Maximum sum subsequence with values differing by at least 2

Given a positive integer array arr[] of size N, the task is to find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent by value i.e if arr[i] is taken into the answer, then neither occurrences of arr[i]-1 nor arr[i]+1 can be selected.

Examples:

Input: arr[] = {2, 2, 2}
Output: 6
Explanation:
The max sum subsequence will be [2, 2, 2] as it does not contain any occurrence of 1 or 3. Hence sum = 2 + 2 + 2 = 6

Input: arr[] = {2, 2, 3}
Output: 6
Explanation:
Subsequence 1: [2, 2] as it does not contain any occurrence of 1 or 3. Hence sum = 2 + 2 = 4
Subsequence 2: [3] as it does not contain any occurrence of 2 or 4. Hence sum = 3
Therefore max sum = 4

Solution Approach: The idea is to use Dynamic Programming, similar to this article</a>.



  1. Create a map to store the number of times the element i appears in the sequence.
  2. To find the answer, it would be easy to first break the problem down into smaller problems. In this case, break the sequence into smaller sequences and find an optimal solution for it.
  3. For the sequence of numbers containing only 0, the answer would be 0. Similarly, if a sequence contains only the number 0 and 1, then the solution would be count[1]*1.
  4. Now build a recursive solution for this problem. For a sequence of numbers containing only the numbers, 0 to n, the choice is to either pick the Nth element or not.

    dp[i] = max(dp[i – 1], dp[i – 2] + i*freq[i] )

    dp[i-1] represent not picking the ith number, then the number before it can be considered.

    dp[i – 2] + i*freq[i] reprent picking the ith number, then the number before it is eliminated, hence the number before that is considered

C++

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// C++ progrm to find maximum sum
// subsequence with values
// differing by at least 2
  
#include <bits/stdc++.h>
using namespace std;
  
// function to find maximum sum
// subsequence such that two
// adjacent values elements are
// not selected
int get_max_sum(int arr[], int n)
{
    // map to store the frequency
    // of array elements
    unordered_map<int, int> freq;
  
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
  
    // make a dp arrray to store
    // answer upto i th value
    int dp[100001];
    memset(dp, 0, sizeof(dp));
  
    // base cases
    dp[0] = 0;
    dp[1] = freq[0];
  
    // iterate for all possible
    // values  of arr[i]
    for (int i = 2; i <= 100000; i++) {
        dp[i] = max(dp[i - 1],
                    dp[i - 2] + i * freq[i]);
    }
  
    // return the last value
    return dp[100000];
}
  
// Driver function
int main()
{
  
    int N = 3;
    int arr[] = { 2, 2, 3 };
    cout << get_max_sum(arr, N);
    return 0;
}

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Python3

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# Python3 program to find maximum sum 
# subsequence with values 
# differing by at least 2 
from collections import defaultdict
  
# Function to find maximum sum 
# subsequence such that two 
# adjacent values elements are 
# not selected
def get_max_sum(arr, n):
      
    # Map to store the frequency 
    # of array elements
    freq = defaultdict(lambda : 0)
      
    for i in range(n):
        freq[arr[i]] += 1
      
    # Make a dp arrray to store 
    # answer upto i th value
    dp = [0] * 100001
      
    # Base cases
    dp[0] = 0
    dp[1] = freq[0]
      
    # Iterate for all possible 
    # values of arr[i] 
    for i in range(2, 100000 + 1):
        dp[i] = max(dp[i - 1], 
                    dp[i - 2] + i * freq[i])
          
    # Return the last value
    return dp[100000]
  
# Driver code
N = 3
arr = [ 2, 2, 3 ]
      
print(get_max_sum(arr, N))
  
# This code is contributed by stutipathak31jan

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Output:

4

Time Complexity:O(N)
Auxillary Space:O(N)

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