Maximize sum of remaining elements after every removal of the array half with greater sum

Given an array arr[] consisting of N integers, the task is to maximize the resultant sum obtained after adding remaining elements after every removal of the array half with maximum sum.

Array can be divided into two non-empty halves left[] and right[] where left[] contains the elements from the indices [0, N/2) and right[] contains the elements from the indices [N/2, N)

Examples:

Input: arr[] = {6, 2, 3, 4, 5, 5} 
Output: 18 
Explanation: 
The given array is arr[] = {6, 2, 3, 4, 5, 5} 
Step 1: 
Sum of left half = 6 + 2 + 3 = 11 
Sum of right half = 4 + 5 + 5 = 12 
Resultant sum S = 11.
Step 2: 
Modified array is arr[] = {6, 2, 3} 
Sum of left half = 6 
Sum of right half = 2 + 3 = 5 
Resultant sum S = 11 + 5 = 16
Step 3: 
Modified array is arr[] = {2, 3} 
Sum of left half = 2 
Sum of right half = 3 
Resultant sum S = 16 + 2 = 18.
Therefore, the resultant sum is 18.

Input: arr[] = {4} 
Output: 0



Naive Approach: The simplest approach to solve the problem is to use recursion. Below are the steps:

  1. Use the concept of prefix sum and initialize a variable, say res to store the final result.
  2. Create a dictionary.
  3. Traverse the array and store all the prefix sum in the dictionary.
  4. Now, iterate over the range [0, N] and store the prefix sum of left and right halves of the array as left and right respectively.
  5. Now, there are three possible conditions: 
    • left > right
    • left < right
    • left == right
  6. For all the above conditions, ignore the maximum sum and add the minimum among left and right sum to the resultant sum and continue the recursive calls.
  7. After all the recursive call ends, print the maximum value of the resultant sum.

Below is the implementation of the above approach:

C++14

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// C++14 program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum sum
int maxweight(int s, int e,
  unordered_map<int, int>& pre)
{
     
    // Base case
    // len of array is 1
    if (s == e)
        return 0;
 
    // Stores the final result
    int ans = 0;
 
    // Traverse the array
    for(int i = s; i < e; i++)
    {
         
        // Store left prefix sum
        int left = pre[i] - pre[s - 1];
 
        // Store right prefix sum
        int right = pre[e] - pre[i];
 
        // Compare the left and right
        if (left < right)
            ans = max(ans, left +
                      maxweight(s, i, pre));
 
        // If both are equal apply
        // the optimal method
        if (left == right)
        {
            // Update with minimum
            ans = max({ans, left +
                       maxweight(s, i, pre),
                             right +
                       maxweight(i + 1, e, pre)});
        }
 
        if (left > right)
            ans = max(ans, right +
                     maxweight(i + 1, e, pre));
    }    
 
    // Return the final ans
    return ans;
}
 
// Function to print maximum sum
void maxSum(vector<int> arr)
{
     
    // Dicitionary to store prefix sums
    unordered_map<int, int> pre;
    pre[-1] = 0;
    pre[0] = arr[0];
 
    // Traversing the array
    for(int i = 1; i < arr.size(); i++)
    {
         
        // Add prefix sum of the array
        pre[i] = pre[i - 1] + arr[i];
    }
    cout << maxweight(0, arr.size() - 1, pre);
}
 
// Driver Code
int main()
{
    vector<int> arr = { 6, 2, 3, 4, 5, 5 };
     
    // Function call
    maxSum(arr);
     
    return 0;
}
 
// This code is contributed by mohit kumar 29

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Java

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// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to find the maximum sum
static int maxweight(int s, int e,
                     Map<Integer, Integer> pre)
{
     
    // Base case
    // len of array is 1
    if (s == e)
        return 0;
 
    // Stores the final result
    int ans = 0;
 
    // Traverse the array
    for(int i = s; i < e; i++)
    {
         
        // Store left prefix sum
        int left = pre.get(i) - pre.get(s - 1);
 
        // Store right prefix sum
        int right = pre.get(e) - pre.get(i);
 
        // Compare the left and right
        if (left < right)
            ans = Math.max(ans, left +
                           maxweight(s, i, pre));
 
        // If both are equal apply
        // the optimal method
        if (left == right)
        {
             
            // Update with minimum
            ans = Math.max(ans, Math.max(left +
                           maxweight(s, i, pre),
                           right + maxweight(i + 1,
                                             e, pre)));
        }
 
        if (left > right)
            ans = Math.max(ans, right +
                           maxweight(i + 1, e, pre));
    }    
     
    // Return the final ans
    return ans;
}
 
// Function to print maximum sum
static void maxSum(List<Integer> arr)
{
     
    // To store prefix sums
    Map<Integer, Integer> pre = new HashMap<>();
    pre.put(-1, 0);
    pre.put(0, arr.get(0));
 
    // Traversing the array
    for(int i = 1; i < arr.size(); i++)
    {
         
        // Add prefix sum of the array
        pre.put(i, pre.getOrDefault(i - 1, 0) +
                   arr.get(i));
    }
    System.out.println(maxweight(0,
                arr.size() - 1, pre));
}
 
