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Maximize Array sum except elements from [i, i+X] for all i such that arr[i] > K

  • Difficulty Level : Expert
  • Last Updated : 14 Jan, 2022

Given an array arr[] of size N and two integers X and K, the task is to find the maximum score that can be achieved by rearranging the elements of the array where the score is calculated as the sum of elements of the array except the next X elements from the index i such arr[i] > K for all possible values of i in the range [0, N).

Example:

Input: arr[] = {9, 13, 16, 21, 6}, X = 2, K = 15
Output: 50
Explanation: The given array can be rearranged as arr[] = {16, 6, 9, 13, 21}. The indices such that arr[i] > K are {0, 4}. Therefore, the next two elements from index 0 and index 4 will be skipped. Therefore, the sum of remaining elements becomes 16 + 13 + 21 = 50, which is the maximum possible.

Input: arr[] = {31, 20, 19, 23, 34, 21, 37}, X = 3, K= 22
Output: 112

 

Approach: The given problem can be solved using a greedy approach. Create two arrays, big[] containing integers greater than K and small[] containing integers smaller than K. It can be observed that if the number of elements greater than K that must be included in the sum is i, then there must exist at least (i − 1)(X + 1) + 1 elements in the array. Hence for each possible i, exclude the ((i − 1)(X + 1) + 1) – i smallest elements of the small[] array from the total sum. The maximum value over all possible sums for each i is the required result.

Below is the implementation of the above approach:

C++




// C++ program, for the above approach
#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 1e5;
 
// Utility function to calculate the
// sum of elements as a prefix array
void calc(int arr[], int N)
{
    sort(arr + 1, arr + N + 1);
    reverse(arr + 1, arr + N + 1);
 
    for (int i = 1; i <= N; i++) {
        arr[i] += arr[i - 1];
    }
}
 
// Function to find the maximum score
// that can be achieved by rearranging
// the elements of given array
int maxScore(int arr[], int X, int K, int N)
{
    // Arrays to store small and big elements
    int small[maxn + 5], big[maxn + 5];
    int k = 0, l = 0;
 
    // Iterate and segregate big and
    // small elements
    for (int i = 0; i < N; i++) {
        if (arr[i] > K) {
            big[++k] = arr[i];
        }
        else {
            small[++l] = arr[i];
        }
    }
    // If k = 0, return the sum
    // of small[]
    if (k == 0) {
        int sum = 0;
        for (int i = 1; i <= N; i++) {
            sum += small[i];
        }
        return sum;
    }
    // Prefix sums of small[]
    // and big[]
    calc(big, k);
    calc(small, l);
 
    // Initialize small[l] within the range
    fill(small + l + 1, small + N + 1, small[l]);
 
    // Variable to store the answer
    int res = 0;
    for (int i = (k + X) / (1 + X); i <= k; i++) {
        if (1ll * (i - 1) * (X + 1) + 1 <= N) {
 
            // Update res with maximum one
            res = max(
                res,
                big[i]
                    + small[N - 1ll
                           * (i - 1)
                           * (X + 1) - 1]);
        }
    }
    // Return res
    return res;
}
 
// Driver Code
int main()
{
    int arr[] = { 9, 13, 16, 21, 6 };
    int X = 2;
    int K = 15;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << maxScore(arr, X, K, N);
 
    return 0;
}

Java




// Java program, for the above approach
import java.util.*;
 
class GFG{
 
static int maxn = (int)1e5;
 
// Utility function to calculate the
// sum of elements as a prefix array
static void calc(int arr[], int N)
{
    Arrays.sort(arr);
    arr = reverse(arr);
 
    for(int i = 1; i <= N; i++)
    {
        arr[i] += arr[i - 1];
    }
}
 
static int[] reverse(int a[])
{
    int i, n = a.length + 1, t;
    for(i = 1; i < n / 2; i++)
    {
        t = a[i];
        a[i] = a[n - i - 1];
        a[n - i - 1] = t;
    }
    return a;
}
 
