[Coding Made Simple] Maximum Sum Subsequence Non-adjacent

Posted 2020-10-03 Push your limit!

Given an array of positive number, find maximum sum subsequence such that elements in this subsequence are not adjacent to each other.

 

Recursive formula: f(n) = Math.max{f(n - 1), f(n - 2) + arr[n - 1]}.

Dynamic programming is used to get rid of the overlapping subproblems.

State: T[i]: the max sum of non-adjacent subsequence from arr[0.... n - 1];

Function: T[i] = Math.max(T[i - 1], T[i - 2] + arr[i - 1]);

Init: T[0] = 0, T[1] = arr[0];

Answer: T[arr.length].

 

 1 import java.util.ArrayList;
 2 
 3 public class MaxSumSubsequence {
 4     private ArrayList<Integer> subseq;
 5     public int getMaxSumSubseqNonAdj(int[] arr) {
 6         if(arr == null || arr.length == 0) {
 7             return 0;
 8         }
 9         int[] T = new int[arr.length + 1];
10         T[0] = 0; T[1] = arr[0];
11         for(int i = 2; i < T.length; i++) {
12             T[i] = Math.max(T[i - 1], T[i - 2] + arr[i - 1]);
13         }
14         subseq = new ArrayList<Integer>();
15         int currSum = T[arr.length];
16         int idx = arr.length;
17         while(currSum != 0) {
18             if(idx >= 2) {
19                 if(T[idx - 1] <= T[idx - 2] + arr[idx - 1]) {
20                     subseq.add(arr[idx - 1]);
21                     currSum -= arr[idx - 1];
22                     idx -= 2;
23                 }
24                 else {
25                     idx--;
26                 }
27             }
28             else {
29                 subseq.add(arr[idx - 1]);
30                 currSum -= arr[idx - 1];
31                 idx--;
32             }
33         }
34         return T[arr.length];
35     }
36 }