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

C++Server Side ProgrammingProgramming

In this problem, we are given an array arr[]. Our task is to create a program to find the maximum sum Bi-tonic subsequence in C++.

Bi-tonic subsequence is a special sequence whose elements first increase and then decrease.

Let’s take an example to understand the problem,

Input

arr[] = {4, 2, 3, 7, 9, 6, 3, 5, 1}

Output

33

Explanation

The Bi-tonic subsequence which gives the largest sum is {2, 3, 7, 9, 6, 5, 1} Sum = 2 + 3 + 7 + 9 + 6 + 5 + 1 = 33

Solution Approach

To find the maximum sum bitonic subsequence, we will create two arrays, incSeq[] and decSeq[] in such a way that for an element i at index, incSeq[i] has sum of all elements from arr[0…i] strictly increasing and decSeq[i] has sum of all elements from arr[i…n] strictly decreasing.

At the end, we will return the maxSum as maximum value from (incSeq[i] + decSeq[i] - arr[i]).

Example

Program to illustrate the wording of our solution,

 Live Demo

#include <iostream>
using namespace std;
int calcMaxVal(int a, int b){
   if(a > b)
      return a;
      return b;
}
int findMaxSumBiTonicSubSeq(int arr[], int N){
   int maxSum = -1;
   int incSeq[N], decSeq[N];
   for (int i = 0; i < N; i++){
      decSeq[i] = arr[i];
      incSeq[i] = arr[i];
   }
   for (int i = 1; i < N; i++)
      for (int j = 0; j < i; j++)
         if (arr[i] > arr[j] && incSeq[i] < incSeq[j] + arr[i]) incSeq[i] = incSeq[j] + arr[i];
   for (int i = N - 2; i >= 0; i--)
      for (int j = N - 1; j > i; j--)
         if (arr[i] > arr[j] && decSeq[i] < decSeq[j] + arr[i])
         decSeq[i] = decSeq[j] + arr[i];
   for (int i = 0; i < N; i++)
      maxSum = calcMaxVal(maxSum, (decSeq[i] + incSeq[i] - arr[i]));
   return maxSum;
}
int main(){
   int arr[] = {4, 2, 3, 7, 9, 6, 3, 5, 1};
   int N = sizeof(arr) / sizeof(arr[0]);
   cout<<"The Maximum Sum of Bi-tonic subsequence is : "<<findMaxSumBiTonicSubSeq(arr, N);
   return 0;
}

Output

The Maximum Sum of Bi-tonic subsequence is : 33
raja
Updated on 15-Oct-2020 13:56:52

Advertisements