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# Maximize sum of given array by rearranging array such that the difference between adjacent elements is atmost 1

• Difficulty Level : Medium
• Last Updated : 20 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to maximize the sum of the array element such that the first element of the array is 1 and the difference between the adjacent elements of the array is at most 1 after performing the following operations:

• Rearrange the array elements in any way.
• Reduce any element to any number that is at least 1.

Examples:

Input: arr[] = {3, 5, 1}
Output: 6
Explanation:
One possible arrangement is {1, 2, 3} having maximum possible sum 6.

Input: arr[] = {1, 2, 2, 2, 3, 4, 5}
Output: 19
Explanation:
One possible arrangement is {1, 2, 2, 2, 3, 4, 5} having maximum possible sum 19.

Naive Approach: The simplest approach is to sort the given array then traverse in the sorted array and reduced the element that doesn’t satisfy the given condition.

Time Complexity: O(N * log N), where N is the size of the given array.
Auxiliary Space: O(N)

Efficient Approach: The idea is to use the Hashing concept of storing the frequencies of the elements of the given array. Follow the below steps to solve the problem:

• Create an auxiliary array count[] of size (N+1) to store frequency of arr[i].
• While storing the frequency in count[] and if arr[i] greater than N then increment count[N].
• Initialize the size and ans as 0 that stores the previously selected integer and maximum possible sum respectively.
• Traverse the given array count[] array using variable K and do the following:
• Iterate while a loop for each K until count[K] > 0 and size < K.
• Increment size by 1 and ans by size and reduce count[K] by 1 inside while loop.
• Increment ans with K*count[K] after the while loop ends.
• After the above steps, print the value of ans as the maximum possible sum.

Below is the implementation of the above approach:

## C++

 // C++ program for the above approach#include using namespace std; // Function to find maximum possible// sum after changing the array elements// as per the given constraintslong maxSum(int a[], int n){         // Stores the frequency of    // elements in given array    int count[n + 1] = {0};     // Update frequency    for(int i = 0; i < n; i++)        count[min(a[i], n)]++;     // Stores the previously    // selected integer    int size = 0;     // Stores the maximum possible sum    long ans = 0;     // Traverse over array count[]    for(int k = 1; k <= n; k++)    {                 // Run loop for each k        while (count[k] > 0 && size < k)        {            size++;            ans += size;            count[k]--;        }         // Update ans        ans += k * count[k];    }     // Return maximum possible sum    return ans;} // Driver Codeint main(){         // Given array arr[]    int arr[] = { 3, 5, 1 };     // Size of array    int n = sizeof(arr) / sizeof(arr[0]);     // Function Call    cout << (maxSum(arr, n));    return 0;} // This code is contributed by akhilsaini

## Java

 // Java program for the above approach import java.util.*; class GFG {     // Function to find maximum possible    // sum after changing the array elements    // as per the given constraints    static long maxSum(int[] a)    {        // Length of given array        int n = a.length;         // Stores the frequency of        // elements in given array        int[] count = new int[n + 1];         // Update frequency        for (int x : a)            count[Math.min(x, n)]++;         // stores the previously        // selected integer        int size = 0;         // Stores the maximum possible sum        long ans = 0;         // Traverse over array count[]        for (int k = 1; k <= n; k++) {             // Run loop for each k            while (count[k] > 0 && size < k) {                size++;                ans += size;                count[k]--;            }             // Update ans            ans += k * count[k];        }         // Return maximum possible sum        return ans;    }     // Driver Code    public static void main(String[] args)    {        // Given array arr[]        int[] arr = { 3, 5, 1 };         // Function Call        System.out.println(maxSum(arr));    }}

## Python3

 # Python3 program for the above approach # Function to find maximum possible# sum after changing the array elements# as per the given constraintsdef maxSum(a, n):     # Stores the frequency of    # elements in given array    count = [0] * (n + 1)     # Update frequency    for i in range(0, n):        count[min(a[i], n)] += 1     # stores the previously    # selected integer    size = 0     # Stores the maximum possible sum    ans = 0     # Traverse over array count[]    for k in range(1, n + 1):                 # Run loop for each k        while (count[k] > 0 and size < k):            size += 1            ans += size            count[k] -= 1         # Update ans        ans += k * count[k]     # Return maximum possible sum    return ans # Driver Codeif __name__ == '__main__':     # Given array arr[]    arr = [ 3, 5, 1 ]     # Size of array    n = len(arr)     # Function Call    print(maxSum(arr, n)) # This code is contributed by akhilsaini

## C#

 // C# program for the above approachusing System; class GFG{ // Function to find maximum possible// sum after changing the array elements// as per the given constraintsstatic long maxSum(int[] a){         // Length of given array    int n = a.Length;     // Stores the frequency of    // elements in given array    int[] count = new int[n + 1];     // Update frequency    for(int i = 0; i < n; i++)        count[Math.Min(a[i], n)]++;     // stores the previously    // selected integer    int size = 0;     // Stores the maximum possible sum    long ans = 0;     // Traverse over array count[]    for(int k = 1; k <= n; k++)    {                 // Run loop for each k        while (count[k] > 0 && size < k)        {            size++;            ans += size;            count[k]--;        }         // Update ans        ans += k * count[k];    }     // Return maximum possible sum    return ans;} // Driver Codepublic static void Main(){         // Given array arr[]    int[] arr = { 3, 5, 1 };     // Function call    Console.Write(maxSum(arr));}} // This code is contributed by akhilsaini

## Javascript



Output:
6

Time Complexity: O(N), where N is the size of the given array.
Auxiliary Space: O(N)

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