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++
// C++ implementation of the approach#include <bits/stdc++.h>#define arrSize 51using namespace std;// To store the states of dpint dp[arrSize];bool v[arrSize];// Function to return the maximized sumint 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 codeint main(){ int arr[] = { 1, 1, 1, 1 }; int n = sizeof(arr) / sizeof(int); cout << sumMax(0, arr, n); return 0;} |
Java
// Java implementation of the approachimport java.io.*;class GFG {static int arrSize = 51;// To store the states of dpstatic int dp[] = new int[arrSize];static boolean v[] = new boolean[arrSize];// Function to return the maximized sumstatic 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 codepublic 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.. |
Python3
# Python 3 implementation of the approacharrSize = 51# To store the states of dpdp = [0 for i in range(arrSize)]v = [False for i in range(arrSize)]# Function to return the maximized sumdef 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 codeif __name__ == '__main__': arr = [1, 1, 1, 1] n = len(arr) print(sumMax(0, arr, n))# This code is contributed by# Surendra_Gangwar |
C#
// C# implementation of the approachusing System;class GFG {static int arrSize = 51;// To store the states of dpstatic int []dp = new int[arrSize];static bool []v = new bool[arrSize];// Function to return the maximized sumstatic 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 codepublic 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.. |
Javascript
<script>// Javascript implementation of the approachvar arrSize = 51;// To store the states of dpvar dp = Array(arrSize);var v = Array(arrSize);// Function to return the maximized sumfunction 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 codevar 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++
// C++ implementation of the approach#include <bits/stdc++.h>#define arrSize 51using namespace std;// To store the states of dpint dp[arrSize];// Function to return the maximized sumint 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 Codeint main(){ int arr[] = {1,1,1,1}; int n = sizeof(arr) / sizeof(int); // function call cout << sumMax(arr, n); return 0;} |
Java
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
# Python implementation of the approachimport syssys.setrecursionlimit(10**7)# To store the states of dpdp = [0] * 51# Function to return the maximized sumdef 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 Codeif __name__ == '__main__': arr = [1,1,1,1] n = len(arr) # function call print(sumMax(arr, n)) |
C#
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 |
Javascript
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 Codeconst arr = [1, 1, 1, 1];const n = arr.length;// Function callconsole.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.
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