Dada una array arr[] que consiste en N enteros positivos y dos enteros positivos A y B , la tarea es reemplazar cada elemento de la array con la diferencia absoluta de las potencias más cercanas de A y B . Si existen dos potencias más cercanas, entonces elija el máximo de los dos.
Ejemplos:
Entrada: arr[] = {5, 12, 25}, A = 2, B = 3
Salida: 1 7 5
Explicación:
- Para el elemento arr[0]( = 5): La potencia de 2 más cercana es 4 y la potencia de 3 más cercana es 3. Por lo tanto, la diferencia absoluta de 4 y 3 es 1.
- Para arr[1]( = 12): La potencia más cercana de 2 es 16 y la potencia más cercana de 3 es 9. Por lo tanto, la diferencia absoluta de 16 y 9 es 7.
- Para arr[2]( = 5): La potencia más cercana de 2 es 32 y la potencia más cercana de 3 es 27. Por lo tanto, la diferencia absoluta de 27 y 32 es 5.
Por lo tanto, la array modificada es {1, 7, 5}.
Entrada: arr[] = {32, 3, 7}, a = 2, b = 3
Salida: 0 1 1
Enfoque: el problema dado se puede resolver encontrando la potencia más cercana de A y B para cada elemento de la array y actualizándola a la diferencia absoluta de los dos valores obtenidos.
Sigue los pasos para resolver el problema
- Recorra la array dada arr[] y realice los siguientes pasos:
- Encuentra los dos valores que son las potencias perfectas de a menor que y mayor que arr[i] y encuentra el más cercano de los dos valores.
- Encuentra los dos valores que son las potencias perfectas de b menor que y mayor que arr[i] y encuentra el más cercano de los dos valores.
- Encuentre la diferencia absoluta entre los dos valores más cercanos obtenidos y actualice arr[i] con ella.
- Después de completar los pasos anteriores, imprima la array modificada arr[] .
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to print the array
void printArray(int arr[], int N)
{
// Traverse the array
for (int i = 0; i < N; i++) {
cout << arr[i] << " ";
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
void nearestPowerDiff(int arr[], int N,
int a, int b)
{
// Traverse the array arr[]
for (int i = 0; i < N; i++) {
// Find the log a of arr[i]
int log_a = log(arr[i]) / log(a);
// Find the power of a less
// than and greater than a
int A = pow(a, log_a);
int B = pow(a, log_a + 1);
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
int log_b = log(arr[i]) / log(b);
// Find the power of b less than
// and greater than b
A = pow(b, log_b);
B = pow(b, log_b + 1);
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver Code
int main()
{
int arr[] = { 5, 12, 25 };
int A = 2, B = 3;
int N = sizeof(arr) / sizeof(arr[0]);
nearestPowerDiff(arr, N, A, B);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG{
// Function to print the array
static void printArray(int[] arr, int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
System.out.print(arr[i] + " ");
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
static void nearestPowerDiff(int[] arr, int N,
int a, int b)
{
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Find the log a of arr[i]
int log_a = (int)(Math.log(arr[i]) /
Math.log(a));
// Find the power of a less
// than and greater than a
int A = (int)(Math.pow(a, log_a));
int B = (int)(Math.pow(a, log_a + 1));
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
int log_b = (int)(Math.log(arr[i]) /
Math.log(b));
// Find the power of b less than
// and greater than b
A = (int)(Math.pow(b, log_b));
B = (int)(Math.pow(b, log_b + 1));
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = Math.abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver Code
public static void main(String[] args)
{
int[] arr = { 5, 12, 25 };
int A = 2, B = 3;
int N = arr.length;
nearestPowerDiff(arr, N, A, B);
}
}
// This code is contributed by subham348
Python3