// Function to find the maximum score
// that can be achieved by rearranging
// the elements of given array
static int maxScore(int arr[], int X, int K, int N)
{
     
    // Arrays to store small and big elements
    int []small = new int[maxn + 5];
    int big[] = new int[maxn + 5];
    int k = 0, l = 0;
 
    // Iterate and segregate big and
    // small elements
    for(int i = 0; i < N; i++)
    {
        if (arr[i] > K)
        {
            big[++k] = arr[i];
        }
        else
        {
            small[++l] = arr[i];
        }
    }
     
    // If k = 0, return the sum
    // of small[]
    if (k == 0)
    {
        int sum = 0;
        for(int i = 1; i <= N; i++)
        {
            sum += small[i];
        }
        return sum;
    }
     
    // Prefix sums of small[]
    // and big[]
    calc(big, k);
    calc(small, l);
 
    // Initialize small[l] within the range
    // fill(small + l + 1, small + N + 1, small[l]);
    for(int i = l + 1; i <= N; i++)
    {
        small[i] = small[l];
    }
     
    // Variable to store the answer
    int res = 0;
    for(int i = (k + X) / (1 + X); i <= k; i++)
    {
        if (1 * (i - 1) * (X + 1) + 1 <= N)
        {
             
            // Update res with maximum one
            res = Math.max(
                res, big[i] + small[N - 1 * (i - 1)  *
                                   (X + 1) - 1]);
        }
    }
     
    // Return res
    return res;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 9, 13, 16, 21, 6 };
    int X = 2;
    int K = 15;
    int N = arr.length;
 
    System.out.print(maxScore(arr, X, K, N));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 program, for the above approach
maxn = int(1e5)
     
# Utility function to calculate the
# sum of elements as a prefix array
def calc(arr, N) :
     
    arr.sort()
    arr[::-1]
 
    for i in range(1, N+1):
        arr[i] += arr[i - 1]
     
# Function to find the maximum score
# that can be achieved by rearranging
# the elements of given array
def maxScore(arr, X, K, N) :
     
    # Arrays to store small and big elements
    small = [10] * (maxn + 5)
    big = [10] * (maxn + 5)
    k = 0
    l = 0
 
    # Iterate and segregate big and
    # small elements
    for i in range(N):
        if (arr[i] > K) :
            big[++k] = arr[i]
         
        else :
            small[++l] = arr[i]
         
     
    # If k = 0, return the sum
    # of small[]
    if (k == 0) :
        sum = 0
        for i in range(1, N+1):
            sum += small[i]
         
        return sum
     
    # Prefix sums of small[]
    # and big[]
    calc(big, k)
    calc(small, l)
 
    # Initialize small[l] within the range
    for i in range(l + 1, N+1):
        small[i] = small[l]
         
 
    # Variable to store the answer
    res = 0
    for i in range((k + X) / (1 + X), k+1):
        if (1 * (i - 1) * (X + 1) + 1 <= N) :
 
            # Update res with maximum one
            res = max(
                res,
                big[i]
                    + small[N - 1 * (i - 1)
                           * (X + 1) - 1])
         
    # Return res
    return res
 
# Driver Code
arr = [ 9, 13, 16, 21, 6 ]
X = 2
K = 15
N = len(arr)
 
print(maxScore(arr, X, K, N))
 
# This code is contributed by sanjoy_62.

C#




// C# program, for the above approach
using System;
class GFG {
 
    static int maxn = (int)1e5;
 
    // Utility function to calculate the
    // sum of elements as a prefix array
    static void calc(int[] arr, int N)
    {
        Array.Sort(arr);
        arr = reverse(arr);
 
        for (int i = 1; i <= N; i++) {
            arr[i] += arr[i - 1];
        }
    }
 
    static int[] reverse(int[] a)
    {
        int i, n = a.Length + 1, t;
        for (i = 1; i < n / 2; i++) {
            t = a[i];
            a[i] = a[n - i - 1];
            a[n - i - 1] = t;
        }
        return a;
    }
 
