Given an array arr[] of positive numbers, The task is to find the maximum sum of a subsequence such that no 2 numbers in the sequence should be adjacent in the array.
Examples:
Input: arr[] = {5, 5, 10, 100, 10, 5}
Output: 110
Explanation: Pick the subsequence {5, 100, 5}.
The sum is 110 and no two elements are adjacent. This is the highest possible sum.Input: arr[] = {3, 2, 7, 10}
Output: 13
Explanation: The subsequence is {3, 10}. This gives sum = 13.
This is the highest possible sum of a subsequence following the given criteriaInput: arr[] = {3, 2, 5, 10, 7}
Output: 15
Explanation: Pick the subsequence {3, 5, 7}. The sum is 15.
Naive Approach: Below is the idea to solve the problem:
Each element has two choices: either it can be the part of the subsequence with the highest sum or it cannot be part of the subsequence. So to solve the problem, build all the subsequences of the array and find the subsequence with the maximum sum such that no two adjacent elements are present in the subsequence.
- The robber has two options: to rob the current house (option “a”) or not to rob it (option “b”).
- If the robber chooses option “a”, they cannot rob the previous house (i-1), but can proceed to the one before that (i-2) and receive all the accumulated loot.
- If option “b” is selected, the robber will receive all the potential loot from the previous house (i-1) and all the houses before it.
- The decision of which option to choose ultimately comes down to determining which is more profitable: the loot obtained from the current house and previous houses before the one before the last, or the loot obtained from the previous house and all prior houses.
- The formula used to make this calculation is: rob(i) = Math.max( rob(i – 2) + currentHouseValue, rob(i – 1) ).
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum sumint rec(vector<int>& nums, int idx){ if (idx >= nums.size()) return 0; return max(nums[idx] + rec(nums, idx + 2), rec(nums, idx + 1));}int findMaxSum(vector<int> arr, int N){ return rec(arr, 0);}// Driver Codeint main(){ // Creating the array vector<int> arr = { 5, 5, 10, 100, 10, 5 }; int N = arr.size(); // Function call cout << findMaxSum(arr, N) << endl; return 0;} |
C
// C code to implement the approach#include <stdio.h>// Function to find the maximum sumint rec(int* nums, int idx, int size){ if (idx >= size) return 0; return (nums[idx] + rec(nums, idx + 2, size) > rec(nums, idx + 1, size)) ? nums[idx] + rec(nums, idx + 2, size) : rec(nums, idx + 1, size);}int findMaxSum(int* arr, int N) { return rec(arr, 0, N); }// Driver Codeint main(){ // Creating the array int arr[] = { 5, 5, 10, 100, 10, 5 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call printf("%d", findMaxSum(arr, N)); return 0;} |
Java
/*package whatever //do not write package name here */import java.util.*;class GFG{ // Function to find the maximum sum static int rec(int nums[], int idx,int N) { if (idx >= N) return 0; return Math.max(nums[idx] + rec(nums, idx + 2, N), rec(nums, idx + 1, N)); } // Function to find the maximum sum static int findMaxSum(int[] arr, int N) { return rec(arr, 0, N); } // Driver Code public static void main(String args[]) { // Creating the array int[] arr = { 5, 5, 10, 100, 10, 5 }; int N = 6; // Function call System.out.println(findMaxSum(arr, N)); }}// This code is contributed by shubhamrajput6156 |
Python3
# Function to find the maximum sumdef rec(nums, idx): if idx >= len(nums): return 0 return max(nums[idx] + rec(nums, idx + 2), rec(nums, idx + 1))def findMaxSum(arr, N): return rec(arr, 0)# Driver Codeif __name__ == "__main__": # Creating the array arr = [5, 5, 10, 100, 10, 5] N = len(arr) # Function call print(findMaxSum(arr, N)) |
C#
// C# code to implement the approachusing System;using System.Collections.Generic;class Program{// Function to find the maximum sumstatic int Rec(List<int> nums, int idx){if (idx >= nums.Count)return 0;return Math.Max(nums[idx] + Rec(nums, idx + 2),Rec(nums, idx + 1));} static int FindMaxSum(List<int> arr, int N){ return Rec(arr, 0);}static void Main(string[] args){ // Creating the list List<int> arr = new List<int>() { 5, 5, 10, 100, 10, 5 }; int N = arr.Count; // Function call Console.WriteLine(FindMaxSum(arr, N));}}//This code is contributed by shivamsharma215 |
Javascript
// Javascript code to implement the approach// Function to find the maximum sumfunction rec(nums, idx){ if (idx >= nums.length) return 0; return Math.max(nums[idx] + rec(nums, idx + 2), rec(nums, idx + 1));}function findMaxSum(arr, N){ return rec(arr, 0);}// Driver Code// Creating the arrayarr = [5, 5, 10, 100, 10, 5];N = arr.length;// Function callconsole.log(findMaxSum(arr, N));// The code is contributed by Nidhi goel. |
Output
110
Time Complexity: O(2N) where N is the number of elements in A. At each index, we have two choices of either robbing or not robbing the current house. Thus this leads to a time complexity of 222…n times ≈ O(2^N)
Auxiliary Space: O(N) It is recursive stack space.
