Given an array of n positive integers, write a program to find the maximum sum of increasing subsequence from prefix till i-th index and also including a given kth element which is after i, i.e., k > i .

Examples:

Input : arr[] = {1, 101, 2, 3, 100, 4, 5}

i-th index = 4 (Element at 4th index is 100)

K-th index = 6 (Element at 6th index is 5.)

Output : 11

So we need to calculate the maximum sum of subsequence (1 101 2 3 100 5) such that 5 is necessarily included in the subsequence, so answer is 11 by subsequence (1 2 3 5).Input : arr[] = {1, 101, 2, 3, 100, 4, 5}

i-th index = 2 (Element at 2nd index is 2)

K-th index = 5 (Element at 5th index is 4.)

Output : 7

So we need to calculate the maximum sum of subsequence (1 101 2 4) such that 4 is necessarily included in the subsequence, so answer is 7 by subsequence (1 2 4).

**Prerequisite :** Maximum Sum Increasing Subsequence

**Simple Approach:**

- Construct a new array containing elements till ith index and the kth element.
- Recursively calculate all the increasing subsequences.
- Discard all the subsequences not having kth element included.
- Calculate the maximum sum from the left over subsequences and display it.

Time Complexity: O(2^{n})

**Efficient Approach:** Use a dynamic approach to maintain a table dp[][]. The value of dp[i][k] stores the maximum sum of increasing subsequence till ith index and containing the kth element.

// CPP program to find maximum sum increasing // subsequence till i-th index and including // k-th index. #include <bits/stdc++.h> #define ll long long int using namespace std; ll pre_compute(ll a[], ll n, ll index, ll k) { ll dp[n][n] = { 0 }; // Initializing the first row of the dp[][]. for (int i = 0; i < n; i++) { if (a[i] > a[0]) dp[0][i] = a[i] + a[0]; else dp[0][i] = a[i]; } // Creating the dp[][] matrix. for (int i = 1; i < n; i++) { for (int j = 0; j < n; j++) { if (a[j] > a[i] && j > i) { if (dp[i - 1][i] + a[j] > dp[i - 1][j]) dp[i][j] = dp[i - 1][i] + a[j]; else dp[i][j] = dp[i - 1][j]; } else dp[i][j] = dp[i - 1][j]; } } // To calculate for i=4 and k=6. return dp[index][k]; } int main() { ll a[] = { 1, 101, 2, 3, 100, 4, 5 }; ll n = sizeof(a) / sizeof(a[0]); ll index = 4, k = 6; printf("%lld", pre_compute(a, n, index, k)); return 0; }

**Output:**

11

Time Complexity: O(n^{2})

**Note:** This approach is very useful if you have to answer multiple such queries of i and k because using the pre calculated dp matrix you can answer such query in O(1) time.

To try similar problem, give this article a read: Maximum product of an increasing subsequence

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