Maximizar el resto de la suma de un par de elementos de array con diferente módulo de paridad K

Dada una array arr[] de tamaño N, que consta de N/2 enteros pares e impares cada uno, y un entero K , la tarea es encontrar el resto máximo de la suma de un par de elementos de la array de diferente paridad módulo K.

Ejemplos:

Entrada: arr[] = {3, 2, 4, 11, 6, 7}, K = 7
Salida: 6
Explicación: 
Suma de un par de elementos de array = 2 + 11
Suma % K = 13 % 7 = 6.
Por lo tanto , el resto máximo posible es 6.

Entrada: arr[] = {8, 11, 17, 16}, K = 13
Salida: 12 

Enfoque: siga los pasos a continuación para resolver el problema:

• Inicialice un HashSet , digamos even , para almacenar todos los elementos de array pares.
• Inicialice un TreeSet , digamos impar, para almacenar todos los elementos de array impares.
• Inicialice una variable, digamos max_rem, para almacenar el resto máximo posible.
• Recorra el HashSet y para cada elemento, encuentre su complemento y búsquelo en el conjunto impar , que es menor que igual a su complemento.
• Actualice max_rem con la suma de los elementos y su complemento.
• Imprime el resto máximo, es decir, el valor de max_rem .

A continuación se muestra la implementación del enfoque anterior:

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the maximum
// remainder of sum of a pair
// of array elements modulo K
void maxRemainder(int A[], int N, int K)
{
     
    // Stores all even numbers
    unordered_set<int> even;
 
    // Stores all odd numbers
    set<int> odd;
 
    // Segregate remainders of even
    // and odd numbers in respective sets
    for(int i = 0; i < N; i++)
    {
        int num = A[i];
         
        if (num % 2 == 0)
            even.insert(num % K);
        else
            odd.insert(num % K);
    }
 
    // Stores the maximum
    // remainder obtained
    int max_rem = 0;
 
    // Find the complement of remainder
    // of each even number in odd set
    for(int x : even)
    {
         
        // Find the complement
        // of remainder x
        int y = K - 1 - x;
 
        auto it = odd.upper_bound(y);
        if (it != odd.begin())
        {
            it--;
            max_rem = max(max_rem, x + *it);
        }
    }
 
    // Print the answer
    cout << max_rem;
}
 
// Driver code
int main()
{
     
    // Given array
    int arr[] = { 3, 2, 4, 11, 6, 7 };
 
    // Size of the array
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Given value of K
    int K = 7;
 
    maxRemainder(arr, N, K);
 
    return 0;
}
 
// This code is contributed by Kingash

// Java program for the above approach
 
import java.util.*;
 
class GFG {
 
    // Function to find the maximum
    // remainder of sum of a pair
    // of array elements modulo K
    static void maxRemainder(int A[],
                             int N, int K)
    {
        // Stores all even numbers
        HashSet<Integer> even
          = new HashSet<>();
 
        // Stores all odd numbers
        TreeSet<Integer> odd
          = new TreeSet<>();
 
        // Segregate remainders of even
        // and odd numbers in respective sets
        for (int num : A) {
            if (num % 2 == 0)
                even.add(num % K);
            else
                odd.add(num % K);
        }
 
        // Stores the maximum
        // remainder obtained
        int max_rem = 0;
 
        // Find the complement of remainder
        // of each even number in odd set
        for (int x : even) {
 
            // Find the complement
            // of remainder x
            int y = K - 1 - x;
            if (odd.floor(y) != null)
                max_rem
                    = Math.max(
              max_rem,
              x + odd.floor(y));
        }
 
        // Print the answer
        System.out.print(max_rem);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given array
        int arr[] = { 3, 2, 4, 11, 6, 7 };
 
        // Size of the array
        int N = arr.length;
 
        // Given value of K
        int K = 7;
 
        maxRemainder(arr, N, K);
    }
}

# Python3 program for the above approach
from bisect import bisect_left
 
# Function to find the maximum
# remainder of sum of a pair
# of array elements modulo K
def maxRemainder(A, N, K):
     
    # Stores all even numbers
    even = {}
 
    # Stores all odd numbers
    odd = {}
 
    # Segregate remainders of even
    # and odd numbers in respective sets
    for i in range(N):
        num = A[i]
 
        if (num % 2 == 0):
            even[num % K] = 1
        else:
            odd[num % K] = 1
 
    # Stores the maximum
    # remainder obtained
    max_rem = 0
 
    # Find the complement of remainder
    # of each even number in odd set
    for x in even:
         
        # Find the complement
        # of remainder x
        y = K - 1 - x
        od = list(odd.keys())
        it = bisect_left(od, y)
         
        if (it != 0):
            max_rem = max(max_rem, x + od[it])
             
    # Print the answer
    print (max_rem)
 
# Driver code
if __name__ == '__main__':
     
    # Given array
    arr = [3, 2, 4, 11, 6, 7]
 
    # Size of the array
    N = len(arr)
 
    # Given value of K
    K = 7
 
    maxRemainder(arr, N, K)
     
# This code is contributed by mohit kumar 29

// JavaScript program for the above approach
 
// Function to find the upper bound of an array.
function upper_bound(arr, X)
{
    let mid;
  
    // Initialise starting index and
    // ending index
    let low = 0;
    let high = arr.length;
  
    // Till low is less than high
    while (low < high) {
        // Find the middle index
        mid = low + Math.floor((high - low) / 2);
  
        // If X is greater than or equal
        // to arr[mid] then find
        // in right subarray
        if (X >= arr[mid]) {
            low = mid + 1;
        }
  
        // If X is less than arr[mid]
        // then find in left subarray
        else {
            high = mid;
        }
    }
    
    // if X is greater than arr[n-1]
    if(low < N && arr[low] <= X) {
       low++;
    }
  
    // Return the upper_bound index
    return low;
}
 
// Function to find the maximum
// remainder of sum of a pair
// of array elements modulo K
function maxRemainder(A, N, K)
{
     
    // Stores all even numbers
    let even = new Array();
 
    // Stores all odd numbers
    let odd = new Array();
 
    // Segregate remainders of even
    // and odd numbers in respective sets
    for(let i = 0; i < N; i++)
    {
        let num = A[i];
         
        if (num % 2 == 0 && !even.includes(num%K)){
            even.push(num % K);
        }    
        else if(num%2 != 0 && !odd.includes(num%K)){
            odd.push(num % K);
        }  
    }
 
    odd.sort();
    // Stores the maximum
    // remainder obtained
    let max_rem = 0;
 
    // Find the complement of remainder
    // of each even number in odd set
    for(let i = 0;i < even.length; i++){
        let x = even[i];
        // Find the complement
        // of remainder xd
        // console.log(x);
        let y = K - 1 - x;
 
        let it = upper_bound(odd, y);
        if (it != 0)
        {
            it--;
            max_rem = Math.max(max_rem, x + odd[it]);
        }
    }
 
    // Print the answer
    console.log(max_rem);
}
 
// Driver code
// Given array
let arr = [3, 2, 4, 11, 6, 7];
 
// Size of the array
let N = arr.length;
 
// Given value of K
let K = 7;
 
maxRemainder(arr, N, K);
 
 
// This code is contributed by Nidhi goel

6

Complejidad temporal: O(N * logN)
Espacio auxiliar: O(N)