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Problem-solving using the decrease and conquer technique

Sarfraz Raza

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Divide and conquer vs. decrease and conquer The three strategies of decrease and conquer Decrease and conquer strategy I (by a constant)Example 1: Insertion sort (recursive implementation)Example 2: Recursive traversal of a linked list Exercise: Printing the linked list in reverse order Decrease and conquer strategy II (by a constant factor, with constant manipulation)Example: Binary search Exercise: Ternary search Further applications Decrease and conquer strategy III (by a constant factor, with some polynomial manipulation)Example: KthK^{th}Kth smallest element Conclusion

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Picture yourself confronting a challenging problem, akin to tackling a jigsaw puzzle. You meticulously dissect it into smaller, more manageable fragments, carefully selecting the most critical piece to address initially while consistently diminishing the problem's size. The decrease and conquer technique adheres to this strategy, systematically segmenting and progressively simplifying the problem. This simplification leads to a more concise version of the problem, ultimately guiding you to the complete solution. Consider it akin to navigating a maze, step by step—beginning by eliminating possibilities through the careful tracing of specific regions to reduce the problem's size—until you conquer the entire puzzle.

Before we delve into three distinct strategies for applying the decrease and conquer technique, it’s essential to grasp the fundamental differences between the decrease and conquer and the divide and conquer approach.

Divide and conquer vs. decrease and conquer#

The decrease and conquer technique closely resembles divide and conquer, differing slightly in its methodology.

In the divide and conquer approach, a problem is subdivided into smaller, identical subproblems. These subproblems undergo recursive resolution until they reach a minimal size (the base case), after which they are directly solved. The final step combines their solutions to address the original, larger problem.

On the other hand, decrease and conquer also simplifies complex problems by partitioning them into smaller instances. But contrary to divide and conquer, which tackles multiple subproblems, decrease and conquer concentrates on resolving a single subproblem, deriving inspiration from mathematical induction. This technique is also known as the incremental or inductive approach.

The three strategies of decrease and conquer#

Let's discuss the following three strategies one by one:

Decrease by a constant size.
Decrease by a constant factor (with constant manipulation on each recursive step).
Decrease by a constant factor (with a polynomial manipulation on each recursive step).

Decrease and conquer strategy I (by a constant)#

In this variant of the decrease and conquer approach, the algorithm reduces the problem size by a constant $c$ with each iteration or recursive step. This signifies that at every stage, the problem is trimmed down into a more manageable subproblem, shrinking to a size of $N - c$ , where $N$ represents the initial problem size and $c$ denotes the portion that has been pruned or separated from the original problem. On each recursive step, it takes $N^k$ steps to process and trim the problem size to $N-c$ , where $c \ge 1 , k\ge 1$ .

The recurrence relation governing this approach adheres to the following pattern:

It is not hard to see that the above recurrence has an overall time complexity of $O(N \times N^k) \approx O(N^{k+1}).$

There are around $N/c$ number of terms, and every term is bounded above by $N^k$ , therefore $T(N) \leq N \times N^k.$
For the lower bound, we can easily say that half of the terms (i.e., $\frac{N}{2c}$ terms) are at least $(N/2)^k$ , as $k$ is a constant, therefore, $2^k$ is just a constant and can be ignored, and the lower bound will be $T(N) \geq \frac{N}{2c} \times (\frac{N}{2})^k \approx \Omega(N^{k+1})$ by ignoring the constants and only taking the terms with base $N.$

Let’s look at some examples for applying this decrease and conquer strategy.

Example 1: Insertion sort (recursive implementation)#

Insertion sort is a sorting algorithm that builds the final sorted array one item at a time. It iterates through an input array, taking one element at a time and inserting it into its correct position within the sorted portion of the array. This process continues until all elements are sorted.

Look at the following animation of insertion sort.

