Maximum sum increasing subsequence from a prefix and a given element after prefix is must

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).

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Prerequisite : Maximum Sum Increasing Subsequence

Simple Approach:

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

Time Complexity: O(2n)

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.

C++

 // CPP program to find maximum sum increasing // subsequence till i-th index and including // k-th index. #include #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)              dp[i] = a[i] + a;                 else              dp[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);     ll index = 4, k = 6;     printf("%lld", pre_compute(a, n, index, k));     return 0; }

Java

 // Java program to find maximum sum increasing // subsequence tiint i-th index and including // k-th index. class GFG {            static int pre_compute(int a[], int n,                              int index, int k)     {         int dp[][] = new int[n][n];                // Initializing the first row of          // the dp[][].         for (int i = 0; i < n; i++) {             if (a[i] > a)                  dp[i] = a[i] + a;              else                 dp[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];     }            // Driver code     public static void main(String[] args)     {         int a[] = { 1, 101, 2, 3, 100, 4, 5 };         int n = a.length;         int index = 4, k = 6;         System.out.println(                   pre_compute(a, n, index, k));     } }    // This code is contributed by Smitha.

Python3

 # Python program to find maximum  # sum increasing subsequence till   # i-th index and including k-th index.     def pre_compute(a, n, index, k):            dp = [[0 for i in range(n)]               for i in range(n)]                     # Initializing the first      # row of the dp[][]     for i in range(n):         if a[i] > a:             dp[i] = a[i] + a         else:             dp[i] = a[i]                    # Creating the dp[][] matrix.     for i in range(1, n):         for j in range(n):             if a[j] > a[i] and 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]    # Driver code a = [1, 101, 2, 3, 100, 4, 5 ] n = len(a) index = 4 k = 6 print(pre_compute(a, n, index, k))    # This code is contributed # by sahilshelangia

C#

 // C# program to find maximum  // sum increasing subsequence  // till i-th index and including // k-th index. using System;    class GFG {     static int pre_compute(int []a, int n,                            int index, int k)     {     int [,]dp = new int[n, n];        // Initializing the first      // row of the dp[][].     for (int i = 0; i < n; i++)      {         if (a[i] > a)              dp[0, i] = a[i] + a;          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]; }    // Driver code static public void Main () {     int []a = {1, 101, 2,                 3, 100, 4, 5};     int n = a.Length;     int index = 4, k = 6;     Console.WriteLine(pre_compute(a, n,                                    index, k)); } }    // This code is contributed by @ajit

PHP

 \$a)              \$dp[\$i] = \$a[\$i] + \$a;              else             \$dp[\$i] = \$a[\$i];          }        // Creating the dp[][] matrix.     for (\$i = 1; \$i < \$n; \$i++)     {         for (\$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]; }    // Driver Code \$a = array( 1, 101, 2, 3, 100, 4, 5 ); \$n = sizeof(\$a); \$index = 4; \$k = 6; echo pre_compute(\$a, \$n, \$index, \$k);    // This code is contributed by ita_c ?>

Output:

11

Time Complexity : O(n2)

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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