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Maximize sum of all elements which are not a part of the Longest Increasing Subsequence

  • Difficulty Level : Hard
  • Last Updated : 29 Jun, 2021

Given an array arr[], the task is to find the maximum sum of all the elements which are not a part of the longest increasing subsequence. 

Examples: 

Input: arr[] = {4, 6, 1, 2, 3, 8} 
Output: 10 
Explanation: 
Elements are 4 and 6 

Input: arr[] = {5, 4, 3, 2, 1} 
Output: 14 
Explanation: 
Elements are 5, 4, 3, 2  

Approach: 



  • The idea is to find the longest increasing subsequence with the minimum sum and then subtract it from the sum of all elements.
  • To do this we will use the concept of LIS using Dynamic Programming and store the sum along with the length of the subsequences and update the minimum sum accordingly.

Below is the implementation of the above approach.

C++




// C++ program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find maximum sum
int findSum(int* arr, int n)
{
    int totalSum = 0;
 
    // Find total sum of array
    for (int i = 0; i < n; i++) {
        totalSum += arr[i];
    }
 
    // Maintain a 2D array
    int dp[2][n];
    for (int i = 0; i < n; i++) {
        dp[0][i] = 1;
        dp[1][i] = arr[i];
    }
 
    // Update the dp array along
    // with sum in the second row
    for (int i = 1; i < n; i++) {
        for (int j = 0; j < i; j++) {
            if (arr[i] > arr[j]) {
                // In case of greater length
                // Update the length along
                // with sum
                if (dp[0][i] < dp[0][j] + 1) {
                    dp[0][i] = dp[0][j] + 1;
                    dp[1][i] = dp[1][j]
                               + arr[i];
                }
 
                // In case of equal length
                // find length update length
                // with minimum sum
                else if (dp[0][i]
                         == dp[0][j] + 1) {
                    dp[1][i]
                        = min(dp[1][i],
                              dp[1][j]
                                  + arr[i]);
                }
            }
        }
    }
    int maxm = 0;
    int subtractSum = 0;
 
    // Find the sum that need to
    // be subtracted from total sum
    for (int i = 0; i < n; i++) {
        if (dp[0][i] > maxm) {
            maxm = dp[0][i];
            subtractSum = dp[1][i];
        }
        else if (dp[0][i] == maxm) {
            subtractSum = min(subtractSum,
                              dp[1][i]);
        }
    }
 
    // Return the sum
    return totalSum - subtractSum;
}
 
// Driver code
int main()
{
    int arr[] = { 4, 6, 1, 2, 3, 8 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << findSum(arr, n);
 
    return 0;
}

Java




// Java program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
class GFG{
 
// Function to find maximum sum
static int findSum(int []arr, int n)
{
    int totalSum = 0;
 
    // Find total sum of array
    for(int i = 0; i < n; i++)
    {
       totalSum += arr[i];
    }
 
    // Maintain a 2D array
    int [][]dp = new int[2][n];
    for(int i = 0; i < n; i++)
    {
       dp[0][i] = 1;
       dp[1][i] = arr[i];
    }
 
    // Update the dp array along
    // with sum in the second row
    for(int i = 1; i < n; i++)
    {
       for(int j = 0; j < i; j++)
       {
          if (arr[i] > arr[j])
          {
               
              // In case of greater length
              // Update the length along
              // with sum
              if (dp[0][i] < dp[0][j] + 1)
              {
                  dp[0][i] = dp[0][j] + 1;
                  dp[1][i] = dp[1][j] + arr[i];
              }
               
              // In case of equal length
              // find length update length
              // with minimum sum
              else if (dp[0][i] == dp[0][j] + 1)
              {
                  dp[1][i] = Math.min(dp[1][i],
                                      dp[1][j] + arr[i]);
              }
          }
       }
    }
    int maxm = 0;
    int subtractSum = 0;
 
    // Find the sum that need to
    // be subtracted from total sum
    for(int i = 0; i < n; i++)
    {
       if (dp[0][i] > maxm)
       {
           maxm = dp[0][i];
           subtractSum = dp[1][i];
       }
       else if (dp[0][i] == maxm)
       {
           subtractSum = Math.min(subtractSum, dp[1][i]);
       }
    }
 
    // Return the sum
    return totalSum - subtractSum;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 4, 6, 1, 2, 3, 8 };
    int n = arr.length;
 
    System.out.print(findSum(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Python3




# Python3 program to find the maximum sum 
# of all elements which are not a part of
# longest increasing sub sequence
 
# Function to find maximum sum
def findSum(arr, n):
 
    totalSum = 0
 
    # Find total sum of array
    for i in range(n):
        totalSum += arr[i]
 
    # Maintain a 2D array
    dp = [[0] * n for i in range(2)]
 
    for i in range(n):
        dp[0][i] = 1
        dp[1][i] = arr[i]
 
    # Update the dp array along
    # with sum in the second row
    for i in range(1, n):
        for j in range(i):
            if (arr[i] > arr[j]):
 
