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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:
Step 1: arr[] = 7 1 8 3 6 6, Score:
Step 2: arr[] = 7 1 8 6 6, Score:
Step 3: arr[] = 7 8 6 6, Score:
Step 4: arr[] = 8 6, Score: 13
Step 5: arr[] = 8 Score: 19
Step 6: arr[] = [] Score: 27
Input: arr[] = 1 2 3
Output:

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 ` `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 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

 ``

Output

```4

```

Time Complexity: O(N)

Auxiliary Space: O(N)

Another Approach Using Memoization:

C++

 `// C++ implementation to find the``// maximum score of the deleting a``// element from an array` `#include ` `using` `namespace` `std;` `// Function to find the maximum``// score of the deleting an element``// from an array``int` `solve(vector<``int``> &arr, ``int` `i)``    ``{``        ``if``(i >= arr.size()) ``// if i is greater than size of array``        ``{``            ``return` `0; ``// then simply returnn zero``        ``}``        ` `        ``// current 'i' on which we are standing``        ``int` `currValue = arr[i];  ``// current value``        ``int` `currSum = arr[i]; ``// intial make sum as same as value``        ``int` `index = i + 1; ``// index to take elemets, so  i + 1``        ` `        ``// while it is the same as the current value, include in our sum``        ``while``(index < arr.size() && arr[index] == currValue)``        ``{``            ``currSum += arr[i];``            ``index++;``        ``}``        ` `        ``// Now, we have to skip all the elements, whose value is equal to``        ``// currValue + 1``        ``while``(index < arr.size() && arr[index] == currValue + 1)``        ``{``            ``index++;``        ``}``        ` `        ``//And lastly, we have two choices-``        ``//whether to include the sum of this current element``        ``// in our answer``        ``// or not include the sum of current element in our answer``        ``// so we explore all possibility and take maximum of them``        ` `        ``return` `max(currSum + solve(arr, index), solve(arr, i + 1));``        ` `        ``// If we decide to take the curr element in our answer, then upto the``        ``// elemet we skip the next value, we paas that index``        ``// but if decided no to make this vurrent element then simply paas``        ``// i + 1``    ``}``    ``int` `findMaximumScore(vector<``int``>& arr) {``        ``int` `n = arr.size(); ``// take the size of the array``        ` `        ``// sort the array to get rid of all arr[i] - 1 elements``        ``sort(arr.begin(), arr.end());``        ` `        ``// solve function which give us our final answer``        ``return` `solve(arr, 0);``        ``//                ↑``        ``//                we start from zero index``    ``}` `// Driver Code``int` `main()``{``    ``int` `n;``    ``n = 3;``    ``vector<``int``> a{ 1, 2, 3 };` `    ``// Function Call``    ``cout << findMaximumScore(a);``    ``return` `0;``}`

Java

 `import` `java.util.ArrayList;``import` `java.util.Arrays;``import` `java.util.Collections;` `public` `class` `Main {` `    ``// Function to find the maximum score of deleting an``    ``// element from an array``    ``static` `int` `solve(ArrayList arr, ``int` `i)``    ``{``        ``if` `(i >= arr.size()) ``// if i is greater than size of array``        ``{``            ``return` `0``;``        ``}``        ` `        ``// current 'i' on which we are standing``        ``int` `currValue = arr.get(i);``        ``int` `currSum = arr.get(i);``        ``int` `index = i + ``1``;``        ` `        ``// while it is the same as the current value, include in our sum``        ``while` `(index < arr.size()``               ``&& arr.get(index) == currValue) {``            ``currSum += arr.get(i);``            ``index++;``        ``}``         ` `        ``// Now, we have to skip all the elements, whose value is equal to``        ``// currValue + 1``        ``while` `(index < arr.size()``               ``&& arr.get(index) == currValue + ``1``) {``            ``index++;``        ``}``        ` `        ``//And lastly, we have two choices-``        ``//whether to include the sum of this current element``        ``// in our answer``        ``// or not include the sum of current element in our answer``        ``// so we explore all possibility and take maximum of them``        ``return` `Math.max(currSum + solve(arr, index),``                        ``solve(arr, i + ``1``));``    ``}``    ` `    ``// If we decide to take the curr element in our answer, then upto the``    ``// elemet we skip the next value, we paas that index``    ``// but if decided no to make this vurrent element then simply paas``    ``// i + 1``    ``static` `int` `findMaximumScore(ArrayList arr)``    ``{``        ``int` `n = arr.size();``        ` `        ``// sort the array to get rid of all arr[i] - 1 elements``        ``Collections.sort(arr);` `        ``return` `solve(arr, ``0``);``    ``}``    ` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``3``;``        ``ArrayList a``            ``= ``new` `ArrayList<>(Arrays.asList(``1``, ``2``, ``3``));` `        ``// Function Call``        ``System.out.println(findMaximumScore(a));``    ``}``}`

