# Replace all array elements with the nearest power of its previous element

• Last Updated : 16 Apr, 2021

Given a circular array arr[] consisting of N integers, the task is to replace all the array elements with the nearest possible power of its previous adjacent element

Examples:

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Input: arr[] = {2, 4, 6, 3, 11}
Output: 1 4 4 6 9
Explanation:
Power of 11 which is nearest to 2 —> 110 = 1
Power of 2 which is nearest to 4 —> 22 = 4
Power of 4 which is nearest to 6 —> 41 = 4
Power of 6 which is nearest to 3 —> 61 = 6
Power of 3 which is nearest to 11—> 32 = 9

Input: arr[] = {3, 2, 4, 3}
Output: 3 3 4 4
Explanation:
Power of 3 which is nearest to 3 —> 31 = 3
Power of 3 which is nearest to 2 —> 31 = 3
Power of 2 which is nearest to 4 —> 22 = 4
Power of 4 which is nearest to 3 —> 41 = 4

Approach: The idea to solve this problem is to traverse the array and for each array element arr[i], find the power of the previous element, say X, which is nearest to arr[i], i.e. XK which is closest to arr[i]. Follow the steps:

• Calculate the value of K, which is equal to the floor value of logx(Y).
• Therefore, K and (K + 1) will be the two integers for which the power could be nearest.
• Calculate XK and X(K + 1) and check which is nearer to Y. Then, print that value.

Below is the implementation of the above approach:

## C++

 `// CPP program for the above approach``#include ``#include ``using` `namespace` `std;` `// Function to calculate log x for given base``int` `LOG(``int` `a, ``int` `b) {``    ``return` `log``(a) / ``log``(b);``}` `// Function to replace all array``// elements with the nearest power``// of previous adjacent nelement``void` `repbyNP(``int` `*arr,``int` `n)``{``  ` `    ``// For the first element, set the last``    ``// array element to its previous element``    ``int` `x = arr[n - 1];``  ` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``      ` `        ``// Find K for which x ^ k is``        ``// nearest to arr[i]``        ``int` `k = LOG(arr[i], x);``        ``int` `temp = arr[i];``      ` `        ``// Find the power to x nearest to arr[i]``        ``if` `(``abs``(``pow``(x,k) - arr[i]) < ``abs``(``pow``(x,k+1) - arr[i]))``            ``arr[i] = ``pow``(x, k);``        ``else``            ``arr[i] = ``pow``(x, k + 1);``      ` `        ``// Update x``        ``x = temp;``    ``}``}``int` `main()``{``  ` `    ``// Driver Code``    ``int` `arr = {2, 4, 6, 3, 11};``    ``int` `n = ``sizeof``(arr)/``sizeof``(arr);``  ` `    ``// Function Call``    ``repbyNP(arr,n);``  ` `    ``// Display the array``    ``for``(``int` `i = 0; i < n; i++)``        ``cout<

## Java

 `// Java program for the above approach``import` `java.util.*;`` ` `class` `GFG``{`` ` `// Function to calculate log x for given base``static` `int` `LOG(``int` `a, ``int` `b) {``    ``return` `(``int``)(Math.log(a) / Math.log(b));``}`` ` `// Function to replace all array``// elements with the nearest power``// of previous adjacent nelement``static` `void` `repbyNP(``int``[] arr,``int` `n)``{``   ` `    ``// For the first element, set the last``    ``// array element to its previous element``    ``int` `x = arr[n - ``1``];``   ` `    ``// Traverse the array``    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``       ` `        ``// Find K for which x ^ k is``        ``// nearest to arr[i]``        ``int` `k = LOG(arr[i], x);``        ``int` `temp = arr[i];``       ` `        ``// Find the power to x nearest to arr[i]``        ``if` `(Math.abs(Math.pow(x,k) - arr[i]) < Math.abs(Math.pow(x,k+``1``) - arr[i]))``            ``arr[i] = (``int``)Math.pow(x, k);``        ``else``            ``arr[i] = (``int``)Math.pow(x, k + ``1``);``       ` `        ``// Update x``        ``x = temp;``    ``}``}`` ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = {``2``, ``4``, ``6``, ``3``, ``11``};``    ``int` `n = arr.length;``   ` `    ``// Function Call``    ``repbyNP(arr,n);``   ` `    ``// Display the array``    ``for``(``int` `i = ``0``; i < n; i++)``        ``System.out.print(arr[i] + ``" "``);``}``}` `// This code is contributed by sanjoy_62.`

## Python3

 `# Python3 program for the above approach``import` `math` `# Function to calculate log x for given base``def` `LOG(x, base):``    ``return` `int``(math.log(x)``/``math.log(base))` `# Function to replace all array``# elements with the nearest power``# of previous adjacent nelement``def` `repbyNP(arr):` `    ``# For the first element, set the last``    ``# array element to its previous element``    ``x ``=` `arr[``-``1``]` `    ``# Traverse the array``    ``for` `i ``in` `range``(``len``(arr)):` `        ``# Find K for which x ^ k is``        ``# nearest to arr[i]``        ``k ``=` `LOG(arr[i], x)``        ``temp ``=` `arr[i]` `        ``# Find the power to x nearest to arr[i]``        ``if` `abs``(x``*``*``k ``-` `arr[i]) < ``abs``(x``*``*``(k ``+` `1``) ``-` `arr[i]):``            ``arr[i] ``=` `x``*``*``k``        ``else``:``            ``arr[i] ``=` `x``*``*``(k ``+` `1``)` `        ``# Update x``        ``x ``=` `temp` `    ``# Return the array``    ``return` `arr`  `# Driver Code``arr ``=` `[``2``, ``4``, ``6``, ``3``, ``11``]` `# Function Call``print``(repbyNP(arr))`

## C#

 `// C# program for the above approach``using` `System;` `class` `GFG{` `// Function to calculate log x for given base``static` `int` `LOG(``int` `a, ``int` `b) {``    ``return` `(``int``)(Math.Log(a) / Math.Log(b));``}``  ` `// Function to replace all array``// elements with the nearest power``// of previous adjacent nelement``static` `void` `repbyNP(``int``[] arr,``int` `n)``{``    ` `    ``// For the first element, set the last``    ``// array element to its previous element``    ``int` `x = arr[n - 1];``    ` `    ``// Traverse the array``    ``for``(``int` `i = 0; i < n; i++)``    ``{``        ` `        ``// Find K for which x ^ k is``        ``// nearest to arr[i]``        ``int` `k = LOG(arr[i], x);``        ``int` `temp = arr[i];``        ` `        ``// Find the power to x nearest to arr[i]``        ``if` `(Math.Abs(Math.Pow(x, k) - arr[i]) <``            ``Math.Abs(Math.Pow(x, k + 1) - arr[i]))``            ``arr[i] = (``int``)Math.Pow(x, k);``        ``else``            ``arr[i] = (``int``)Math.Pow(x, k + 1);``        ` `        ``// Update x``        ``x = temp;``    ``}``}` `// Driver Code``public` `static` `void` `Main(``string``[] args)``{``    ``int``[] arr = {2, 4, 6, 3, 11};``    ``int` `n = arr.Length;``    ` `    ``// Function Call``    ``repbyNP(arr,n);``    ` `    ``// Display the array``    ``for``(``int` `i = 0; i < n; i++)``        ``Console.Write(arr[i] + ``" "``);``}``}` `// This code is contributed by code_hunt.`

## Javascript

 ``
Output:
`[1, 4, 4, 1, 9]`

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

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