Maximum sum such that no two elements are adjacent using Dynamic Programming (Memoization):
We can cache the overlapping subproblems
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum sumint rec(vector<int>& nums, int idx, vector<int>& dp){ if (idx >= nums.size()) return 0; if (dp[idx] != -1) return dp[idx]; return dp[idx] = max(rec(nums, idx + 1, dp), nums[idx] + rec(nums, idx + 2, dp));}int findMaxSum(vector<int>& nums, int N){ vector<int> dp(N + 1, -1); return rec(nums, 0, dp);}// Driver Codeint main(){ // Creating the array vector<int> arr = { 5, 5, 10, 100, 10, 5 }; int N = arr.size(); // Function call cout << findMaxSum(arr, N) << endl; return 0;} |
C
// C code to implement the approach#include <stdio.h>#include <stdlib.h>// Function to find the maximum sumint rec(int* nums, int idx, int* dp, int size){ if (idx >= size) return 0; if (dp[idx] != -1) return dp[idx]; return dp[idx] = (rec(nums, idx + 1, dp, size) > nums[idx] + rec(nums, idx + 2, dp, size)) ? rec(nums, idx + 1, dp, size) : nums[idx] + rec(nums, idx + 2, dp, size);}int findMaxSum(int* nums, int N){ int* dp = (int*)malloc((N + 1) * sizeof(int)); for (int i = 0; i < N + 1; i++) dp[i] = -1; int result = rec(nums, 0, dp, N); free(dp); return result;}// Driver Codeint main(){ // Creating the array int arr[] = { 5, 5, 10, 100, 10, 5 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call printf("%d\n", findMaxSum(arr, N)); return 0;} |
Java
import java.util.*;public class Main { // Function to find the maximum sum public static int rec(List<Integer> nums, int idx, int[] dp) { // base case: if the current index is out of bounds, return 0 if (idx >= nums.size()) { return 0; } // if the result for the current index has already been computed, return it if (dp[idx] != -1) { return dp[idx]; } // recursive case: compute the maximum sum by either skipping the current element or including it dp[idx] = Math.max(rec(nums, idx + 1, dp), nums.get(idx) + rec(nums, idx + 2, dp)); return dp[idx]; } public static int findMaxSum(List<Integer> nums, int N) { int[] dp = new int[N + 1]; Arrays.fill(dp, -1); return rec(nums, 0, dp); } public static void main(String[] args) { // Creating the array List<Integer> arr = Arrays.asList(5, 5, 10, 100, 10, 5); int N = arr.size(); // Function call System.out.println(findMaxSum(arr, N)); }} |
Python3
def findMaxSum(nums, N): dp = [-1] * (N + 1) def rec(idx, dp): if idx >= len(nums): return 0 if dp[idx] != -1: return dp[idx] dp[idx] = max(rec(idx + 1, dp), nums[idx] + rec(idx + 2, dp)) return dp[idx] return rec(0, dp)# Driver Codearr = [5, 5, 10, 100, 10, 5]N = len(arr)print(findMaxSum(arr, N)) |
C#
using System;using System.Collections.Generic;class GFG{ // Function to find the maximum sum static int Rec(List<int> nums, int idx, List<int> dp) { if (idx >= nums.Count) return 0; if (dp[idx] != -1) return dp[idx]; return dp[idx] = Math.Max(Rec(nums, idx + 1, dp), nums[idx] + Rec(nums, idx + 2, dp)); } static int FindMaxSum(List<int> nums, int N) { List<int> dp = new List<int>(N + 1); for (int i = 0; i <= N; i++) { dp.Add(-1); } return Rec(nums, 0, dp); } static void Main(string[] args) { // Creating the list List<int> arr = new List<int>() { 5, 5, 10, 100, 10, 5 }; int N = arr.Count; // Function call Console.WriteLine(FindMaxSum(arr, N)); }} |
Javascript
function findMaxSum(nums, N) { const dp = new Array(N + 1).fill(-1); function rec(idx, dp) { if (idx >= nums.length) { return 0; } if (dp[idx] !== -1) { return dp[idx]; } dp[idx] = Math.max(rec(idx + 1, dp), nums[idx] + rec(idx + 2, dp)); return dp[idx]; } return rec(0, dp);}// Driver Codeconst arr = [5, 5, 10, 100, 10, 5];const N = arr.length;console.log(findMaxSum(arr, N)); |
Output
110
Time complexity: O(N) (recursion)
Space complexity: O(N)+O(N), one is recursive stack space and another O(N) is for dp array.