#include <iostream>
#include <vector>
using namespace std;
void recursiveInsertionSort(vector<int>& arr, int n) 
{
    if (n <= 1)
        return;
    // Sorting the first n-1 elements, assuming the last element is pruned
    recursiveInsertionSort(arr, n - 1);
    // Inserting the nth element into the sorted subarray of n-1 elements
    int j = n - 1;
    while (j-1 >= 0 && arr[j] < arr[j-1]) 
    {
        swap(arr[j-1] , arr[j]); // Shifting the j-1'th element forward
        j--;
    }
}
void print(string msg, vector<int>& arr)
{
    cout << msg;
    for (int i = 0; i < arr.size(); i++)
        cout << arr[i] << " ";
    cout << endl;
}
int main() 
{
    vector<int> arr = {12, 11, 13, 5, -6};
    int n = arr.size();
    print("Before Sorting: ", arr);
    recursiveInsertionSort(arr, n);
    print("After Sorting:  ", arr);
    return 0;
}

#include <iostream>
using namespace std;
struct ListNode 
{
    int data;
    ListNode* next;
    ListNode(int val) : data(val), next(nullptr) 
    {}
};
void recursiveLinkedListTraversal(ListNode* node) 
{
    if (node == nullptr)
        return;
    cout << node->data << " ";
    recursiveLinkedListTraversal(node->next);
}
int main() 
{
    // Creating a sample linked list: 10 -> 20 -> 30 -> 40 -> 50
    ListNode* head = new ListNode(10);
    head->next = new ListNode(20);
    head->next->next = new ListNode(30);
    head->next->next->next = new ListNode(40);
    head->next->next->next->next = new ListNode(50);
    recursiveLinkedListTraversal(head);
    return 0;
}

In such a case where $c$ is a constant the total time of the problem solution will take $c\times log_b N$ which is just $O(log_b N)$ , if $c$ is just a constant.

Example: Binary search#

One famous strategy for such a case is binary search; it works in the following fashion:

Our task is to search a given value $T$ in a contiguous sorted array $V$ . We take the middle index $i$ and see if the value at the middle index $V_i$ is equal to $T.$ Then, we find the index of the value, i.e., $i.$
However, if the value $V_i$ is greater than $T$ then it shrinks the search space to the first half (that is it searches in the range from the beginning of the range up to $i-1$ ). Otherwise, it searches in the second half (that is, it searches in the range from $i+1$ up to the end of the range).

The above idea of binary searching follows this recurrence:

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
// Recursive binary search function
int BinarySearch(vector<int>& arr, int target, int left, int right, int & indexCount) 
{
    if (left <= right) 
    {
        indexCount++;
        int mid = left + (right - left) / 2; // This is to avoid overflow
        // If the target is found at the middle element
        if (arr[mid] == target) 
            return mid;
        
        // If the target is in the left subarray
        if (arr[mid] > target) 
            return BinarySearch(arr, target, left, mid - 1, indexCount);
        
        // If the target is in the right subarray
        return BinarySearch(arr, target, mid + 1, right, indexCount);
    }
    // If the target is not found
    return -1;
}
// Wrapper function for binary search
int BinarySearch(vector<int>& arr, int target, int & indexCount) 
{
    indexCount = 0;
    return BinarySearch(arr, target, 0, arr.size() - 1, indexCount);
}
void print(string msg, vector<int>& arr)
{
    cout << msg << ": { ";
    for(int i=0; i<arr.size(); i++)
        cout << arr[i] << " ";
    cout << "}"<<endl;
}
void randomFill(vector<int> &arr, int si, int ei)
{
    // Filling random values between the given range [si-ei]
    int range = ei-si+1; // Both ends included
    srand(time(0));
    for(int i=0; i<arr.size(); i++)
    {
        arr[i] = rand()%range + si; // Assigning a random value in a given range
    }
}
int main() {
    vector<int> arr(1<<20); 
        //  = {1, 3, 5, 7, 9, 11, 13, 15}; // You may test on small number too
    
    // print("Array ", arr); // Do not print an array of 1 million size :)
    randomFill(arr, 0, 1<<30); // Filling the vector with values between 0-2^{30}
    sort(arr.begin(), arr.end()); // sorting the array
    int target = arr[789]; 
            // Element to be searched after sorting present at 789 index. You can 
            // change it with any other index.
    int fi = -1;
    int indexCount=0;
    if ((fi = BinarySearch(arr, target, indexCount))!=-1) 
    {
        cout << target << " is found in the array at index: " << fi 
             << ".  It took " << indexCount << " steps to do this search" << endl;
    } 
    else 
    {
        cout << target << " is not found in the array." 
        << ". It took " << indexCount << " steps to find that"
        << endl;
    }
    return 0;
}

Implementation details

In the provided code, a main() function and a binary search function are implemented in C++. The main function is responsible for testing the binary search algorithm on a large array.

