based indexing). This process can be seen in the code with:

   alls = sorted(set(nums))
   m = {v: i for i, v in enumerate(alls, 1)}

   Here, alls is the array of unique, sorted values and m maps each original value to its rank in the sorted array.

2. Implementation of BIT: A BIT is a data structure that supports updating individual elements and calculating prefix sums efficiently. The Python class BinaryIndexedTree includes methods for updating the tree (update) and querying prefix sums (query), utilizing a helper method lowbit to isolate the rightmost bit of a binary representation.

   class BinaryIndexedTree:
       def __init__(self, n):
           self.n = n
           self.c = [0] * (n + 1)
   
       @staticmethod
       def lowbit(x):
           return x & -x
   
       def update(self, x, delta):
           while x <= self.n:
               self.c[x] += delta
               x += BinaryIndexedTree.lowbit(x)
   
       def query(self, x):
           s = 0
           while x > 0:
               s += self.c[x]
               x -= BinaryIndexedTree.lowbit(x)
           return s

   • self.c is the internal array of the BIT, initialized with zeros.
   • lowbit method isolates the lowest one bit of x, which is essential for navigating the tree structure.
   • update method adds delta to the value at index x, affecting all subsequent sums that include this element.
   • query method sums all the elements up to index x (exclusive), allowing us to get the count of smaller elements seen so far.

3. Processing Elements in Reverse: We initialize a BIT with the size equal to the number of unique elements in nums. Then, we traverse the nums array in reverse, translating each element to its discretized form and using the update and query operations of BIT to build the counts array.

   tree = BinaryIndexedTree(len(m))
   ans = []
   for v in nums[::-1]:
       x = m[v]
       tree.update(x, 1)
       ans.append(tree.query(x - 1))
   return ans[::-1]

   • For each element v encountered, we look up its discretized index x.
   • Before updating x in BIT, we perform a query to get the count of elements smaller than x.
   • The result of each query is added to ans.
   • Since we've traversed the array in reverse, we need to reverse ans before returning it to get the correct order.

This implementation ensures that each update and query call is handled in O(log n) time, leading to an overall time complexity of O(n log n) due to processing each of the n elements once. It's a significant improvement over the brute force method and is well-suited for handling large input arrays.

Example Walkthrough

Let's consider a small example with the input array nums = [3, 4, 2, 7, 5]. Using the solution approach outlined, we want to find the number of elements to the right that are smaller than each element in nums.

1. Discretization of Values: First, we discretize the array. The sorted unique values will be [2, 3, 4, 5, 7]. Our mapping m will be {2: 1, 3: 2, 4: 3, 5: 4, 7: 5}.

2. Implementation of BIT: We create a Binary Indexed Tree for the five unique values.

   tree = BinaryIndexedTree(5)

3. Processing Elements in Reverse: We process the elements from right to left, updating the tree and querying for counts:

   • Starting with the last element 5, its discretized value is 4. Before we update, there are no smaller elements, so ans[] starts with [0].
   • Next is 7, with discretized value 5. The query(4) will find 0 elements smaller, so ans[] becomes [0, 0].
   • Now 2 is encountered with a discretized value of 1. Previous smaller elements don't exist (query(0)), thus ans[] is [0, 0, 0].
   • The fourth element is 4, corresponding to discretized 3. The query(2) shows there is one smaller (2), so ans[] is [0, 0, 1, 0].
   • The first element in reverse is 3, with discretized value 2. A query(1) will find there is one element smaller (2), adding up to ans[] as [0, 1, 1, 0, 0].

After processing, we have ans[] = [0, 1, 1, 0, 0], and we reverse it to get the final output: [0, 0, 1, 1, 0].

This corresponds to:

• 3 (no smaller elements to the right),
• 4 (no smaller elements to the right),
• 2 (one smaller element, 5, to the right),
• 7 (one smaller element, 5, to the right),
• 5 (no smaller elements to the right).

Final Output: [0, 0, 1, 1, 0].

The Binary Indexed Tree speeds up the process where each update and query happens in logarithmic time relative to the number of unique elements. Overall, the time complexity of processing the array in this manner is O(n log n).

Solution Implementation

Time and Space Complexity

The given Python code implements a Binary Indexed Tree (also known as a Fenwick Tree) to solve the problem of finding the count of smaller numbers after each element for a given list of integers.

Time Complexity:

1. Sorting unique values (sorted(set(nums))):

   • This operation involves extracting unique elements from nums and then sorting them, which has a time complexity of O(k log k) where k is the number of unique elements.

2. Building a map for value to index (m = {v: i for i, v in enumerate(alls, 1)}):

   • This takes O(k) time, as it is iterating over all unique elements once.

3. Binary Indexed Tree operations within the main loop (for v in nums[::-1]: ...):

   • For each v in the list nums, we perform an update and a query operation on the tree. Since each update and query takes O(log n) time and we are doing it for all n elements, this part takes a total of O(n log n) time.

4. Reversing the answer list (return ans[::-1]):

   • This operation takes O(n) time for a list of n elements.

Combining these parts, the overall time complexity is O(n log n) since the sorting can be considered a preprocessing step with a negligible O(k log k) complexity compared to the O(n log n) complexity of the main operations with the Binary Indexed Tree.

Space Complexity:

1. Storage for the Binary Indexed Tree (self.c = [0] * (n + 1)):

   • The space used by the tree is O(n) where n is the number of elements in the input nums.

2. Space for the sorted unique values and map (alls and m):

   • The sorted list of unique values alls and the map m both take O(k) space where k is the number of unique elements.

3. Space for the answer (ans):

   • Since we store the count of smaller elements for each input element, this also takes O(n) space.

The dominant space complexity for the entire algorithm is O(n), which takes into account the space for the Binary Indexed Tree and the results list.

In conclusion, the code has a time complexity of O(n log n) and a space complexity of O(n).

Learn more about how to find time and space complexity quickly using problem constraints.