Maximum sum such that no two elements are adjacent using Dynamic Programming:
- As seen above, each element has two choices. If one element is picked then its neighbours cannot be picked. Otherwise, its neighbours may be picked or may not be.
- So the maximum sum till ith index has two possibilities: the subsequence includes arr[i] or it does not include arr[i].
- If arr[i] is included then the maximum sum depends on the maximum subsequence sum till (i-1)th element excluding arr[i-1].
- Otherwise, the maximum sum is the same as the maximum subsequence sum till (i-1) where arr[i-1] may be included or excluded.
So build a 2D dp[N][2] array where dp[i][0] stores maximum subsequence sum till ith index with arr[i] excluded and dp[i][1] stores the sum when arr[i] is included.
The values will be obtained by the following relations: dp[i][1] = dp[i-1][0] + arr[i] and dp[i][0] = max(dp[i-1][0], dp[i-1][1])
Follow the steps mentioned below to implement the above idea:
- If the size of the array is 1, then the answer is arr[0].
- Initialize the values of dp[0][0] = 0 and dp[0][1] = arr[0].
- Iterate from i = 1 to N-1:
- Fill the dp array as per the relation shown above.
- Return the maximum between dp[N-1][1] and dp[N-1][0] as that will be the answer.
Below is the implementation of the above approach.
C++
// C++ code to implement the approach#include <bits/stdc++.h>using namespace std;// Function to find the maximum sumint findMaxSum(vector<int> arr, int N){ // Declare dp array int dp[N][2]; if (N == 1) { return arr[0]; } // Initialize the values in dp array dp[0][0] = 0; dp[0][1] = arr[0]; // Loop to find the maximum possible sum for (int i = 1; i < N; i++) { dp[i][1] = dp[i - 1][0] + arr[i]; dp[i][0] = max(dp[i - 1][1], dp[i - 1][0]); } // Return the maximum sum return max(dp[N - 1][0], dp[N - 1][1]);}// Driver Codeint main(){ // Creating the array vector<int> arr = { 5, 5, 10, 100, 10, 5 }; int N = arr.size(); // Function call cout << findMaxSum(arr, N) << endl; return 0;} |
C
// C code to implement the approach#include <stdio.h>// Function to find the maximum sumint findMaxSum(int arr[], int N){ // Declare dp array int dp[N][2]; if (N == 1) { return arr[0]; } // Initialize the values in dp array dp[0][0] = 0; dp[0][1] = arr[0]; // Loop to find the maximum possible sum for (int i = 1; i < N; i++) { dp[i][1] = dp[i - 1][0] + arr[i]; dp[i][0] = (dp[i - 1][1] > dp[i - 1][0]) ? dp[i - 1][1] : dp[i - 1][0]; } // Return the maximum sum return (dp[N - 1][0] > dp[N - 1][1]) ? dp[N - 1][0] : dp[N - 1][1];}// Driver Codeint main(){ // Creating the array int arr[] = { 5, 5, 10, 100, 10, 5 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call printf("%d\n", findMaxSum(arr, N)); return 0;} |
Java
/*package whatever //do not write package name here */import java.io.*;class GFG{ // Function to find the maximum sum static int findMaxSum(int[] arr, int N) { // Declare dp array int[][] dp = new int[N][2]; if (N == 1) { return arr[0]; } // Initialize the values in dp array dp[0][0] = 0; dp[0][1] = arr[0]; // Loop to find the maximum possible sum for (int i = 1; i < N; i++) { dp[i][1] = dp[i - 1][0] + arr[i]; dp[i][0] = Math.max(dp[i - 1][1], dp[i - 1][0]); } // Return the maximum sum return Math.max(dp[N - 1][0], dp[N - 1][1]); } // Driver Code public static void main(String args[]) { // Creating the array int[] arr = { 5, 5, 10, 100, 10, 5 }; int N = arr.length; // Function call System.out.println(findMaxSum(arr, N)); }}// This code is contributed by shinjanpatra |
Python3
