Suma de la diferencia absoluta de todos los pares elevados a la potencia K

Dada una array arr[] de N enteros y un número K , la tarea es encontrar la suma de la diferencia absoluta de todos los pares elevados a la potencia K en una array dada, es decir, $\sum_{i=1}^{N} \sum_{j=1}^{N} \ |arr[i]-arr[j]|^K$ .
Ejemplos:

Entrada: arr[] = {1, 2, 3}, K = 1
Salida: 8
Explicación:
Suma de |1-1|+|1-2|+|1-3|+|2-1|+|2 -2|+|2-3|+|3-1|+|3-2|+|3-3| = 8
Entrada: arr[] = {1, 2, 3}, K = 3
Salida: 20
Explicación:
Suma de |1 – 1| ³ + |1 – 2| ³ + |1 – 3| ³ + |2 – 1| ³ + |2 – 2| ³ + |2 – 3| ³ + |3 – 1| ³ + |3 – 2| ³ + |3 – 3| ³ = 20

Enfoque ingenuo: la idea es generar todos los pares posibles y encontrar la diferencia absoluta de cada par elevado a la potencia K y sumarlos.
Complejidad de tiempo: O((log K)*N ² )
Espacio auxiliar: O(1)
Enfoque eficiente: Podemos mejorar la complejidad de tiempo del enfoque ingenuo con los siguientes cálculos:
Para todos los pares posibles, tenemos que encontrar el valor de

$Sum = \sum_{i=1}^{N} \sum_{j=1}^{N} |arr[i] - arr[j]|^{K}$

Dado que para pares (arr[i], arr[j]) el valor de $|arr[i]-arr[j]|^{K}$ se calcula dos veces. Entonces la ecuación anterior también se puede escribir como:

$Sum = 2 * \sum_{i=1}^{N} \sum_{j=1}^{i-1} |arr[i] - arr[j]|^{K}$

Escribiendo $|arr[i]-arr[j]|^{K}$ en términos de ecuación binomial:

$Sum = 2 * \sum_{i=1}^{N} \sum_{j=1}^{i-1} \sum_{a=0}^{K} \binom{K}{a}arr[i]^{K}*(-arr[j])^{K - a}$ (Ecuación 1)

Sea Pre[i][a] = $\sum_{j=1}^{i-1}(-arr[j])^{K - a}$ (Ecuación 2)
De la Ecuación 1 y la Ecuación 2, tenemos

$Sum = 2 * \sum_{i=1}^{N} \sum_{a=0}^{K} \binom{K}{a}Pre[i][a]*arr[i]^{K}$

El valor de Pre[i][a] se puede calcular como:

Pre[i][a] = {(-arr[1]) ^{K – a} + (-arr[2]) ^{K – a} … . . +(-arr[i – 1]) ^{K – a} }.
Entonces, Pre[i+1][a] = Pre[i][a]+(-arr[i]) ^{K – a}

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
#define ll long long
using namespace std;
 
class Solution {
 
public:
    // Since K can be 100 max
    ll ncr[101][101];
    int n, k;
    vector<ll> A;
 
    // Constructor
    Solution(int N, int K, vector<ll> B)
    {
        // Initializing with -1
        memset(ncr, -1, sizeof(ncr));
 
        n = N;
        k = K;
        A = B;
 
        // Making vector A as 1-Indexing
        A.insert(A.begin(), 0);
    }
 
    ll f(int N, int K);
 
    ll pairsPower();
};
 
// To Calculate the value nCk
ll Solution::f(int n, int k)
{
    if (k == 0)
        return 1LL;
 
    if (n == k)
        return 1LL;
 
    if (n < k)
        return 0;
 
    if (ncr[n][k] != -1)
        return ncr[n][k];
 
    // Since nCj = (n-1)Cj + (n-1)C(j-1);
    return ncr[n][k] = f(n - 1, k)
                       + f(n - 1, k - 1);
}
 
// Function that summation of absolute
// differences of all pairs raised
// to the power k
ll Solution::pairsPower()
{
    ll pre[n + 1][k + 1];
    ll ans = 0;
 
    // Sort the given array
    sort(A.begin() + 1, A.end());
 
    // Precomputation part, O(n*k)
    for (int i = 1; i <= n; ++i) {
        pre[i][0] = 1LL;
 
        for (int j = 1; j <= k; j++) {
            pre[i][j] = A[i]
                        * pre[i][j - 1];
        }
 
        if (i != 1) {
            for (int j = 0; j <= k; ++j)
                pre[i][j] = pre[i][j]
                            + pre[i - 1][j];
        }
    }
 
    // Traverse the array arr[]
    for (int i = n; i >= 2; --i) {
 
        // For each K
        for (int j = 0; j <= k; j++) {
 
            ll val = f(k, j);
 
            ll val1 = pow(A[i], k - j)
                      * pre[i - 1][j];
 
            val = val * val1;
 
            if (j % 2 == 0)
                ans = (ans + val);
 
            else
                ans = (ans - val);
        }
    }
 
    ans = 2LL * ans;
 
    // Return the final answer
    return ans;
}
 
// Driver Code
int main()
{
    // Given N and K
    int N = 3;
    int K = 3;
 
    // Given array
    vector<ll> arr = { 1, 2, 3 };
 
