Maximum score of deleting an element from an Array based on given condition
Given an array arr[], the task is to find the maximum score of deleting an element where each element of the array can be deleted with the score of the element, but the constraint is if we delete arr[i], then arr[i] + 1 and arr[i] – 1 is gets automatically deleted with 0 scores.
Examples:
Input: arr[] = {7, 2, 1, 8, 3, 3, 6, 6}
Output: 27
Explanation:
Step 0: arr[] = 7 2 1 8 3 3 6 6, Score: 0
Step 1: arr[] = 7 1 8 3 6 6, Score: 3
Step 2: arr[] = 7 1 8 6 6, Score: 6
Step 3: arr[] = 7 8 6 6, Score: 7
Step 4: arr[] = 8 6, Score: 13
Step 5: arr[] = 8 Score: 19
Step 6: arr[] = [] Score: 27
Input: arr[] = 1 2 3
Output: 4
Approach: The idea is to use Dynamic Programming to solve this problem. The key observation of the problem is that for removing any element from the array the occurrence of the element and the value itself is the important factor.
Let’s take an example to understand the observation if the sequence was 4 4 5. Then, we have two choices to choose from 4 to 5. Now, on choosing 4, his score would be 4*2 = 8. On the other hand, if we choose 5, his score would be 5*1 = 5. Clearly, the maximal score is 8.
Hence, for the above sequence 4 4 5, freq[4] = 2, and freq[5] = 1.
Finally, to find the optimal score, it would be easy to first break the problem down into a smaller problem. In this case, we break the sequence into smaller sequences and find an optimal solution for it. 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.
Recurrence Relation:
dp[i] = max(dp[i – 1], dp[i – 2] + i*freq[i])
Basically, we have 2 cases, either to pick an ith element and other is not to pick an ith element.
Case 1: If we pick the ith element, Maximum score till the ith element will be dp[i-2] + i*freq[i] (choosing ith element means deleting (i-1)th element)
Case 2: If we don’t pick the ith element, the Maximum Score till ith element will dp[i-1]
Now as We have to maximize the score, We will take the maximum of both.
Below is the implementation of the above approach:
C++
// C++ implementation to find the // maximum score of the deleting a // element from an array #include <bits/stdc++.h> using namespace std; // Function to find the maximum // score of the deleting an element // from an array int findMaximumScore(vector< int > a, int n) { // Creating a map to keep // the frequency of numbers unordered_map< int , int > freq; // Loop to iterate over the // elements of the array for ( int i = 0; i < n; i++) { freq[a[i]]++; } // Creating a DP array to keep // count of max score at ith element // and it will be filled // in the bottom Up manner vector< int > dp(*max_element(a.begin(), a.end()) + 1, 0); dp[0] = 0; dp[1] = freq[1]; // Loop to choose the elements of the // array to delete from the array for ( int i = 2; i < dp.size(); i++) dp[i] = max( dp[i - 1], dp[i - 2] + freq[i] * i); return dp[dp.size() - 1]; } // Driver Code int main() { int n; n = 3; vector< int > a{ 1, 2, 3 }; // Function Call cout << findMaximumScore(a, n); return 0; } |
Java
// Java implementation to find the // maximum score of the deleting a // element from an array import java.util.*; class GFG{ // Function to find the maximum // score of the deleting an element // from an array static int findMaximumScore( int []a, int n) { // Creating a map to keep // the frequency of numbers @SuppressWarnings ( "unchecked" ) HashMap<Integer, Integer> freq = new HashMap(); // Loop to iterate over the // elements of the array for ( int i = 0 ; i < n; i++) { if (freq.containsKey(a[i])) { freq.put(a[i], freq.get(a[i]) + 1 ); } else { freq.put(a[i], 1 ); } } // Creating a DP array to keep // count of max score at ith element // and it will be filled // in the bottom Up manner int []dp = new int [Arrays.stream(a).max().getAsInt() + 1 ]; dp[ 0 ] = 0 ; dp[ 1 ] = freq.get( 1 ); // Loop to choose the elements of the // array to delete from the array for ( int i = 2 ; i < dp.length; i++) dp[i] = Math.max(dp[i - 1 ], dp[i - 2 ] + freq.get(i) * i); return dp[dp.length - 1 ]; } // Driver Code public static void main(String[] args) { int n; n = 3 ; int []a = { 1 , 2 , 3 }; // Function call System.out.print(findMaximumScore(a, n)); } } // This code is contributed by 29AjayKumar |