In the main function, a vector arr of size $2^{20}$ is created to hold integer values. The randomFill() function is called to populate this vector with random integers between $0$ and $2^{30}$ . After filling, the vector is sorted in ascending order using the sort() function from the C++ Standard Library.
Next, a target value is selected from the sorted array. In this example, the target is set to the element at index 789, but we can change it to any other index for testing. The BinarySearch() function is called to find the target value within the array. If the target is found, the index of the target in the sorted array is printed along with the number of steps taken for the binary search. If the target is not found, a message indicating that it wasn’t found is printed along with the number of steps taken.
The BinarySearch() function itself is a recursive binary search algorithm. It takes a sorted array, a target value, and two indices representing the left and right boundaries of the search range. It recursively divides the search range in half until it finds the target or determines that the target is not present in the array. The number of steps taken during the search is tracked using the indexCount parameter, and this count is returned along with the index of the target element or -1 if the target is not found.

Overall, the main function demonstrates how the binary search algorithm can efficiently locate a target value in a sorted array and measure the number of steps required for the search.

Activity: Play the guessing game in the following interactive widget. First, you need to choose a particular number that you want the binary search to search for; the animated shrinking mechanism will show how it will search for the element in the minimum number of steps by following the binary search algorithm.

Food for thought: What will happen if $N$ is not a perfect power of $3$ . Can you still solve the problem in a similar bound?

Further applications #

The concept of strategy II pruning has a broad spectrum of applications, serving as the bedrock for numerous robust data structures. One notable example is the B-tree, characterized by nodes that form a balanced b-ary tree, where each node can house a maximum of $b$ branches and at least $b/2$ branches. This structural balance results in an efficient searching cost of approximately $O(log_b N),$ making B-trees a cornerstone in optimizing database data retrieval operations.

For example, if we assume $c = 2$ , the above recurrence seems like a recurrence of the binary search only if $k = 0$ because in that case, $N^k$ will be a constant. For this strategy, we assume that $k\ge1,$ that is, we have at least a linear function, i.e., $O(N^k)$ where $k\geq1$ meaning that we scan through the entire input search space at least once.

In such a case, the recurrence becomes a decreasing geometric series as seen in the expanded expression for $T(N)$ :

This entire expression is $O(N^k),$ as $(1 + \frac{1}{c^k}+\frac{1}{c^{2k}}+\frac{1}{c^{3k}}+\ldots ) \approx O(1),$ a decreasing geometric series is always bounded by the first term, which is $1$ in this case.

Prune and search: This approach can be thought of as a strategy where, after pruning, the problem size reduces by a factor of $c$ and the computation for smaller recursive calls is asymptomatically achieved for free.

Let’s look at this strategy in action for finding the K’th smallest element in expected linear time.

Example: KthK^{th}Kth smallest element#

In this problem, we aim to find the $K^{th}$ smallest element from an unsorted array. This means we want to determine the element that would be at the $K^{th}$ position if the array were sorted in ascending order. For example, if we have an array [12, 5, 23, 9, 42, 1, 3, 7, 19] and we want to find the 4th smallest element. The algorithm should return 7.

Strategy: Decrease and conquer approach

The idea is again a pruning strategy; it randomly selects an element, and based on that, it partitions all the elements smaller than the chosen element in the given array to the left of the array, and those values that are bigger than the partition element to the right of the array. This can be done by the famous partition() function, which is the foundation of the quicksort algorithm.

Observation: The probability that element we randomly select lies in the sorted middle range from 25% to 75% is 0.5.

