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 numbersint 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 codeint main(){ int a[] = {1, 2, 3}; int n = sizeof(a) / sizeof(a[0]); cout << maximizeSum(a, n); return 0;} |
Java
// Java implementation of the approachimport java.util.*;import java.math.*;class GFG{// Function to maximise the sum of selected nummbers//by deleting occurences of Ai-1 and Ai+1static int getMaximumSum(int []arr){ // Number of elements in the array int n = arr.length; // Largest element in the array int max = -1; for(int i = 0; i < n; i++) { max = Math.max(max, arr[i]); } // An array to count the occurence of each element int []ans = new int[max + 1]; for(int i = 0; i < n; i++) { ans[arr[i]]++; } // Using the above mentioned approach for(int i = 2; i <= max; i++) { ans[i] = Math.max(ans[i - 1], ans[i - 2] + i* ans[i]); } return ans[max]; }// Driver codepublic static void main(String[] args){ int []a = {1, 2, 3}; System.out.println(getMaximumSum(a));}}// This code is contributed by Sooraj Kumar (NITS CSE 2020) |
Python3
# Python3 program to Maximize the sum of selected# numbers by deleting three consecutive numbers.# function to maximize the sum of# selected numbersdef 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 codeif __name__ == "__main__" : a = [1, 2, 3] n = len(a) print(maximizeSum(a, n))# This code is contributed by Ryuga |
C#
// C# implementation of the approachusing System;class GFG{// Function to maximise the sum of selected nummbers//by deleting occurences of Ai-1 and Ai+1static int getMaximumSum(int []arr){ // Number of elements in the array int n = arr.Length; // Largest element in the array int max = -1; for(int i = 0; i < n; i++) { max = Math.Max(max, arr[i]); } // An array to count the occurence of each element int []ans = new int[max + 1]; for(int i = 0; i < n; i++) { ans[arr[i]]++; } // Using the above mentioned approach for(int i = 2; i <= max; i++) { ans[i] = Math.Max(ans[i - 1], ans[i - 2] + i* ans[i]); } return ans[max];}// Driver codepublic static void Main(string[] args){ int []a = {1, 2, 3}; Console.Write(getMaximumSum(a));}}// This code is contributed by rock_cool |
Javascript
<script> // Javascript implementation of the approach // Function to maximise the sum of selected nummbers //by deleting occurences of Ai-1 and Ai+1 function getMaximumSum(arr) { // Number of elements in the array let n = arr.length; // Largest element in the array let max = -1; for(let i = 0; i < n; i++) { max = Math.max(max, arr[i]); } // An array to count the occurence of each element let ans = new Array(max + 1); ans.fill(0); for(let i = 0; i < n; i++) { ans[arr[i]]++; } // Using the above mentioned approach for(let i = 2; i <= max; i++) { ans[i] = Math.max(ans[i - 1], ans[i - 2] + i* ans[i]); } return ans[max]; } let a = [1, 2, 3]; document.write(getMaximumSum(a));</script> |
Output:
4
Time Complexity: O(Amax), where Amax is the maximum element present in the array A[].
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