Given an array sequence [A1, A2 …An], the task is to find the maximum possible sum of increasing subsequence S of length k such that S1<=S2<=S3………<=Sk.
Examples:
Input :
n = 8 k = 3
A=[8 5 9 10 5 6 21 8]
Output : 40
Possible Increasing subsequence of Length 3 with maximum possible sum is 9 10 21Input :
n = 9 k = 4
A=[2 5 3 9 15 33 6 18 20]
Output : 62
Possible Increasing subsequence of Length 4 with maximum possible sum is 9 15 18 20
One thing that is clearly visible that it can be easily solved with dynamic programming and this problem is a simple variation of Longest Increasing Subsequence. If you are unknown of how to calculate the longest increasing subsequence then see the implementation going to the link.
Naive Approach:
In the brute force approach, first we will try to find all the subsequences of length k and will check whether they are increasing or not. There could be nCk such sequences in the worst case when all elements are in increasing order. Now we will find the maximum possible sum for such sequences.
Time Complexity would be O((nCk)*n).
Efficient Approach:
We will be using a two-dimensional dp array in which dp[i][l] means that maximum sum subsequence of length l taking array values from 0 to i and the subsequence is ending at index ‘i’. Range of ‘l’ is from 0 to k-1. Using the approach of longer increasing subsequence on the inner loop when j<i we will check if arr[j] < arr[i] for checking if subsequence increasing.
This problem can be divided into its subproblems:
dp[i][1]=arr[i] for length 1 , maximum increasing subsequence is equal to the array value
dp[i][l+1]= max(dp[i][l+1], dp[j][l]+arr[i]) for any length l between 1 to k-1
This means that if for ith position and subsequence of length l+1 , there exists some subsequence at j (j < i) of length l for which sum of dp[j][l] + arr[i] is more than its initial calculated value then update that value.
Then finally we will find the maximum value of dp[i][k] i.e for every ‘i’ if subsequence of k length is causing more sum then update the required ans.
Below is the implementation code:
/*C++ program to calculate the maximum sum of increasing subsequence of length k*/#include <bits/stdc++.h>using namespace std;
int MaxIncreasingSub(int arr[], int n, int k)
{ // In the implementation dp[n][k] represents
// maximum sum subsequence of length k and the
// subsequence is ending at index n.
int dp[n][k + 1], ans = -1;
// Initializing whole multidimensional
// dp array with value -1
memset(dp, -1, sizeof(dp));
// For each ith position increasing subsequence
// of length 1 is equal to that array ith value
// so initializing dp[i][1] with that array value
for (int i = 0; i < n; i++) {
dp[i][1] = arr[i];
}
// Starting from 1st index as we have calculated
// for 0th index. Computing optimized dp values
// in bottom-up manner
for (int i = 1; i < n; i++) {
for (int j = 0; j < i; j++) {
// check for increasing subsequence
if (arr[j] < arr[i]) {
for (int l = 1; l <= k - 1; l++) {
// Proceed if value is pre calculated
if (dp[j][l] != -1) {
// Check for all the subsequences
// ending at any j<i and try including
// element at index i in them for
// some length l. Update the maximum
// value for every length.
dp[i][l + 1] = max(dp[i][l + 1],
dp[j][l] + arr[i]);
}
}
}
}
}
// The final result would be the maximum
// value of dp[i][k] for all different i.
for (int i = 0; i < n; i++) {
if (ans < dp[i][k])
ans = dp[i][k];
}
// When no subsequence of length k is
// possible sum would be considered zero
return (ans == -1) ? 0 : ans;
}// Driver functionint main()
{ int n = 8, k = 3;
int arr[n] = { 8, 5, 9, 10, 5, 6, 21, 8 };
int ans = MaxIncreasingSub(arr, n, k);
cout << ans << "\n";
return 0;
} |
/*Java program to calculate the maximum sum of increasing subsequence of length k*/import java.util.*;
class GFG
{ static int MaxIncreasingSub(int arr[], int n, int k)
{ // In the implementation dp[n][k] represents
// maximum sum subsequence of length k and the
// subsequence is ending at index n.
int dp[][]=new int[n][k + 1], ans = -1;
// Initializing whole multidimensional
// dp array with value -1
for(int i = 0; i < n; i++)
for(int j = 0; j < k + 1; j++)
dp[i][j]=-1;
// For each ith position increasing subsequence
// of length 1 is equal to that array ith value
// so initializing dp[i][1] with that array value
for (int i = 0; i < n; i++)
{
dp[i][1] = arr[i];
}
// Starting from 1st index as we have calculated
// for 0th index. Computing optimized dp values
// in bottom-up manner
for (int i = 1; i < n; i++)
{
for (int j = 0; j < i; j++)
{
// check for increasing subsequence
if (arr[j] < arr[i])
{
for (int l = 1; l <= k - 1; l++)
{
// Proceed if value is pre calculated
if (dp[j][l] != -1)
{
// Check for all the subsequences
// ending at any j<i and try including
// element at index i in them for
// some length l. Update the maximum
// value for every length.
dp[i][l + 1] = Math.max(dp[i][l + 1],
dp[j][l] + arr[i]);
}
}
}
}
}
// The final result would be the maximum
// value of dp[i][k] for all different i.