# Python code to implement the approach# Function to find the maximum sumdef findMaxSum(arr, N): # Declare dp array dp = [[0 for i in range(2)] for j in range(N)] if (N == 1): return arr[0] # Initialize the values in dp array dp[0][0] = 0 dp[0][1] = arr[0] # Loop to find the maximum possible sum for i in range(1,N): dp[i][1] = dp[i - 1][0] + arr[i] dp[i][0] = max(dp[i - 1][1], dp[i - 1][0]) # Return the maximum sum return max(dp[N - 1][0], dp[N - 1][1])# Driver Code# Creating the arrayarr = [ 5, 5, 10, 100, 10, 5 ]N = len(arr)# Function callprint(findMaxSum(arr, N))# This code is contributed by shinjanpatra |
C#
// C# program for above approachusing System;class GFG{ // Function to find the maximum sumint findMaxSum(int[] arr,int n){ // Declare dp array int [,] dp = new int [n,2]; if (n == 1) { return arr[0]; } // Initialize the values in dp array dp[0,0] = 0; dp[0,1] = arr[0]; // Loop to find the maximum possible sum for (int i = 1; i < n; i++) { dp[i,1] = dp[i - 1,0] + arr[i]; dp[i,0] = Math.Max(dp[i - 1,1], dp[i - 1,0]); } // Return the maximum sum return Math.Max(dp[n - 1,0], dp[n - 1,1]);}// Driver codestatic public void Main (){ GFG small = new GFG(); int[] arr = {5, 5, 10, 100, 10, 5}; int n = arr.Length; // Function Call Console.WriteLine(small.findMaxSum(arr,n));}}// This code is contributed by Aarti_Rathi |
Javascript
<script>// JavaScript code to implement the approach// Function to find the maximum sumfunction findMaxSum(arr, N){ // Declare dp array let dp = new Array(N); for(let i = 0; i < N; i++) { dp[i] = new Array(2); } if (N == 1) { return arr[0]; } // Initialize the values in dp array dp[0][0] = 0; dp[0][1] = arr[0]; // Loop to find the maximum possible sum for (let i = 1; i < N; i++) { dp[i][1] = dp[i - 1][0] + arr[i]; dp[i][0] = Math.max(dp[i - 1][1], dp[i - 1][0]); } // Return the maximum sum return Math.max(dp[N - 1][0], dp[N - 1][1]);}// Driver Code// Creating the arraylet arr = [ 5, 5, 10, 100, 10, 5 ];let N = arr.length;// Function calldocument.write(findMaxSum(arr, N),"</br>");/*This code is contribute by shinjanpatra */</script> |
Output
110
Time Complexity: O(N)
Auxiliary Space: O(N)
Space Optimized Approach: The above approach can be optimized to be done in constant space based on the following observation:
As seen from the previous dynamic programming approach, the value of current states (for ith element) depends upon only two states of the previous element. So instead of creating a 2D array, we can use only two variables to store the two states of the previous element.
- Say excl stores the value of the maximum subsequence sum till i-1 when arr[i-1] is excluded and
- incl stores the value of the maximum subsequence sum till i-1 when arr[i-1] is included.
- The value of excl for the current state( say excl_new) will be max(excl ,incl). And the value of incl will be updated to excl + arr[i].
Illustration:
Consider arr[] = {5, 5, 10, 100, 10, 5}
Initially at i = 0: incl = 5, excl = 0
For i = 1: arr[i] = 5
=> excl_new = 5
=> incl = (excl + arr[i]) = 5
=> excl = excl_new = 5For i = 2: arr[i] = 10
=> excl_new = max(excl, incl) = 5
=> incl = (excl + arr[i]) = 15
=> excl = excl_new = 5For i = 3: arr[i] = 100
=> excl_new = max(excl, incl) = 15
=> incl = (excl + arr[i]) = 105
=> excl = excl_new = 15For i = 4: arr[i] = 10
=> excl_new = max(excl, incl) = 105
=> incl = (excl + arr[i]) = 25
=> excl = excl_new = 105For i = 5: arr[i] = 5
=> excl_new = max(excl, incl) = 105
=> incl = (excl + arr[i]) = 110
=> excl = excl_new = 105So, answer is max(incl, excl) = 110
Follow the steps mentioned below to implement the above approach:
- Initialize incl and excl with arr[0] and 0 respectively.
- Iterate from i = 1 to N-1:
- Update the values of incl and excl as mentioned above.
- Return the maximum of incl and excl after the iteration is over as the answer.
Below is the implementation of the above approach.