#include <iostream>
#include <vector>
#include <cstdlib>
#include <ctime>
using namespace std;
// Function to partition the vector and return the index of the pivot element
int partition(vector<int>& arr, int low, int high) 
{
    // Choosing a random pivot within the range
    int randomIndex = low + rand() % (high - low + 1);
    swap(arr[randomIndex], arr[high]); // Moving pivot to the end
    int pivot = arr[high];
    // Initializing indices for elements to be compared
    int i = low - 1;
    for (int j = low; j <= high - 1; j++) 
    {
        // If the current element is smaller than or equal to the pivot
        if (arr[j] <= pivot) 
        {
            i++;
            swap(arr[i], arr[j]); // Swapping arr[i] and arr[j]
        }
    }
    
    // Swapping arr[i+1] and arr[high] (pivot)
    swap(arr[i + 1], arr[high]);
    return i + 1; // Returning the index of the pivot element
}
// Function to find the K'th smallest element recursively
int findKthSmallest(std::vector<int>& arr, int low, int high, int k) 
{
    if (k > 0 && k <= high - low + 1) 
    {
        // Partitioning the array and getting the index of the pivot
        int pivotIndex = partition(arr, low, high);
        // If the pivot is the K'th element, we return it
        if (pivotIndex - low == k - 1)
            return arr[pivotIndex];
        // If the K'th element is in the left subarray, we recurse on the left subarray
        if (pivotIndex - low > k - 1)
            return findKthSmallest(arr, low, pivotIndex - 1, k);
        // If the K'th element is in the right subarray, we recurse on the right subarray
        return findKthSmallest(arr, pivotIndex + 1, high, k - (pivotIndex - low + 1));
    }
    // If K is out of range, we return -1
    return -1;
}
int main() 
{
    vector<int> arr = {12, 5, 23, 9, 42, 1, 3, 7, 19};
    int k = 4; // Finding the 4th smallest element
    // Seeding the random number generator
    srand(time(0));
    int result = findKthSmallest(arr, 0, arr.size() - 1, k);
    if (result != -1)
        cout << "The " << k << "th smallest element is: " << result << std::endl;
    else
        cout << "Invalid value of K." << std::endl;
    return 0;
}

How is the $K^{th}$ element searched recursively?

The probability that a randomly chosen pivot point lies within the range of the middle sorted array is 50%, i.e., 0.5. Therefore, the expectation is that every 2nd element selection will partition the array, in the worst case, $\frac{3}{4}\times N$ elements on one side and $\frac{1}{4} \times N$ elements on the other side of the array. The worst case could be that the $K^{th}$ element to be searched lies in the bigger range.

Applications: This approach of finding $K^{th}$ smallest element is the backbone of randomized quicksort. It calls this subroutine to find the middle element in expected linear time, then partitions the array based on the middle element, ensuring that the two recursive calls are almost the same. Therefore, making it into a recurrence of something like this:
$T(N) = T(N/c) + T((1-1/c) N) + O(N), \space where \space c>1$
This recurrence has the time complexity of $O(N \log_c N) \approx O(N \log N)$ making it one of the fastest known algorithms for sorting $N$ elements.

For further details about the 3^rd strategy (and details about the geometric series and its bounds), read the following blog Geometric series and classic search and prune.

Conclusion#

The decrease and conquer stratagies are very powerful tools in algorithmic recursive problem solving, and a topic that often comes up in coding interviews. It draws inspiration from backward mathematical induction, a fundamental concept in mathematics. Systematically reducing complex problems into smaller, more manageable parts provides a structured approach to finding solutions.

In addition to its relevance in algorithm design, this approach finds applications in various fields, including game theory, where it aids in analyzing and solving complex strategic scenarios. Its versatility extends further into the realm of dynamic programming, enabling the efficient computation of optimal solutions in diverse problem domains. With its roots in mathematical principles and wide-ranging applications, decrease and conquer is a valuable strategy at the intersection of mathematics, computer science, and problem-solving.

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Problem-solving using the decrease and conquer technique

Divide and conquer vs. decrease and conquer#

The three strategies of decrease and conquer#

Decrease and conquer strategy I (by a constant)#

Example 1: Insertion sort (recursive implementation)#

Example 2: Recursive traversal of a linked list#

Exercise: Printing the linked list in reverse order#

Decrease and conquer strategy II (by a constant factor, with constant manipulation)#

Example: Binary search#

Exercise: Ternary search#

Further applications #

Decrease and conquer strategy III (by a constant factor, with some polynomial manipulation)#

Example: KthK^{th}Kth smallest element#

Conclusion#