    // Creation of Object of class
    Solution obj(N, K, arr);
 
    // Function Call
    cout << obj.pairsPower() << endl;
    return 0;
}

Python3

# Python3 program for the above approach
class Solution:
     
    def __init__(self, N, K, B):
 
        self.ncr = []
         
        # Since K can be 100 max
        for i in range(101):
            temp = []
            for j in range(101):
                 
                # Initializing with -1
                temp.append(-1)
                 
            self.ncr.append(temp)
 
        self.n = N
        self.k = K
 
        # Making vector A as 1-Indexing
        self.A = [0] + B
 
    # To Calculate the value nCk
    def f(self, n, k):
         
        if k == 0:
            return 1
        if n == k:
            return 1
        if n < k:
            return 0
             
        if self.ncr[n][k] != -1:
            return self.ncr[n][k]
 
        # Since nCj = (n-1)Cj + (n-1)C(j-1);
        self.ncr[n][k] = (self.f(n - 1, k) +
                          self.f(n - 1, k - 1))
                           
        return self.ncr[n][k]
 
    # Function that summation of absolute
    # differences of all pairs raised
    # to the power k
    def pairsPower(self):
         
        pre = []
         
        for i in range(self.n + 1):
            temp = []
            for j in range(self.k + 1):
                temp.append(0)
                 
            pre.append(temp)
 
        ans = 0
 
        # Sort the given array
        self.A.sort()
 
        # Precomputation part, O(n*k)
        for i in range(1, self.n + 1):
            pre[i][0] = 1
 
            for j in range(1, self.k + 1):
                pre[i][j] = (self.A[i] *
                                pre[i][j - 1])
 
            if i != 1:
                for j in range(self.k + 1):
                    pre[i][j] = (pre[i][j] +
                                 pre[i - 1][j])
 
        # Traverse the array arr[]
        for i in range(self.n, 1, -1):
             
            # For each K
            for j in range(self.k + 1):
                val = self.f(self.k, j)
                val1 = (pow(self.A[i],
                            self.k - j) *
                             pre[i - 1][j])
 
                val = val * val1
 
                if j % 2 == 0:
                    ans = ans + val
                else:
                    ans = ans - val
 
        ans = 2 * ans
 
        # Return the final answer
        return ans
 
# Driver code
if __name__ == '__main__':
 
    # Given N and K
    N = 3
    K = 3
     
    # Given array
    arr = [ 1, 2, 3 ]
 
    # Creation of object of class
    obj = Solution(N, K, arr)
     
    # Function call
    print(obj.pairsPower())
 
# This code is contributed by Shivam Singh

C#

// C# program to implement above approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG
{
 
    // Since K can be 100 max
    static long[][] ncr = new long[101][];
    static int n, k;
    static List<long> A;
 
    // To Calculate the value nCk
    static long f(int n, int k)
    {
        if (k == 0)
            return (long)1;
 
        if (n == k)
            return (long)1;
 
        if (n < k)
            return 0;
 
        if (ncr[n][k] != -1)
            return ncr[n][k];
 
        ncr[n][k] = f(n - 1, k) + f(n - 1, k - 1);
        // Since nCj = (n-1)Cj + (n-1)C(j-1);
        return ncr[n][k];
    }
 
    // Function that summation of absolute
    // differences of all pairs raised
    // to the power k
    static long pairsPower()
    {
        long[][] pre = new long[n + 1][];
        long ans = 0;
        for(int i = 0 ; i <= n ; i++){
            pre[i] = new long[k + 1];
        }
 
        // Sort the given array
        A.Sort(1, A.Count - 1, Comparer<long>.Default);
 
        // Precomputation part, O(n*k)
        for (int i = 1 ; i <= n ; ++i) {
            pre[i][0] = (long)1;
 
            for (int j = 1 ; j <= k ; j++) {
                pre[i][j] = A[i] * pre[i][j - 1];
            }
 
            if (i != 1) {
                for (int j = 0 ; j <= k ; ++j){
                    pre[i][j] = pre[i][j] + pre[i - 1][j];
                }
            }
        }
 
        // Traverse the array arr[]
        for (int i = n ; i >= 2 ; --i) {
 
            // For each K
            for (int j = 0 ; j <= k ; j++) {
 
                long val = f(k, j);
 
                long val1 = (long)(Math.Pow(A[i], k - j)) * pre[i - 1][j];
 
                val = val * val1;
 
                if (j % 2 == 0)
                    ans = (ans + val);
 
                else
                    ans = (ans - val);
            }
        }
 
        ans = (long)(2) * ans;
 
        // Return the final answer
        return ans;
    }
 
    // Driver code
    public static void Main(string[] args){
 
        // Given N and K
        n = 3;
        k = 3;
 
        // Initializing with -1
        for(int i = 0 ; i <= 100 ; i++){
            ncr[i] = new long[101];
            for(int j = 0 ; j <= 100 ; j++){
                ncr[i][j] = -1;
            }
        }
 
        // Given array
        // Making vector A as 1-Indexing
        // So adding 0 in the beginning
        A = new List<long>{ 0, 1, 2, 3 };
 
        // Function Call
        Console.WriteLine(pairsPower());
 
    }
}

Producción:

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

Publicación traducida automáticamente

Artículo escrito por ssatyanand7 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

C++

Python3

C#

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