Maximize the sum of selected numbers from an array to make it empty
Given an array of N numbers, we need to maximize the sum of selected numbers. At each step, you need to select a number Ai, delete one occurrence of it and delete all occurrences of Ai-1 and Ai+1 (if they exist) in the array. Repeat these steps until the array gets empty. The problem is to maximize the sum of the selected numbers.
Note: We have to delete all the occureneces of Ai+1 and Ai-1 elements if they are present in the array and not Ai+1 and Ai-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:
C++
// 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; } |
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Python3
# Python3 program to Maximize the sum of selected # numbers by deleting three consecutive numbers. # function to maximize the sum of # selected numbers def maximizeSum(a, n) : # stores the occurrences of the numbers ans = dict.fromkeys(range(0, n + 1), 0) # marks the occurrence of every # number in the sequence for i in range(n) : ans[a[i]] += 1 # maximum in the sequence maximum = max(a) # traverse till maximum and apply # the recurrence relation for i in range(2, maximum + 1) : 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 if __name__ == "__main__" : a = [1, 2, 3] n = len(a) print(maximizeSum(a, n)) # This code is contributed by Ryuga |
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Output:
4
Time Complexity: O(Amax), where Amax is the maximum element present in the array A[].
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