                # In case of greater length
                # update the length along
                # with sum
                if (dp[0][i] < dp[0][j] + 1):
                    dp[0][i] = dp[0][j] + 1
                    dp[1][i] = dp[1][j] + arr[i]
 
                # In case of equal length
                # find length update length
                # with minimum sum
                elif (dp[0][i] == dp[0][j] + 1):
                    dp[1][i] = min(dp[1][i],
                                   dp[1][j] +
                                     arr[i])
 
    maxm = 0
    subtractSum = 0
 
    # Find the sum that need to
    # be subtracted from total sum
    for i in range(n):
        if (dp[0][i] > maxm):
            maxm = dp[0][i]
            subtractSum = dp[1][i]
 
        elif (dp[0][i] == maxm):
            subtractSum = min(subtractSum,
                              dp[1][i])
 
    # Return the sum
    return totalSum - subtractSum
 
# Driver code
arr = [ 4, 6, 1, 2, 3, 8 ]
n = len(arr)
 
print(findSum(arr, n))
 
# This code is contributed by himanshu77

C#




// C# program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
using System;
class GFG{
 
// Function to find maximum sum
static int findSum(int []arr, int n)
{
    int totalSum = 0;
 
    // Find total sum of array
    for(int i = 0; i < n; i++)
    {
        totalSum += arr[i];
    }
 
    // Maintain a 2D array
    int [,]dp = new int[2, n];
    for(int i = 0; i < n; i++)
    {
        dp[0, i] = 1;
        dp[1, i] = arr[i];
    }
 
    // Update the dp array along
    // with sum in the second row
    for(int i = 1; i < n; i++)
    {
        for(int j = 0; j < i; j++)
        {
            if (arr[i] > arr[j])
            {
                     
                // In case of greater length
                // Update the length along
                // with sum
                if (dp[0, i] < dp[0, j] + 1)
                {
                    dp[0, i] = dp[0, j] + 1;
                    dp[1, i] = dp[1, j] + arr[i];
                }
                     
                // In case of equal length
                // find length update length
                // with minimum sum
                else if (dp[0, i] == dp[0, j] + 1)
                {
                    dp[1, i] = Math.Min(dp[1, i],
                                        dp[1, j] + arr[i]);
                }
            }
        }
    }
    int maxm = 0;
    int subtractSum = 0;
 
    // Find the sum that need to
    // be subtracted from total sum
    for(int i = 0; i < n; i++)
    {
        if (dp[0, i] > maxm)
        {
            maxm = dp[0, i];
            subtractSum = dp[1, i];
        }
        else if (dp[0, i] == maxm)
        {
            subtractSum = Math.Min(subtractSum, dp[1, i]);
        }
    }
 
    // Return the sum
    return totalSum - subtractSum;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 4, 6, 1, 2, 3, 8 };
    int n = arr.Length;
 
    Console.Write(findSum(arr, n));
}
}
 
// This code is contributed by sapnasingh4991

Javascript




<script>
 
// Javascript program to find the Maximum sum of
// all elements which are not a part of
// longest increasing sub sequence
 
// Function to find maximum sum
function findSum(arr, n)
{
    var totalSum = 0;
 
    // Find total sum of array
    for(var i = 0; i < n; i++)
    {
        totalSum += arr[i];
    }
 
    // Maintain a 2D array
    var dp = Array.from(Array(2), () => Array(n));
    for(var i = 0; i < n; i++)
    {
        dp[0][i] = 1;
        dp[1][i] = arr[i];
    }
 
    // Update the dp array along
    // with sum in the second row
    for(var i = 1; i < n; i++)
    {
        for(var j = 0; j < i; j++)
        {
            if (arr[i] > arr[j])
            {
                 
                // In case of greater length
                // Update the length along
                // with sum
                if (dp[0][i] < dp[0][j] + 1)
                {
                    dp[0][i] = dp[0][j] + 1;
                    dp[1][i] = dp[1][j] + arr[i];
                }
 
                // In case of equal length
                // find length update length
                // with minimum sum
                else if (dp[0][i] == dp[0][j] + 1)
                {
                    dp[1][i] = Math.min(dp[1][i],
                               dp[1][j] + arr[i]);
                }
            }
        }
    }
    var maxm = 0;
    var subtractSum = 0;
 
    // Find the sum that need to
    // be subtracted from total sum
    for(var i = 0; i < n; i++)
    {
        if (dp[0][i] > maxm)
        {
            maxm = dp[0][i];
            subtractSum = dp[1][i];
        }
        else if (dp[0][i] == maxm)
        {
            subtractSum = Math.min(subtractSum,
                                   dp[1][i]);
        }
    }
 
    // Return the sum
    return totalSum - subtractSum;
}
 
// Driver code
var arr = [ 4, 6, 1, 2, 3, 8 ];
var n = arr.length;
 
document.write( findSum(arr, n));
 
// This code is contributed by rrrtnx
 
</script>
Output: 
10

 

Time Complexity: O(N2) where N is the length of the array arr[]
Auxiliary Space: O(N)

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