Monday, December 19, 2016

Maximum sum alternating subsequence


http://www.geeksforgeeks.org/maximum-sum-alternating-subsequence-sum
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.
This problem is similar to Longest Increasing Subsequence (LIS) problem. and can be solved using Dynamic Programming.
We maintain two arrays
dp[i] : Stores sum of maximum sum alternating subsequence
        starting with first element and ending with arr[i].
smaller[i] : True if next expected element in the maximum sum 
             alternating subsequence starting with first 
             element and ending with arr[i] is smaller than
             arr[i].
             False otherwise.

dp[i] can be written as:
dp[i] = arr[i] + max( dp[j] ) where 0 < j < i and 
                         (arr[j] < arr[i] & flag[j] = True)
                                  OR
                         (arr[j] > arr[i] & flag[j] = False)

Final result is maximum of all dp[] values.
int maxAlternateSum(int arr[], int n)
{
    // dp[i] is going to sum of maximum of alternating
    // sequence that must end with arr[i]
    int dp[n];
 
    // smaller[i] is going to store expected sign of
    // next element in the alternating subsequence that
    // includes arr[i]  smaller[i] = 1 indicates next
    // expected element is smaller and smaller[0] = 0
    // indicates that next expected element is greater.
    int smaller[n];
 
    dp[0] = arr[0];  // As per question, first element must
                     // be part of solution.
 
    smaller[0] = 1;  // As per question, the alternating seq
                     // must begin with decreasing part.
 
 
    // Traverse remaining elements and compute dp[i] for them
    for (int i = 1; i < n ; i++)
    {
        // Since we want max sum, we initialize current max
        // ending with i as minus infinite.
        dp[i] = INT_MIN;
 
        // Repeatedly check from first element as we can
        // get maximum sum by including any element
        for (int j=0; j<i; j++)
        {
            // If next expected element for arr[j] is smaller
            if (smaller[j] == 1)
            {
                // Check if element is actually smaller
                if (arr[i] < arr[j])
                {
                    // Store the  max sum till  that element
                    dp[i] = max(dp[i], dp[j] + arr[i]);
 
                    // Store the complement of flag as we want
                    // next element larger than current
                    smaller[i] = ! smaller[j];
                }
            }
            else
            {
                // If next expected element for arr[j] is greater
                if (arr[j] < arr[i])
                {
                    // Store the max sum till  that element
                    dp[i] = max (dp[i], dp[j] + arr[i]);
 
                    // Store the complement of flag as we want next
                    // element smaller than current
                    smaller[i] = ! smaller[j];
                }
            }
        }
    }
 
    // Result is maximum of all values in dp[]
    int result = dp[0];
    for (int i=1; i<n; i++)
        if (result < dp[i])
            result = dp[i];
    return result;
}




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