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Maximum sum Bi-tonic sub-sequence in C++

Introduction of Bi-tonic Subsequence:

A sequence of elements within an array that follows a particular pattern, which first increasing and then decreasing, is referred to as a bi-tonic subsequence in C++. Formally speaking, a subsequence arr[i1], arr[i2],..., arr[ik] of an array arr[] of size n is deemed bi-tonic if and only if:

  • i1 < i2 < ... < ik (elements maintain their relative order in the array)
  • arr[i1] <= arr[i2] <= ... <= arr[ik-1] >= arr[ik] (elements increase and then decrease)

Problem Statement:

Finding the Maximum Sum Bi-tonic Subsequence:

Finding the bi-tonic subsequence with the largest sum of its elements within a given array is the issue we are trying to solve. This issue has applications in several fields, such as:

  • Analysis of the stock market (finding peak-and-dip patterns).
  • Signal processing: identifying upward and downward trends.
  • Optimization issues (determining the best schedules or distribution of resources).

Problem Definition

Given: A n-integer array arr[].

  • The task is to find the bi-tonic subsequence's maximum sum inside arr[].
  • The maximum sum of a bi-tonic subsequence is the output.

Constraints:

  • Positive and negative integers can be contained in the array arr[].
  • There might or might not be a bi-tonic subsequence in the array.
  • The output should be 0 (or any suitable value indicating no such subsequence) if there is no bi-tonic subsequence.

Input and Output Requirements:

Input:

  • An array arr[] of n integers represents the data to be analyzed.
  • The number n represents the array's size.

Output:

The maximum sum of a bi-tonic subsequence within arr[] is represented by a single integer.

For instance:

  • Input: arr[] = {1, 15, 51, 45, 33, 100, 12, 18, 9}, n = 9
  • Output: 194 (corresponding to the bi-tonic subsequence 1, 15, 51, 100, 18, 9)

Approach and Algorithm

Algorithm: Dynamic Programming

Steps:

Initialization:

  • First create two arrays: MSIBS[n] and MDSB[n].
    • The maximum sum of an increasing subsequence ending at the index i will be stored in MSIBS[i].
    • Starting at the index i, MDSB[i] will hold the maximum sum of a decreasing subsequence.
    • Set all other MSIBS and MDSB elements to 0 and MSIBS[0] to arr[0].

Calculate Increasing Subsequence Sums:

  • Iterate from i = 1 to n-1:
  • For each i, iterate from j = 0 to i-1:
  • If arr[i] > arr[j] and MSIBS[i] < MSIBS[j] + arr[i]:
  • Update MSIBS[i] with MSIBS[j] + arr[i].

Calculate Decreasing Subsequence Sums (in reverse order):

  • Iterate from i = n-2 to 0:
  • For each i, iterate from j = n-1 down to i+1:
  • If arr[i] > arr[j] and MDSB[i] < MDSB[j] + arr[i]:
  • Update MDSB[i] with MDSB[j] + arr[i].

Return maxSum.

Intuition:

  • Finding the maximum increasing and decreasing subsequences is the problem's decomposition.
  • For these subproblems, intermediate results are efficiently stored by MSIBS and MDSB.
  • Building the bi-tonic subsequence by layering increasing and decreasing parts is possible.
  • The right values from MSIBS and MDSB are combined to get the ultimate maximum sum.
  • Due to avoiding redundant calculations, dynamic programming results in O(n^2) time complexity.

Implementation in C++

#include <iostream>

#include <vector>

int maxSumBitonicSubsequence(const std::vector<int>& arr) {

    int n = arr.size();

    std::vector<int> MSIBS(n, 0);

    std::vector<int> MSDSB(n, 0);

    MSIBS[0] = arr[0];

    for (int i = 1; i < n; i++) {

        for (int j = 0; j < i; j++) {

            if (arr[i] > arr[j] && MSIBS[i] < MSIBS[j] + arr[i]) {

                MSIBS[i] = MSIBS[j] + arr[i];

            }

        }

    }

    MSDSB[n - 1] = arr[n - 1];

    for (int i = n - 2; i >= 0; i--) {

        for (int j = n - 1; j > i; j--) {

            if (arr[i] > arr[j] && MSDSB[i] < MSDSB[j] + arr[i]) {

                MSDSB[i] = MSDSB[j] + arr[i];

            }

        }

    }

    int maxSum = 0;

    for (int i = 0; i < n; i++) {

        int currentSum = MSIBS[i] + MSDSB[i] - arr[i];

        maxSum = std::max(maxSum, currentSum);

    }

    return maxSum;

}

int main() {

    std::vector<int> arr = {1, 15, 51, 45, 33, 100, 12, 18, 9};

    int n = arr.size();

    int maxSum = maxSumBitonicSubsequence(arr);

    std::cout << "Maximum sum bi-tonic subsequence: " << maxSum << std::endl;

    return 0;

}

Output:

Maximum sum Bi-tonic sub-sequence in C++

Time and Space Complexity

Time Complexity:

  • O(n^2) due to the nested loops used to compute MDSB and MSIBS.
  • There are n^2 operations because each loop iterates over n elements.

Space Complexity:

  • O(n), mostly due to using two auxiliary arrays, each with size n: MSIBS and MDSB.
  • Compared to n, additional space for variables and function calls is insignificant.

Enhancement for more expansive arrays:

  • Consider different strategies such as divide-and-conquer or more sophisticated dynamic programming methods that could lower the time complexity to O(n log n) or even O(n).
  • Space optimization: Investigate methods to reduce the space complexity to O(1) if space is a significant constraint. It may involve modifying the input array or employing a single auxiliary array.