Python

 `def` `solve(arr, i):``    ``# Base case: if i is greater than or equal to the size of the array, return 0``    ``if` `i >``=` `len``(arr):``        ``return` `0` `    ``currValue ``=` `arr[i]  ``# current value at index i``    ``currSum ``=` `arr[i]  ``# initialize the current sum as the current value``    ``index ``=` `i ``+` `1`  `# index to take elements, starting from i + 1` `    ``# While the elements have the same value as the current value, include them in our sum``    ``while` `index < ``len``(arr) ``and` `arr[index] ``=``=` `currValue:``        ``currSum ``+``=` `arr[i]``        ``index ``+``=` `1` `    ``# Now, we have to skip all the elements whose value is equal to currValue + 1``    ``while` `index < ``len``(arr) ``and` `arr[index] ``=``=` `currValue ``+` `1``:``        ``index ``+``=` `1` `    ``# Lastly, we have two choices - whether to include the sum of the current element in our answer``    ``# or not include the sum of the current element in our answer. So, we explore all possibilities and take the maximum of them``    ``return` `max``(currSum ``+` `solve(arr, index), solve(arr, i ``+` `1``))`  `def` `find_maximum_score(arr):``    ``arr.sort()  ``# Sort the array to get rid of all arr[i] - 1 elements``    ``# Call the solve function to get our final answer, starting from index 0``    ``return` `solve(arr, ``0``)`  `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``arr ``=` `[``1``, ``2``, ``3``]``    ``print``(find_maximum_score(arr))`

C#

 `// C# implementation to find the``// maximum score of the deleting a``// element from an array``using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `class` `GFG {``    ``// Function to find the maximum score by combining``    ``// adjacent elements in the list``    ``static` `int` `Solve(List<``int``> arr, ``int` `i)``    ``{``        ``// Base case: if we reach the end of the list,``        ``// return 0``        ``if` `(i >= arr.Count) {``            ``return` `0;``        ``}` `        ``// Initialize variables to keep track of the current``        ``// value, current sum, and next index``        ``int` `currValue = arr[i];``        ``int` `currSum = arr[i];``        ``int` `index = i + 1;` `        ``// Calculate the sum of adjacent elements with the``        ``// same value``        ``while` `(index < arr.Count``               ``&& arr[index] == currValue) {``            ``currSum += arr[i];``            ``index++;``        ``}` `        ``// Skip elements that have a value of "currValue +``        ``// 1" and move to the next unique value``        ``while` `(index < arr.Count``               ``&& arr[index] == currValue + 1) {``            ``index++;``        ``}` `        ``// Return the maximum score by either choosing the``        ``// current element and recursively calling the``        ``// function for the next unique element or skipping``        ``// the current element and calling the function for``        ``// the next element.``        ``return` `Math.Max(currSum + Solve(arr, index),``                        ``Solve(arr, i + 1));``    ``}` `    ``// Function to find the maximum score by first sorting``    ``// the list and then calling Solve``    ``static` `int` `FindMaximumScore(List<``int``> arr)``    ``{``        ``arr.Sort(); ``// Sort the list in ascending order``        ``return` `Solve(arr,``                     ``0); ``// Start the recursive calculation``                         ``// from the beginning of the list``    ``}` `    ``static` `void` `Main()``    ``{``        ``List<``int``> a = ``new` `List<``int``>{ 1, 2, 3 };` `        ``// Find the maximum score and print the result``        ``Console.WriteLine(FindMaximumScore(a));``    ``}``}`

Javascript

 `// Function to recursively find the maximum score of``// deleting an element from an array``function` `solve(arr, i) {``    ``if` `(i >= arr.length) {``        ``return` `0;``    ``}` `    ``// current 'i' on which we are standing``    ``let currValue = arr[i]; ``// current value``    ``let currSum = arr[i]; ``// initialize sum as the current value``    ``let index = i + 1; ``// index to take elements, so i + 1` `    ``// while the element is the same as the current value, include``    ``// it in our sum``    ``while` `(index < arr.length && arr[index] === currValue) {``        ``currSum += arr[i];``        ``index++;``    ``}` `    ``// Now, we have to skip all the elements whose value is equal``    ``// to currValue + 1``    ``while` `(index < arr.length && arr[index] === currValue + 1) {``        ``index++;``    ``}` `    ``// Explore two choices:``    ``// 1. Include the sum of the current element in our answer``    ``// and move to the next unique element (index).``    ``// 2. Exclude the current element from our answer and move``    ``// to the next element (i + 1).``    ``// Take the maximum of these two possibilities.``    ``return` `Math.max(currSum + solve(arr, index), solve(arr, i + 1));``}` `// Function to find the maximum score of deleting an element from``// an array``function` `findMaximumScore(arr) {``    ``const n = arr.length;` `    ``// Sort the array in ascending order to get rid of all arr[i] - 1``    ``// elements``    ``arr.sort((a, b) => a - b);` `    ``// Call the solve function to get the final answer``    ``return` `solve(arr, 0);``    ``//        ↑``    ``// Start from the zero index``}` `// Driver Code``const a = [1, 2, 3];``console.log(findMaximumScore(a));`

Output

```4

```

Time complexity: O(n)
Auxiliary Space: O(m)