Given an array, the task is to find sum of maximum sum alternating subsequence starting with first element. Here alternating sequence means first decreasing, then increasing, then decreasing, … For example 10, 5, 14, 3 is an alternating sequence.

Note that the reverse type of sequence (increasing – decreasing – increasing -…) is not considered alternating here.

Examples:

Input :arr[] = {4, 3, 8, 5, 3, 8}Output :28 Explanation: The alternating subsequence (starting with first element) that has above maximum sum is {4, 3, 8, 5, 8}Input :arr[] = {4, 8, 2, 5, 6, 8}Output :14 The alternating subsequence (starting with first element) that has above maximum sum is {4, 2, 8}

This problem is similar to Longest Increasing Subsequence (LIS) problem. and can be solved using Dynamic Programming.

Create two empty array that store result of maximum sum of alternate sub-sequence inc[] : inc[i] stores results of maximum sum alternating subsequence ending with arr[i] such that arr[i] isgreaterthan previous element of the subsequence dec[] : dec[i] stores results of maximum sum alternating subsequence ending with arr[i] such that arr[i] islessthan previous element of the subsequence Include first element of 'arr' in both inc[] and dec[] inc[0] = dec[0] = arr[0] // Maintain a flag i.e. it will makes the greater // elements count only if the first decreasing element // is counted. flag = 0 Traversal two loops i goes from 1 to n-1 j goes 0 to i-1 IF arr[j] > arr[i] dec[j] = max(dec[i], inc[j] + arr[i]) // Denotes first decreasing is found flag = 1 ELSE IF arr[j] < arr[i] && flag == 1 inc[i] = max(inc[i], dec[j]+arr[i]); Final Last Find maximum value inc[] and dec[] .

Below is implementation of above idea.

## C/C++

// C++ program to find sum of maximum // sum alternating sequence starting with // first element. #include<bits/stdc++.h> using namespace std; // Return sum of maximum sum alternating // sequence starting with arr[0] and is first // decreasing. int maxAlternateSum(int arr[], int n) { if (n == 1) return arr[0]; // create two empty array that store result of // maximum sum of alternate sub-sequence // stores sum of decreasing and increasing // sub-sequence int dec[n]; memset(dec, 0, sizeof(dec)); // store sum of increasing and decreasing sun-sequence int inc[n]; memset(inc, 0, sizeof(inc)); // As per question, first element must be part // of solution. dec[0] = inc[0] = arr[0]; int flag = 0 ; // Traverse remaining elements of array for (int i=1; i<n; i++) { for (int j=0; j<i; j++) { // IF current sub-sequence is decreasing the // update dec[j] if needed. dec[i] by current // inc[j] + arr[i] if (arr[j] > arr[i]) { dec[i] = max(dec[i], inc[j]+arr[i]); // Revert the flag , if first decreasing // is found flag = 1; } // If next element is greater but flag should be 1 // i.e. this element should be counted after the // first decreasing element gets counted else if (arr[j] < arr[i] && flag == 1) // If current sub-sequence is increasing // then update inc[i] inc[i] = max(inc[i], dec[j]+arr[i]); } } // find maximum sum in b/w inc[] and dec[] int result = INT_MIN; for (int i = 0 ; i < n; i++) { if (result < inc[i]) result = inc[i]; if (result < dec[i]) result = dec[i]; } // return maximum sum alternate sun-sequence return result; } //Driver program int main() { int arr[]= {8, 2, 3, 5, 7, 9, 10}; int n = sizeof(arr)/sizeof(arr[0]); cout << "Maximum sum = " << maxAlternateSum(arr , n ) << endl; return 0; }

## Java

// Java program to find sum of maximum // sum alternating sequence starting with // first element. public class GFG { // Return sum of maximum sum alternating // sequence starting with arr[0] and is first // decreasing. static int maxAlternateSum(int arr[], int n) { if (n == 1) return arr[0]; // create two empty array that store result of // maximum sum of alternate sub-sequence // stores sum of decreasing and increasing // sub-sequence int dec[] = new int[n]; // store sum of increasing and decreasing sun-sequence int inc[] = new int[n]; // As per question, first element must be part // of solution. dec[0] = inc[0] = arr[0]; int flag = 0 ; // Traverse remaining elements of array for (int i=1; i<n; i++) { for (int j=0; j<i; j++) { // IF current sub-sequence is decreasing the // update dec[j] if needed. dec[i] by current // inc[j] + arr[i] if (arr[j] > arr[i]) { dec[i] = Math.max(dec[i], inc[j]+arr[i]); // Revert the flag , if first decreasing // is found flag = 1; } // If next element is greater but flag should be 1 // i.e. this element should be counted after the // first decreasing element gets counted else if (arr[j] < arr[i] && flag == 1) // If current sub-sequence is increasing // then update inc[i] inc[i] = Math.max(inc[i], dec[j]+arr[i]); } } // find maximum sum in b/w inc[] and dec[] int result = Integer.MIN_VALUE; for (int i = 0 ; i < n; i++) { if (result < inc[i]) result = inc[i]; if (result < dec[i]) result = dec[i]; } // return maximum sum alternate sun-sequence return result; } // Driver Method public static void main(String[] args) { int arr[]= {8, 2, 3, 5, 7, 9, 10}; System.out.println("Maximum sum = " + maxAlternateSum(arr , arr.length)); } }

Output:

Maximum sum = 25

Time Complexity : O(n^{2})

Auxiliary Space : O(n)

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