// Driver code
public static void main (String[] args)
{
    List<Integer> arr = Arrays.asList(6, 2, 3,
                                      4, 5, 5);
     
    // Function call
    maxSum(arr);
}
}
 
// This code is contributed by offbeat

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Python3

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# Python3 program to implement
# the above approach
 
# Function to find the maximum sum
def maxweight(s, e, pre):
 
    # Base case
    # len of array is 1
    if s == e:
        return 0
 
    # Stores the final result
    ans = 0
 
    # Traverse the array
    for i in range(s, e):
 
        # Store left prefix sum
        left = pre[i] - pre[s - 1]
 
        # Store right prefix sum
        right = pre[e] - pre[i]
 
        # Compare the left and right
        if left < right:
            ans = max(ans, left \
            + maxweight(s, i, pre))
 
        # If both are equal apply
        # the optimal method
        if left == right:
 
            # Update with minimum
            ans = max(ans, left \
                  + maxweight(s, i, pre),
                    right \
                  + maxweight(i + 1, e, pre))
 
        if left > right:
            ans = max(ans, right \
                  + maxweight(i + 1, e, pre))
 
    # Return the final ans
    return ans
 
# Function to print maximum sum
def maxSum(arr):
 
    # Dicitionary to store prefix sums
    pre = {-1: 0, 0: arr[0]}
 
    # Traversing the array
    for i in range(1, len(arr)):
 
        # Add prefix sum of the array
        pre[i] = pre[i - 1] + arr[i]
     
    print(maxweight(0, len(arr) - 1, pre))
 
# Drivers Code
 
arr = [6, 2, 3, 4, 5, 5]
 
# Function Call
maxSum(arr)

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C#

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// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
     
// Function to find the maximum sum
static int maxweight(int s, int e,
                     Dictionary<int,
                     int> pre)
{
  // Base case
  // len of array is 1
  if (s == e)
    return 0;
 
  // Stores the
  // readonly result
  int ans = 0;
 
  // Traverse the array
  for(int i = s; i < e; i++)
  {
    // Store left prefix sum
    int left = pre[i] - pre[s - 1];
 
    // Store right prefix sum
    int right = pre[e] - pre[i];
 
    // Compare the left and right
    if (left < right)
      ans = Math.Max(ans, left +
            maxweight(s, i, pre));
 
    // If both are equal apply
    // the optimal method
    if (left == right)
    {
      // Update with minimum
      ans = Math.Max(ans, Math.Max(left +
                          maxweight(s, i, pre),
                          right + maxweight(i + 1,
                                            e, pre)));
    }
 
    if (left > right)
      ans = Math.Max(ans, right +
            maxweight(i + 1, e, pre));
  }    
 
  // Return the readonly ans
  return ans;
}
 
// Function to print maximum sum
static void maxSum(List<int> arr)
{
     
  // To store prefix sums
  Dictionary<int,
             int> pre = new Dictionary<int,
                                       int>();
  pre.Add(-1, 0);
  pre.Add(0, arr[0]);
 
  // Traversing the array
  for(int i = 1; i < arr.Count; i++)
  {
    // Add prefix sum of the array
    if(pre[i - 1] != 0)
      pre.Add(i, pre[i - 1] + arr[i]);
    else
      pre.Add(i, arr[i]);
  }
  Console.WriteLine(maxweight(0,
                    arr.Count - 1, pre));
}
 
// Driver code
public static void Main(String[] args)
{
  List<int> arr = new List<int>();
  arr.Add(6);
  arr.Add(2);
  arr.Add(3);
  arr.Add(4);
  arr.Add(5);
  arr.Add(5);
 
  // Function call
  maxSum(arr);
}
}
 
// This code is contributed by gauravrajput1

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Output: 

18










Time Complexity: O(2N
Auxiliary Space: O(N2)

Efficient Approach: To optimize the above approach, the idea is to observe that there are a number of repeated overlapping subproblems
Therefore, for optimization, use Dynamic Programming. The idea is to use a dictionary and keep track of the result values so that when they are required in further computations it can be accessed without calculating them again.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
int dp[100][100];
 
// Function to find the maximum sum
int maxweight(int s, int e,
          map<int, int> pre)
{
     
    // Base Case
    if (s == e)
        return 0;
 
    // Create a key to map
    // the values
 
    // Check if (mapped key is
    // found in the dictionary
    if (dp[s][e] != -1)
        return dp[s][e];
 
    int ans = 0;
 
    // Traverse the array
    for(int i = s; i < e; i++)
    {
         
        // Store left prefix sum
        int left = pre[i] - pre[s - 1];
 
        // Store right prefix sum
        int right = pre[e] - pre[i];
 
        // Compare the left and
        // right values
        if (left < right)
            ans = max(
                ans, (int)(left +
                           maxweight(s, i, pre)));
 
        if (left == right)
            ans = max(
                ans, max(left + maxweight(s, i,
                                          pre),
                         right + maxweight(i + 1,
                                           e, pre)));
 
        if (left > right)
            ans = max(
                ans, right + maxweight(i + 1, e, pre));
 
        // Store the value in dp array
        dp[s][e] = ans;
    }
 
    // Return the final answer
    return dp[s][e];
}
 
// Function to print maximum sum
void maxSum(int arr[], int n)
{
     
    // Stores prefix sum
    map<int, int> pre;
    pre[-1] = 0;
    pre[0] = arr[0];
 
    // Store results of subproblems
    memset(dp, -1, sizeof dp);
 
    // Traversing the array
    for(int i = 0; i < n; i++)
 
        // Add prefix sum of array
        pre[i] = pre[i - 1] + arr[i];
 
    // Print the answer
    cout << (maxweight(0, n - 1, pre));
}
 
// Driver Code
int main()
{
    int arr[] = { 6, 2, 3, 4, 5, 5 };
 
    // Function call
    maxSum(arr, 6);
}
 
// This code is contributed by grand_master

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Java

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// Java program to implement
// the above approach
import java.util.*;
 
class solution{
    
static int[][] dp = new int[100][100];
 
// Function to find the maximum sum
static int maxweight(int s, int e,
                     HashMap<Integer, Integer> pre)
{
     
    // Base Case
    if (s == e)
        return 0;
 
    // Create a key to map
    // the values
 
    // Check if (mapped key is
    // found in the dictionary
    if (dp[s][e] != -1)
        return dp[s][e];
 
    int ans = 0;
 
    // Traverse the array
    for(int i = s; i < e; i++)
    {
         
        // Store left prefix sum
        int left = pre.get(i) -
                   pre.get(s - 1);
 
        // Store right prefix sum
        int right = pre.get(e) -
                    pre.get(i);
 
        // Compare the left and
        // right values
        if (left < right)
            ans = Math.max(ans, (int)(left +
                           maxweight(s, i, pre)));
 
        if (left == right)
            ans = Math.max(ans,
                           Math.max(left + maxweight(s, i,
                                                     pre),
                                    right + maxweight(i + 1,
                                                      e, pre)));
 
        if (left > right)
            ans = Math.max(ans, right + maxweight(i + 1,
                                                  e, pre));
 
        // Store the value in dp array
        dp[s][e] = ans;
    }
     
    // Return the final answer
    return dp[s][e];
}
 
// Function to print maximum sum
static void maxSum(int arr[], int n)
{
     
    // Stores prefix sum
    HashMap<Integer,
            Integer> pre = new HashMap<Integer,
                                       Integer>();
    pre.put(-1, 0);
    pre.put(0, arr[0]);
 
    // Store results of subproblems
    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 100; j++)
          dp[i][j] = -1;
    }
 
    // Traversing the array
    for(int i = 0; i < n; i++)
 
        // Add prefix sum of array
        pre.put(i, pre.get(i - 1) + arr[i]);
 
    // Print the answer
    System.out.print((maxweight(0, n - 1, pre)));
}
 
// Driver Code
public static void main(String args[])
{
    int []arr = { 6, 2, 3, 4, 5, 5 };
     
    // Function call
    maxSum(arr, 6);
}
}
 
// This code is contributed by Surendra_Gangwar

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Python3

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# Python3 program to implement
# the above approach
 
# Function to find the maximum sum
def maxweight(s, e, pre, dp):
 
    # Base Case 
    if s == e:
        return 0
 
    # Create a key to map
    # the values
    key = (s, e)
    
    # Check if mapped key is
    # found in the dictionary
    if key in dp:
        return dp[key]
 
    ans = 0
 
    # Traverse the array
    for i in range(s, e):
 
         # Store left prefix sum
        left = pre[i] - pre[s-1]
  
        # Store right prefix sum
        right = pre[e] - pre[i]
 
        # Compare the left and
        # right values
        if left < right:
            ans = max(ans, left \
                + maxweight(s, i, pre, dp))
 
        if left == right:
 
            # Update with minimum
            ans = max(ans, left \
                  + maxweight(s, i, pre, dp),
                  right \
                  + maxweight(i + 1, e, pre, dp))
 
        if left > right:
           ans = max(ans, right \
                 + maxweight(i + 1, e, pre, dp))
 
        # Store the value in dp array
        dp[key] = ans
 
    # Return the final answer
    return dp[key]
 
# Function to print maximum sum
def maxSum(arr):
 
    # Stores prefix sum
    pre = {-1: 0, 0: arr[0]}
 
    # Store results of subproblems
    dp = {}
 
    # Traversing the array
    for i in range(1, len(arr)):
         
        # Add prefix sum of array
        pre[i] = pre[i - 1] + arr[i]
 
    # Print the answer
    print(maxweight(0, len(arr) - 1, pre, dp))
 
# Driver Code
 
arr = [6, 2, 3, 4, 5, 5]
 
# Function Call
maxSum(arr)

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Output: 

18










Time Complexity: O(N3
Auxiliary Space: O(N2)
 

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