Given an array of N numbers, we need to maximize the sum of selected numbers. At each step you need to select a number A_{i}, delete one occurrence of **A _{i}-1 (if exists), A_{i}+1 (if exists) and A_{i}** each from the array. Repeat these steps until the array gets empty. The problem is to maximize the sum of selected numbers.

**Note**: We have to delete A_{i}+1 and A_{i}-1 elements if they are present in the array and not A_{i+1} and A_{i-1}.

Examples:

Input : a[] = {1, 2, 3} Output : 4 Explanation: At first step we select 1, so 1 and 2 are deleted from the sequence leaving us with 3. Then we select 3 from the sequence and delete it. So the sum of selected numbers is 1+3 = 4. Input : a[] = {1, 2, 2, 2, 3, 4} Output : 10 Explanation : Select one of the 2's from the array, so 2, 2-1, 2+1 will be deleted and we are left with {2, 2, 4}, since 1 and 3 are deleted. Select 2 in next two steps, and then select 4 in the last step. We get a sum of 2+2+2+4=10 which is the maximum possible.

Our aim is to maximize the sum of selected numbers. The idea is to pre-calculate the occurrence of all numbers x in the array a[] in a hash **ans**. Now our recurrence relation will decide either to select a number or not. If we select the number then we take the occurrences of that number and the value stored at ans[i-2] as ans[i-1] will be deleted and not be taken to count. If we do not select the number then we take ans[i-1] which have been pre-calculated while moving forward.

ans[i] = max(ans[i-1], ans[i-2] + ans[i]*i )

At the end, ans[maximum] will have the maximum sum of selected numbers.

Below is the implementation of above idea:

`// CPP program to Maximize the sum of selected ` `// numbers by deleting three consecutive numbers. ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// function to maximize the sum of selected numbers ` `int` `maximizeSum(` `int` `a[], ` `int` `n) { ` ` ` ` ` `// stores the occurrences of the numbers ` ` ` `unordered_map<` `int` `, ` `int` `> ans; ` ` ` ` ` `// marks the occurrence of every number in the sequence ` ` ` `for` `(` `int` `i = 0; i < n; i++) ` ` ` `ans[a[i]]++; ` ` ` ` ` `// maximum in the sequence ` ` ` `int` `maximum = *max_element(a, a + n); ` ` ` ` ` `// traverse till maximum and apply the recurrence relation ` ` ` `for` `(` `int` `i = 2; i <= maximum; i++) ` ` ` `ans[i] = max(ans[i - 1], ans[i - 2] + ans[i] * i); ` ` ` ` ` `// return the ans stored in the index of maximum ` ` ` `return` `ans[maximum]; ` `} ` ` ` `// Driver code ` `int` `main() ` `{ ` ` ` `int` `a[] = {1, 2, 3}; ` ` ` `int` `n = ` `sizeof` `(a) / ` `sizeof` `(a[0]); ` ` ` `cout << maximizeSum(a, n); ` ` ` `return` `0; ` `} ` |

**Output:**

4

**Time Complexity:** O(A_{max}), where A_{max} is the maximum element present in the array A[].

## Recommended Posts:

- Maximum difference of zeros and ones in binary string | Set 2 (O(n) time)
- Probability of reaching a point with 2 or 3 steps at a time
- Maximum subarray sum in an array created after repeated concatenation
- Count of arrays having consecutive element with different values
- Remove array end element to maximize the sum of product
- Maximum path sum that starting with any cell of 0-th row and ending with any cell of (N-1)-th row
- Maximum path sum for each position with jumps under divisibility condition
- Maximum sum subarray removing at most one element
- Minimum cost to fill given weight in a bag
- Find number of solutions of a linear equation of n variables
- Longest Common Substring | DP-29
- Subset Sum Problem | DP-25
- 0-1 Knapsack Problem | DP-10
- Min Cost Path | DP-6
- Largest Sum Contiguous Subarray

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.