# Python3 program for the above approach import math # Function to print the array def printArray(arr, N): # Traverse the array for i in range(N): print(arr[i], end = " ") # Function to modify array elements # by absolute difference of the # nearest perfect power of a and b def nearestPowerDiff(arr, N, a, b): # Traverse the array arr[] for i in range(N): # Find the log a of arr[i] log_a = int(math.log(arr[i]) / math.log(a)) # Find the power of a less # than and greater than a A = int(pow(a, log_a)) B = int(pow(a, log_a + 1)) if ((arr[i] - A) < (B - arr[i])): log_a = A else: log_a = B # Find the log b of arr[i] log_b = int(math.log(arr[i]) / math.log(b)) # Find the power of b less than # and greater than b A = int(pow(b, log_b)) B = int(pow(b, log_b + 1)) if ((arr[i] - A) < (B - arr[i])): log_b = A else: log_b = B # Update arr[i] with absolute # difference of log_a & log _b arr[i] = abs(log_a - log_b) # Print the modified array printArray(arr, N) # Driver Code arr = [ 5, 12, 25 ] A = 2 B = 3 N = len(arr) nearestPowerDiff(arr, N, A, B) # This code is contributed by sanjoy_62
C#
// C# program for the above approach
using System;
public class GFG{
// Function to print the array
static void printArray(int[] arr, int N)
{
// Traverse the array
for(int i = 0; i < N; i++)
{
Console.Write(arr[i] + " ");
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
static void nearestPowerDiff(int[] arr, int N,
int a, int b)
{
// Traverse the array arr[]
for(int i = 0; i < N; i++)
{
// Find the log a of arr[i]
int log_a = (int)(Math.Log(arr[i]) /
Math.Log(a));
// Find the power of a less
// than and greater than a
int A = (int)(Math.Pow(a, log_a));
int B = (int)(Math.Pow(a, log_a + 1));
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
int log_b = (int)(Math.Log(arr[i]) /
Math.Log(b));
// Find the power of b less than
// and greater than b
A = (int)(Math.Pow(b, log_b));
B = (int)(Math.Pow(b, log_b + 1));
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = Math.Abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver Code
public static void Main(string[] args)
{
int[] arr = { 5, 12, 25 };
int A = 2, B = 3;
int N = arr.Length;
nearestPowerDiff(arr, N, A, B);
}
}
// This code is contributed by AnkThon
Javascript
<script>
// JavaScript implementation of the above approach
// Function to print the array
function printArray(arr, N)
{
// Traverse the array
for(let i = 0; i < N; i++)
{
document.write(arr[i] + " ");
}
}
// Function to modify array elements
// by absolute difference of the
// nearest perfect power of a and b
function nearestPowerDiff(arr, N,
a, b)
{
// Traverse the array arr[]
for(let i = 0; i < N; i++)
{
// Find the log a of arr[i]
let log_a = Math.floor(Math.log(arr[i]) /
Math.log(a));
// Find the power of a less
// than and greater than a
let A = Math.floor(Math.pow(a, log_a));
let B = Math.floor(Math.pow(a, log_a + 1));
if ((arr[i] - A) < (B - arr[i]))
log_a = A;
else
log_a = B;
// Find the log b of arr[i]
let log_b = Math.floor(Math.log(arr[i]) /
Math.log(b));
// Find the power of b less than
// and greater than b
A = Math.floor(Math.pow(b, log_b));
B = Math.floor(Math.pow(b, log_b + 1));
if ((arr[i] - A) < (B - arr[i]))
log_b = A;
else
log_b = B;
// Update arr[i] with absolute
// difference of log_a & log _b
arr[i] = Math.abs(log_a - log_b);
}
// Print the modified array
printArray(arr, N);
}
// Driver code
let arr = [ 5, 12, 25 ];
let A = 2, B = 3;
let N = arr.length;
nearestPowerDiff(arr, N, A, B);
// This code is contributed by susmitakundugoaldanga.
</script>
Producción:
1 7 5
Complejidad temporal: O(N)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por subhammahato348 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA