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++
#include <bits/stdc++.h>
#define arrSize 51
using namespace std;
int dp[arrSize];
bool v[arrSize];
int sumMax(int i, int arr[], int n)
{
if (i >= n - 1)
return 0;
if (v[i])
return dp[i];
v[i] = true;
dp[i] = max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
return dp[i];
}
int main()
{
int arr[] = { 1, 1, 1, 1 };
int n = sizeof(arr) / sizeof(int);
cout << sumMax(0, arr, n);
return 0;
}
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Java
import java.io.*;
class GFG
{
static int arrSize = 51;
static int dp[] = new int[arrSize];
static boolean v[] = new boolean[arrSize];
static int sumMax(int i, int arr[], int n)
{
if (i >= n - 1)
return 0;
if (v[i])
return dp[i];
v[i] = true;
dp[i] = Math.max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
return dp[i];
}
public static void main (String[] args)
{
int arr[] = { 1, 1, 1, 1 };
int n = arr.length;
System.out.println(sumMax(0, arr, n));
}
}
|
Python3
arrSize = 51
dp = [0 for i in range(arrSize)]
v = [False for i in range(arrSize)]
def sumMax(i,arr,n):
if (i >= n - 1):
return 0
if (v[i]):
return dp[i]
v[i] = True
dp[i] = max(arr[i] + arr[i + 1] + sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n))
return dp[i]
if __name__ == '__main__':
arr = [1, 1, 1, 1]
n = len(arr)
print(sumMax(0, arr, n))
|
C#
using System;
class GFG
{
static int arrSize = 51;
static int []dp = new int[arrSize];
static bool []v = new bool[arrSize];
static int sumMax(int i, int []arr, int n)
{
if (i >= n - 1)
return 0;
if (v[i])
return dp[i];
v[i] = true;
dp[i] = Math.Max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
return dp[i];
}
public static void Main ()
{
int []arr = { 1, 1, 1, 1 };
int n = arr.Length;
Console.WriteLine(sumMax(0, arr, n));
}
}
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Javascript
<script>
var arrSize = 51;
var dp = Array(arrSize);
var v = Array(arrSize);
function sumMax(i, arr, n)
{
if (i >= n - 1)
return 0;
if (v[i])
return dp[i];
v[i] = true;
dp[i] = Math.max(arr[i] + arr[i + 1]
+ sumMax(i + 3, arr, n),
sumMax(i + 1, arr, n));
return dp[i];
}
var arr = [1, 1, 1, 1];
var n = arr.length;
document.write( sumMax(0, arr, n));
</script>
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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++
#include <bits/stdc++.h>
#define arrSize 51
using namespace std;
int dp[arrSize];
int sumMax(int arr[], int n) {
dp[n-1] = dp[n] = dp[n+1] = 0;
for(int i=n-2; i>=0; i--) {
dp[i] = max(arr[i] + arr[i+1] + dp[i+3], dp[i+1]);
}
return dp[0];
}
int main()
{
int arr[] = {1,1,1,1};
int n = sizeof(arr) / sizeof(int);
cout << sumMax(arr, n);
return 0;
}
|
Java
import java.util.Arrays;
public class Main {
private static final int arrSize = 51;
private static int[] dp = new int[arrSize];
private static int sumMax(int[] arr, int n) {
dp[n - 1] = dp[n] = dp[n + 1] = 0;
for (int i = n - 2; i >= 0; i--) {
dp[i] = Math.max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
}
return dp[0];
}
public static void main(String[] args) {
int[] arr = {1, 1, 1, 1};
int n = arr.length;
System.out.println(sumMax(arr, n));
}
}
|
Python
import sys
sys.setrecursionlimit(10**7)
dp = [0] * 51
def sumMax(arr, n):
dp[n-1] = dp[n] = dp[n+1] = 0
for i in range(n-2, -1, -1):
dp[i] = max(arr[i] + arr[i+1] + dp[i+3], dp[i+1])
return dp[0]
if __name__ == '__main__':
arr = [1,1,1,1]
n = len(arr)
print(sumMax(arr, n))
|
C#
using System;
public class Program
{
const int arrSize = 51;
static int[] dp = new int[arrSize];
static int sumMax(int[] arr, int n)
{
dp[n - 1] = dp[n] = dp[n + 1] = 0;
for (int i = n - 2; i >= 0; i--) {
dp[i] = Math.Max(
arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
}
return dp[0];
}
public static void Main()
{
int[] arr = { 1, 1, 1, 1 };
int n = arr.Length;
Console.WriteLine(sumMax(arr, n));
}
}
|
Javascript
function sumMax(arr, n) {
const dp = Array(n + 2).fill(0);
for (let i = n - 2; i >= 0; i--) {
dp[i] = Math.max(arr[i] + arr[i + 1] + dp[i + 3], dp[i + 1]);
}
return dp[0];
}
const arr = [1, 1, 1, 1];
const n = arr.length;
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.