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 arrayint 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 Codeint 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 arrayimport java.util.*;
class GFG{
// Function to find the maximum// score of the deleting an element// from an arraystatic 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 Codepublic 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 arrayfrom collections import defaultdict
# Function to find the maximum# score of the deleting an element# from an arraydef 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 Codeif __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 arrayusing System;
using System.Linq;
using System.Collections.Generic;
class GFG{
// Function to find the maximum// score of the deleting an element// from an arraystatic 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 Codepublic 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 arrayfunction 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 Codelet n = 3;let a=[1, 2, 3];// Function calldocument.write(findMaximumScore(a, n));// This code is contributed by avanitrachhadiya2155</script> |
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 <bits/stdc++.h>using namespace std;
// Function to find the maximum// score of the deleting an element// from an arrayint 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 Codeint 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<Integer> 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<Integer> 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<Integer> 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 Codeif __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 arrayusing 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 arrayfunction 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 arrayfunction 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 Codeconst a = [1, 2, 3];console.log(findMaximumScore(a)); |
Output
4
Time complexity: O(n)
Auxiliary Space: O(m)
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