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- Explanation: Sum of {3, 5, 7, 20} ⇒ 35. Here, we have removed 1 and 4 which makes the rest of the array sorted. Then the remaining elements make maximum sum increasing subsequence. Algorithm for Maximum Sum Increasing Subsequence 1. Declare an array say maxSumIS as the same size as the length of an array. 2. Set the output to 0. 3.
- Output: 3 4 5. Explanation: Subsequence 3 4 5 with sum 12 is the subsequence with maximum possible sum. Input: arr [] = {6, 3, 4, 1, 1, 8, 7, -4, 2} Output: 6 3 4 8 7. Approach: The task can be solved using Greedy Approach. The idea is to take the maximum possible elements from arr [] in the subsequence. Follow the steps below to solve the problem.
- Algorithm. Take two array a [] and l [] where a [] is your original array and l [] array keeps the length of increasing subsequence for each element. Initialize l [] with value 1 because every element will also be considered in length of increasing subsequence. Update the value of l [i] for every ith element in array a [] by keeping track of ...
- In general, we have m [i] = max {max (m [j] + a [i], a [i]), j = 1 to i - 4} This will be O (n^2). You can get an O (n) solution from this pretty easily. Notice that it always helps to choose the maximum m [j] in the above recurrence. So, compute the maximum of m as you traverse it, like in the following pseudocode: