Given an array arr[] of N integers, you can select some indexes such that every selected index has exactly one other selected index adjacent to it and the sum of elements at the chosen indexes should be maximum.
In other words, the task is to select elements from an array such that a single element alone is not selected and elements at three consecutive indices are not selected and the sum of selected elements should be maximum.
The task is to print the maximized sum.
Examples:
Input: arr[] = {1, 2, 3, 1, 4}
Output: 8
arr[0] + arr[1] + arr[3] + arr[4] = 1 + 2 + 1 + 4 = 8
Input: arr[] = {1, 1, 1, 1}
Output: 2
Approach: Dynamic programming can be used to solve this problem. This problem can be translated to selecting pairs of adjacent integers such that no two pairs are adjacent or have an element in common
i.e. if (arr[i], arr[i + 1]) is a pair we selected then neither (arr[i + 2], arr[i + 3]) nor (arr[i + 1], arr[i + 2]) can be selected.
Let’s decide the states of the dp according to the above statement.
For every index i, we will either select indexes i and i + 1 i.e. make a pair or not make it. In case, we make a pair, we won’t be able to select the index i + 2 as it will make 2 elements adjacent to i + 1. So, we will have to solve for i + 3 next. If we don’t make a pair, we will simply solve for i + 1.
So the recurrence relation will be.
dp[i] = max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1])
There are N states in total and each state takes O(1) time to solve. Thus, the time complexity will be O(N).
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> #define arrSize 51 using namespace std;
// To store the states of dp int dp[arrSize];
bool v[arrSize];
// Function to return the maximized sum int sumMax( int i, int arr[], int n)
{ // Base case
if (i >= n - 1)
return 0;
// Checks if a state is
// already solved
if (v[i])
return dp[i];
v[i] = true ;
// Recurrence relation
dp[i] = max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
// Return the result
return dp[i];
} // Driver code int main()
{ int arr[] = { 1, 1, 1, 1 };
int n = sizeof (arr) / sizeof ( int );
cout << sumMax(0, arr, n);
return 0;
} |
// Java implementation of the approach import java.io.*;
class GFG
{ static int arrSize = 51 ;
// To store the states of dp static int dp[] = new int [arrSize];
static boolean v[] = new boolean [arrSize];
// Function to return the maximized sum static int sumMax( int i, int arr[], int n)
{ // Base case
if (i >= n - 1 )
return 0 ;
// Checks if a state is
// already solved
if (v[i])
return dp[i];
v[i] = true ;
// Recurrence relation
dp[i] = Math.max(arr[i] + arr[i + 1 ]
+ sumMax(i + 3 , arr, n),
sumMax(i + 1 , arr, n));
// Return the result
return dp[i];
} // Driver code public static void main (String[] args)
{ int arr[] = { 1 , 1 , 1 , 1 };
int n = arr.length;
System.out.println(sumMax( 0 , arr, n));
} } // This code is contributed by anuj_67.. |
# Python 3 implementation of the approach arrSize = 51
# To store the states of dp dp = [ 0 for i in range (arrSize)]
v = [ False for i in range (arrSize)]
# Function to return the maximized sum def sumMax(i,arr,n):
# Base case
if (i > = n - 1 ):
return 0
# Checks if a state is
# already solved
if (v[i]):
return dp[i]
v[i] = True
# Recurrence relation
dp[i] = max (arr[i] + arr[i + 1 ] + sumMax(i + 3 , arr, n),
sumMax(i + 1 , arr, n))
# Return the result
return dp[i]
# Driver code if __name__ = = '__main__' :
arr = [ 1 , 1 , 1 , 1 ]
n = len (arr)
print (sumMax( 0 , arr, n))
# This code is contributed by # Surendra_Gangwar |
// C# implementation of the approach using System;
class GFG
{ static int arrSize = 51;
// To store the states of dp static int []dp = new int [arrSize];
static bool []v = new bool [arrSize];
// Function to return the maximized sum static int sumMax( int i, int []arr, int n)
{ // Base case
if (i >= n - 1)
return 0;
// Checks if a state is
// already solved
if (v[i])
return dp[i];
v[i] = true ;
// Recurrence relation
dp[i] = Math.Max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
// Return the result
return dp[i];
} // Driver code public static void Main ()
{ int []arr = { 1, 1, 1, 1 };
int n = arr.Length;
Console.WriteLine(sumMax(0, arr, n));
} } // This code is contributed by anuj_67.. |
<script> // Javascript implementation of the approach var arrSize = 51;
// To store the states of dp var dp = Array(arrSize);
var v = Array(arrSize);
// Function to return the maximized sum function sumMax(i, arr, n)
{ // Base case
if (i >= n - 1)
return 0;
// Checks if a state is
// already solved
if (v[i])
return dp[i];
v[i] = true ;
// Recurrence relation
dp[i] = Math.max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
// Return the result
return dp[i];
} // Driver code var arr = [1, 1, 1, 1];
var n = arr.length;
document.write( sumMax(0, arr, n)); </script> |
2
Time Complexity: O(N), where N is the given size of the array.
Auxiliary Space: O(51), no extra space is required, so it is a constant.
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a DP to store the solution of the subproblems.
- Initialize the DP with base cases
- Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP
- Return the final solution stored in dp[0].
Implementation :
// C++ implementation of the approach #include <bits/stdc++.h> #define arrSize 51 using namespace std;
// To store the states of dp int dp[arrSize];
// Function to return the maximized sum int sumMax( int arr[], int n) {
// Initializing the base cases
dp[n-1] = dp[n] = dp[n+1] = 0;
// Calculating the dp table
for ( int i=n-2; i>=0; i--) {
dp[i] = max(arr[i] + arr[i+1] + dp[i+3], dp[i+1]);
}
// Return the result
return dp[0];
} // Driver Code int main()
{ int arr[] = {1,1,1,1};
int n = sizeof (arr) / sizeof ( int );
// function call
cout << sumMax(arr, n);
return 0;
} |
import java.util.Arrays;
public class Main {
// To store the states of dp
private static final int arrSize = 51 ;
private static int [] dp = new int [arrSize];
// Function to return the maximized sum
private static int sumMax( int [] arr, int n) {
// Initializing the base cases
dp[n - 1 ] = dp[n] = dp[n + 1 ] = 0 ;
// Calculating the dp table
for ( int i = n - 2 ; i >= 0 ; i--) {
dp[i] = Math.max(arr[i] + arr[i + 1 ] + dp[i + 3 ], dp[i + 1 ]);
}
// Return the result
return dp[ 0 ];
}
// Driver Code
public static void main(String[] args) {
int [] arr = { 1 , 1 , 1 , 1 };
int n = arr.length;
// function call
System.out.println(sumMax(arr, n));
}
} |
# Python implementation of the approach import sys
sys.setrecursionlimit( 10 * * 7 )
# To store the states of dp dp = [ 0 ] * 51
# Function to return the maximized sum def sumMax(arr, n):
# Initializing the base cases
dp[n - 1 ] = dp[n] = dp[n + 1 ] = 0
# Calculating the dp table
for i in range (n - 2 , - 1 , - 1 ):
dp[i] = max (arr[i] + arr[i + 1 ] + dp[i + 3 ], dp[i + 1 ])
# Return the result
return dp[ 0 ]
# Driver Code if __name__ = = '__main__' :
arr = [ 1 , 1 , 1 , 1 ]
n = len (arr)
# function call
print (sumMax(arr, n))
|
using System;
public class Program
{ // To store the states of dp
const int arrSize = 51;
static int [] dp = new int [arrSize];
// Function to return the maximized sum
static int sumMax( int [] arr, int n)
{
// Initializing the base cases
dp[n - 1] = dp[n] = dp[n + 1] = 0;
// Calculating the dp table
for ( int i = n - 2; i >= 0; i--) {
dp[i] = Math.Max(
arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
}
// Return the result
return dp[0];
}
public static void Main()
{
int [] arr = { 1, 1, 1, 1 };
int n = arr.Length;
// function call
Console.WriteLine(sumMax(arr, n));
}
} // This code is contributed by user_dtewbxkn77n |
function sumMax(arr, n) {
// Initializing the base cases
const dp = Array(n + 2).fill(0);
// Calculating the dp table
for (let i = n - 2; i >= 0; i--) {
dp[i] = Math.max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
}
// Return the result
return dp[0];
} // Driver Code const arr = [1, 1, 1, 1]; const n = arr.length; // Function call console.log(sumMax(arr, n)); |
Output:
2
Time Complexity: O(N), where N is the given size of the array.
Auxiliary Space: O(51), no extra space is required, so it is a constant.