    // Function to find the maximum score
    // that can be achieved by rearranging
    // the elements of given array
    static int maxScore(int[] arr, int X, int K, int N)
    {
 
        // Arrays to store small and big elements
        int[] small = new int[maxn + 5];
        int[] big = new int[maxn + 5];
        int k = 0, l = 0;
 
        // Iterate and segregate big and
        // small elements
        for (int i = 0; i < N; i++) {
            if (arr[i] > K) {
                big[++k] = arr[i];
            }
            else {
                small[++l] = arr[i];
            }
        }
 
        // If k = 0, return the sum
        // of small[]
        if (k == 0) {
            int sum = 0;
            for (int i = 1; i <= N; i++) {
                sum += small[i];
            }
            return sum;
        }
 
        // Prefix sums of small[]
        // and big[]
        calc(big, k);
        calc(small, l);
 
        // Initialize small[l] within the range
        // fill(small + l + 1, small + N + 1, small[l]);
        for (int i = l + 1; i <= N; i++) {
            small[i] = small[l];
        }
 
        // Variable to store the answer
        int res = 0;
        for (int i = (k + X) / (1 + X); i <= k; i++) {
            if (1 * (i - 1) * (X + 1) + 1 <= N) {
 
                // Update res with maximum one
                res = Math.Max(
                    res,
                    big[i]
                        + small[N - 1 * (i - 1) * (X + 1)
                                - 1]);
            }
        }
 
        // Return res
        return res;
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        int[] arr = { 9, 13, 16, 21, 6 };
        int X = 2;
        int K = 15;
        int N = arr.Length;
 
        Console.WriteLine(maxScore(arr, X, K, N));
    }
}
 
// This code is contributed by ukasp.

Javascript




<script>
 
       // JavaScript Program to implement
       // the above approach
       let maxn = 1e5;
 
       // Utility function to calculate the
       // sum of elements as a prefix array
       function calc(arr, N)
       {
           let arr1 = arr.slice(0, 1)
           let arr2 = arr.slice(1, N + 1)
           arr2.sort(function (a, b) { return a - b });
           arr2.reverse();
 
           arr = arr1.concat(arr2)
           for (let i = 1; i <= N; i++) {
               arr[i] += arr[i - 1];
           }
           return arr;
       }
 
       // Function to find the maximum score
       // that can be achieved by rearranging
       // the elements of given array
       function maxScore(arr, X, K, N)
       {
        
           // Arrays to store small and big elements
           let small = new Array(maxn + 5).fill(0)
           let big = new Array(maxn + 5).fill(0);
           let k = 0, l = 0;
 
           // Iterate and segregate big and
           // small elements
           for (let i = 0; i < N; i++) {
               if (arr[i] > K) {
                   big[++k] = arr[i];
               }
               else {
                   small[++l] = arr[i];
               }
           }
            
           // If k = 0, return the sum
           // of small[]
           if (k == 0) {
               let sum = 0;
               for (let i = 1; i < N; i++) {
                   sum += small[i];
               }
               return sum;
           }
            
           // Prefix sums of small[]
           // and big[]
           big = calc(big, k);
           small = calc(small, l);
 
           // Initialize small[l] within the range
           for (let i = l + 1; i <= N; i++) {
               small[i] = small[l];
           }
            
           // Variable to store the answer
           let res = 0;
           for (let i = Math.floor((k + X) / (1 + X)); i <= k; i++) {
               if (1 * (i - 1) * (X + 1) + 1 <= N) {
 
                   // Update res with maximum one
                   res = Math.max(
                       res,
                       big[i]
                       + small[N - 1
                       * (i - 1)
                       * (X + 1) - 1]);
               }
           }
            
           // Return res
           return res;
       }
 
       // Driver Code
       let arr = [9, 13, 16, 21, 6];
       let X = 2;
       let K = 15;
       let N = arr.length;
 
       document.write(maxScore(arr, X, K, N));
 
   // This code is contributed by Potta Lokesh
   </script>
Output
50

Time Complexity: O(N*logN)
Auxiliary Space: O(N)

 


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