Python3
# Python3 implementation to find the # maximum score of the deleting a # element from an array from collections import defaultdict # Function to find the maximum # score of the deleting an element # from an array def findMaximumScore(a, n): # Creating a map to keep # the frequency of numbers freq = defaultdict ( int ) # Loop to iterate over the # elements of the array for i in range (n): freq[a[i]] + = 1 # Creating a DP array to keep # count of max score at ith element # and it will be filled # in the bottom Up manner dp = [ 0 ] * ( max (a) + 1 ) dp[ 0 ] = 0 dp[ 1 ] = freq[ 1 ] # Loop to choose the elements of the # array to delete from the array for i in range ( 2 , len (dp)): dp[i] = max (dp[i - 1 ], dp[i - 2 ] + freq[i] * i) return dp[ - 1 ] # Driver Code if __name__ = = "__main__" : n = 3 a = [ 1 , 2 , 3 ] # Function Call print (findMaximumScore(a, n)) # This code is contributed by Chitranayal |
C#
// C# implementation to find the // maximum score of the deleting a // element from an array using System; using System.Linq; using System.Collections.Generic; class GFG{ // Function to find the maximum // score of the deleting an element // from an array static int findMaximumScore( int []a, int n) { // Creating a map to keep // the frequency of numbers Dictionary< int , int > freq = new Dictionary< int , int >(); // Loop to iterate over the // elements of the array for ( int i = 0; i < n; i++) { if (freq.ContainsKey(a[i])) { freq[a[i]] = freq[a[i]] + 1; } else { freq.Add(a[i], 1); } } // Creating a DP array to keep // count of max score at ith element // and it will be filled // in the bottom Up manner int []dp = new int [a.Max() + 1]; dp[0] = 0; dp[1] = freq[1]; // Loop to choose the elements of the // array to delete from the array for ( int i = 2; i < dp.Length; i++) dp[i] = Math.Max(dp[i - 1], dp[i - 2] + freq[i] * i); return dp[dp.Length - 1]; } // Driver Code public static void Main(String[] args) { int n; n = 3; int []a = { 1, 2, 3 }; // Function call Console.Write(findMaximumScore(a, n)); } } // This code is contributed by 29AjayKumar |
Javascript
<script> // Javascript implementation to find the // maximum score of the deleting a // element from an array // Function to find the maximum // score of the deleting an element // from an array function findMaximumScore(a,n) { // Creating a map to keep // the frequency of numbers let freq = new Map(); // Loop to iterate over the // elements of the array for (let i = 0; i < n; i++) { if (freq.has(a[i])) { freq.set(a[i], freq.get(a[i]) + 1); } else { freq.set(a[i], 1); } } // Creating a DP array to keep // count of max score at ith element // and it will be filled // in the bottom Up manner let dp = new Array(Math.max(...a)+1); dp[0] = 0; dp[1] = freq.get(1); // Loop to choose the elements of the // array to delete from the array for (let i = 2; i < dp.length; i++) dp[i] = Math.max(dp[i - 1], dp[i - 2] + freq.get(i) * i); return dp[dp.length - 1]; } // Driver Code let n = 3; let a=[1, 2, 3]; // Function call document.write(findMaximumScore(a, n)); // This code is contributed by avanitrachhadiya2155 </script> |
4
Time Complexity: O(N)
Auxiliary Space: O(N)