for (int i = 0; i < n; i++)
{
if (ans < dp[i][k])
ans = dp[i][k];
}
// When no subsequence of length k is
// possible sum would be considered zero
return (ans == -1) ? 0 : ans;
} // Driver code public static void main(String args[])
{ int n = 8, k = 3;
int arr[] = { 8, 5, 9, 10, 5, 6, 21, 8 };
int ans = MaxIncreasingSub(arr, n, k);
System.out.println(ans );
} } // This code is contributed by Arnab Kundu |
# Python program to calculate the maximum sum# of increasing subsequence of length kdef MaxIncreasingSub(arr, n, k):
# In the implementation dp[n][k] represents
# maximum sum subsequence of length k and the
# subsequence is ending at index n.
dp = [-1]*n
ans = -1
# Initializing whole multidimensional
# dp array with value - 1
for i in range(n):
dp[i] = [-1]*(k+1)
# For each ith position increasing subsequence
# of length 1 is equal to that array ith value
# so initializing dp[i][1] with that array value
for i in range(n):
dp[i][1] = arr[i]
# Starting from 1st index as we have calculated
# for 0th index. Computing optimized dp values
# in bottom-up manner
for i in range(1,n):
for j in range(i):
# check for increasing subsequence
if arr[j] < arr[i]:
for l in range(1,k):
# Proceed if value is pre calculated
if dp[j][l] != -1:
# Check for all the subsequences
# ending at any j < i and try including
# element at index i in them for
# some length l. Update the maximum
# value for every length.
dp[i][l+1] = max(dp[i][l+1],
dp[j][l] + arr[i])
# The final result would be the maximum
# value of dp[i][k] for all different i.
for i in range(n):
if ans < dp[i][k]:
ans = dp[i][k]
# When no subsequence of length k is
# possible sum would be considered zero
return (0 if ans == -1 else ans)
# Driver Codeif __name__ == "__main__":
n, k = 8, 3
arr = [8, 5, 9, 10, 5, 6, 21, 8]
ans = MaxIncreasingSub(arr, n, k)
print(ans)
# This code is contributed by# sanjeev2552 |
/*C# program to calculate the maximum sum of increasing subsequence of length k*/using System;
class GFG
{ static int MaxIncreasingSub(int []arr, int n, int k)
{ // In the implementation dp[n,k] represents
// maximum sum subsequence of length k and the
// subsequence is ending at index n.
int [,]dp=new int[n, k + 1];
int ans = -1;
// Initializing whole multidimensional
// dp array with value -1
for(int i = 0; i < n; i++)
for(int j = 0; j < k + 1; j++)
dp[i, j]=-1;
// For each ith position increasing subsequence
// of length 1 is equal to that array ith value
// so initializing dp[i,1] with that array value
for (int i = 0; i < n; i++)
{
dp[i, 1] = arr[i];
}
// Starting from 1st index as we have calculated
// for 0th index. Computing optimized dp values
// in bottom-up manner
for (int i = 1; i < n; i++)
{
for (int j = 0; j < i; j++)
{
// check for increasing subsequence
if (arr[j] < arr[i])
{
for (int l = 1; l <= k - 1; l++)
{
// Proceed if value is pre calculated
if (dp[j, l] != -1)
{
// Check for all the subsequences
// ending at any j<i and try including
// element at index i in them for
// some length l. Update the maximum
// value for every length.
dp[i, l + 1] = Math.Max(dp[i, l + 1],
dp[j, l] + arr[i]);
}
}
}
}
}
// The final result would be the maximum
// value of dp[i,k] for all different i.
for (int i = 0; i < n; i++)
{
if (ans < dp[i, k])
ans = dp[i, k];
}
// When no subsequence of length k is
// possible sum would be considered zero
return (ans == -1) ? 0 : ans;
} // Driver code public static void Main(String []args)
{ int n = 8, k = 3;
int []arr = { 8, 5, 9, 10, 5, 6, 21, 8 };
int ans = MaxIncreasingSub(arr, n, k);
Console.WriteLine(ans );
} }// This code contributed by Rajput-Ji |
<script>// Javascript program to calculate the// maximum sum of increasing subsequence // of length kfunction MaxIncreasingSub(arr, n, k)
{ // In the implementation dp[n][k]
// represents maximum sum subsequence
// of length k and the subsequence is
// ending at index n.
let dp = new Array(n);
for(let i = 0; i < dp.length; i++)
{
dp[i] = new Array(2);
}
let ans = -1;
// Initializing whole multidimensional
// dp array with value -1
for(let i = 0; i < n; i++)
for(let j = 0; j < k + 1; j++)
dp[i][j] = -1;
// For each ith position increasing
// subsequence of length 1 is equal
// to that array ith value so
// initializing dp[i][1] with that
// array value
for(let i = 0; i < n; i++)
{
dp[i][1] = arr[i];
}
// Starting from 1st index as we
// have calculated for 0th index.
// Computing optimized dp values
// in bottom-up manner
for(let i = 1; i < n; i++)
{
for(let j = 0; j < i; j++)
{
// Check for increasing subsequence
if (arr[j] < arr[i])
{
for(let l = 1; l <= k - 1; l++)
{
// Proceed if value is pre calculated
if (dp[j][l] != -1)
{
// Check for all the subsequences
// ending at any j<i and try including
// element at index i in them for
// some length l. Update the maximum
// value for every length.
dp[i][l + 1] = Math.max(dp[i][l + 1],
dp[j][l] +
arr[i]);
}
}
}
}
}
// The final result would be the maximum
// value of dp[i][k] for all different i.
for(let i = 0; i < n; i++)
{
if (ans < dp[i][k])
ans = dp[i][k];
}
// When no subsequence of length k is
// possible sum would be considered zero
return(ans == -1) ? 0 : ans;
} // Driver Codelet n = 8, k = 3; let arr = [ 8, 5, 9, 10, 5, 6, 21, 8 ]; let ans = MaxIncreasingSub(arr, n, k); document.write(ans);// This code is contributed by code_hunt</script> |
40
Time complexity: O(n^2*k)
Space complexity: O(n^2)