C++
// C++ code to implement the above approach#include <bits/stdc++.h>using namespace std;// Function to return max sum such that // no two elements are adjacent int FindMaxSum(vector<int> arr, int n){ int incl = arr[0]; int excl = 0; int excl_new; int i; for (i = 1; i < n; i++) { // Current max excluding i excl_new = max(incl, excl); // Current max including i incl = excl + arr[i]; excl = excl_new; } // Return max of incl and excl return max(incl, excl);}// Driver codeint main(){ vector<int> arr = { 5, 5, 10, 100, 10, 5 }; int N = arr.size(); // Function call cout << FindMaxSum(arr, N); return 0;}// This approach is contributed by Debanjan |
C
// C code to implement the approach#include <stdio.h>// Function to return max sum such that// no two elements are adjacentint findMaxSum(int arr[], int n){ int incl = arr[0]; int excl = 0; int excl_new; int i; for (i = 1; i < n; i++) { // Current max excluding i excl_new = (incl > excl) ? incl : excl; // Current max including i incl = excl + arr[i]; excl = excl_new; } // Return max of incl and excl return ((incl > excl) ? incl : excl);}// Driver codeint main(){ int arr[] = { 5, 5, 10, 100, 10, 5 }; int N = sizeof(arr) / sizeof(arr[0]); // Function call printf("%d", findMaxSum(arr, N)); return 0;} |
Java
// Java code to implement the approachimport java.lang.*;import java.util.*;class MaximumSum { // Function to return max sum such that // no two elements are adjacent int findMaxSum(int arr[], int n) { int incl = arr[0]; int excl = 0; int excl_new; int i; for (i = 1; i < n; i++) { // Current max excluding i excl_new = Math.max(incl, excl); // Current max including i incl = excl + arr[i]; excl = excl_new; } // Return max of incl and excl return Math.max(incl, excl); } // Driver code public static void main(String[] args) { MaximumSum sum = new MaximumSum(); int arr[] = new int[] { 5, 5, 10, 100, 10, 5 }; int N = arr.length; // Function call System.out.println( sum.findMaxSum(arr, arr.length)); }}// This code has been contributed by Mayank Jaiswal |
Python3
# Python code to implement the approach# Function to return max sum such that # no two elements are adjacentdef findMaxSum(arr, n): incl = 0 excl = 0 for i in arr: # Current max excluding i new_excl = max (excl, incl) # Current max including i incl = excl + i excl = new_excl # Return max of incl and excl return max(excl, incl)# Driver codeif __name__ == "__main__": arr = [5, 5, 10, 100, 10, 5] N = 6 # Function call print (findMaxSum(arr, N))# This code is contributed by Kalai Selvan |
C#
// C# code to implement the approachusing System;class GFG { // Function to return max sum such // that no two elements are adjacent static int findMaxSum(int []arr, int n) { int incl = arr[0]; int excl = 0; int excl_new; int i; for (i = 1; i < n; i++) { // Current max excluding i excl_new = (incl > excl) ? incl : excl; // Current max including i incl = excl + arr[i]; excl = excl_new; } // Return max of incl and excl return ((incl > excl) ? incl : excl); } // Driver code public static void Main() { int []arr = new int[]{5, 5, 10, 100, 10, 5}; int N = arr.Length; // Function call Console.Write(findMaxSum(arr, N)); }}// This code has been contributed by// nitin mittal |
Javascript
<script>// Javascript program for the above approach// Function to return max sum such that// no two elements are adjacent function FindMaxSum(arr, n){ let incl = arr[0]; let excl = 0; let excl_new; let i; for(i = 1; i < n; i++) { // Current max excluding i excl_new = (incl > excl) ? incl : excl; // Current max including i incl = excl + arr[i]; excl = excl_new; } // Return max of incl and excl return ((incl > excl) ? incl : excl);}// Driver codelet arr = [ 5, 5, 10, 100, 10, 5 ];document.write(FindMaxSum(arr, arr.length));// This code is contributed by Mayank Tyagi</script> |
PHP
<?php// PHP code to find Maximum sum // such that no two elements // are adjacent/* Function to return max sum such that no two elements are adjacent */function FindMaxSum($arr, $n){ $incl = $arr[0]; $excl = 0; $excl_new; $i;for ($i = 1; $i <$n; $i++){ // current max excluding i $excl_new = ($incl > $excl)? $incl: $excl; // current max including i $incl = $excl + $arr[$i]; $excl = $excl_new;}// return max of incl and excl return (($incl > $excl)? $incl : $excl);}// Driver Code$arr = array(5, 5, 10, 100, 10, 5);$n = sizeof($arr);echo FindMaxSum($arr, $n); // This code is contributed by Ajit?> |
Output
110
Time Complexity: O(N)
Auxiliary Space: O(1).
Similar problem: Find maximum possible stolen value from houses
Please write comments if you find any bug in the above program/algorithm or other ways to solve the same problem.
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