Find the absolute difference between the nearest powers of two given integers for every array element

Given an array arr[] consisting of N positive integers and two positive integers A and B, the task is to replace each array element with the absolute difference of the nearest powers of A and B. If there exists two nearest powers, then choose the maximum of the two.

Examples:

Input: arr[] = {5, 12, 25}, A = 2, B = 3
Output: 1 7 5
Explanation:

• For element arr[0]( = 5): The nearest power of 2 is 4 and the nearest power of 3 is 3. Therefore, the absolute difference of 4 and 3 is 1.
• For arr[1]( = 12): The nearest power of 2 is 16 and the nearest power of 3 is 9. Therefore, the absolute difference of 16 and 9 is 7.
• For arr[2]( = 5): The nearest power of 2 is 32 and the nearest power of 3 is 27. Therefore, the absolute difference of 27 and 32 is 5.

Therefore, the modified array is {1, 7, 5}.

Input: arr[] = {32, 3, 7}, a = 2, b = 3
Output: 0 1 1

Approach: The given problem can be solved by finding both the nearest power of A and B for every array element and update it to the absolute difference of the two values obtained. 
Follow the steps to solve the problem

• Traverse the given array arr[] and perform the following steps:
  • Find the two values that are the perfect powers of a less than and greater than arr[i] and find the closest of the two values.
  • Find the two values that are the perfect powers of b less than and greater than arr[i] and find the closest of the two values.
  • Find the absolute difference between the two closest values obtained and update arr[i] with it.
• After completing the above steps, print the modified array arr[].

Below is the implementation of the above approach:

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>

Output:

1 7 5

Time Complexity: O(N)
Auxiliary Space: O(1)