Replace array elements that contains K as a digit with the nearest power of K
Last Updated :
28 Jun, 2021
Given an array arr[] of size N and an integer K, the task is to replace every array element consisting of K as a digit, with its nearest power of K.
Note: If there happen to be two nearest powers then take the greater one.
Examples:
Input: arr[] = {432, 953, 232, 333}, K = 3
Output: {243, 729, 243, 243}
Explanation:
arr[0] = 35 = 243.
arr[1] = 36 = 729.
arr[2] = 35 = 243.
arr[3] = 35 = 243.
Input: arr[] = {532, 124, 244, 485}, K = 4
Output: {532, 64, 256, 256}
Approach: The idea is to convert each element of the array into a string and then search for K in the string and if found then replace it with the nearest power of K. Follow the steps below to solve the problem:
- Declare a function to find the nearest power of K:
- Find the value of p for which Xp will be closest to the element.
- For calculating the p takes the floor value of logX(Element).
- Therefore, p and p+1 will be the two integers for which the nearest power can be obtained.
- Calculate Xk and X(K + 1) and check which is nearer to the element and return that element.
- Traverse the array arr[]:
- Convert each element into a string.
- Traverse the string and check for the presence of digit K and if found, then replace the array element with the nearest power of K and break from that point.
- Print the modified array.
Below is the implementation of the above approach:
C++
// C++program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate the
// power of base nearest to x
int nearestPow(int x, int base)
{
// Stores logX to the base K
int k = int(log(x) / log(base));
if (abs(pow(base, k) - x) < abs(pow(base, (k + 1)) - x))
return pow(base, k);
else
return pow(base, (k + 1));
}
// Function to replace array
// elements with nearest power of K
void replaceWithNearestPowerOfK(int arr[], int K, int n)
{
// Traverse the array
for (int i = 0; i < n; i++) {
// Convert integer into a string
string strEle = to_string(arr[i]);
for (int c = 0; c < strEle.length(); c++) {
// If K is found, then replace
// with the nearest power of K
if ((strEle[c]-'0') == K) {
arr[i] = nearestPow(arr[i], K);
break;
}
}
}
// Print the array
for (int i = 0; i < n; i++)
cout << arr[i] << " ";
}
// Driver Code
int main()
{
// Given array
int arr[] = { 432, 953, 232, 333 };
int n = sizeof(arr) / sizeof(arr[0]);
// Given value of K
int K = 3;
// Function call to replace array
// elements with nearest power of K
replaceWithNearestPowerOfK(arr, K, n);
}
// This code is contributed by ukasp.
// Java program for the above approach
import java.io.*;
class GFG {
// Function to calculate the
// power of base nearest to x
static int nearestPow(int x, int base1)
{
// Stores logX to the base K
int k = (int)(Math.log(x) / Math.log(base1));
if (Math.abs(Math.pow(base1, k) - x) <
Math.abs(Math.pow(base1, (k + 1)) - x))
return (int)Math.pow(base1, k);
else
return (int)Math.pow(base1, (k + 1));
}
// Function to replace array
// elements with nearest power of K
static void replaceWithNearestPowerOfK(int []arr, int K,
int n)
{
// Traverse the array
for(int i = 0; i < n; i++)
{
// Convert integer into a string
int num = arr[i];
String strEle = String.valueOf(num);
for(int c = 0; c < strEle.length(); c++)
{
// If K is found, then replace
// with the nearest power of K
if ((strEle.charAt(c) - '0') == K)
{
arr[i] = nearestPow(arr[i], K);
break;
}
}
}
// Print the array
for(int i = 0; i < n; i++)
System.out.print(arr[i] + " ");
}
// Driver Code
public static void main (String[] args)
{
// Given array
int []arr = { 432, 953, 232, 333 };
int n = arr.length;
// Given value of K
int K = 3;
// Function call to replace array
// elements with nearest power of K
replaceWithNearestPowerOfK(arr, K, n);
}
}
// This code is contributed by avanitrachhadiya2155
# Python3 program
# for the above approach
import math
# Function to calculate the
# power of base nearest to x
def nearestPow(x, base):
# Stores logX to the base K
k = int(math.log(x, base))
if abs(base**k - x) < abs(base**(k+1) - x):
return base**k
else:
return base**(k+1)
# Function to replace array
# elements with nearest power of K
def replaceWithNearestPowerOfK(arr, K):
# Traverse the array
for i in range(len(arr)):
# Convert integer into a string
strEle = str(arr[i])
for c in strEle:
# If K is found, then replace
# with the nearest power of K
if int(c) == K:
arr[i] = nearestPow(arr[i], K)
break
# Print the array
print(arr)
# Driver Code
# Given array
arr = [432, 953, 232, 333]
# Given value of K
K = 3
# Function call to replace array
# elements with nearest power of K
replaceWithNearestPowerOfK(arr, K)
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to calculate the
// power of base nearest to x
static int nearestPow(int x, int base1)
{
// Stores logX to the base K
int k = (int)(Math.Log(x) / Math.Log(base1));
if (Math.Abs(Math.Pow(base1, k) - x) < Math.Abs(Math.Pow(base1, (k + 1)) - x))
return (int)Math.Pow(base1, k);
else
return (int)Math.Pow(base1, (k + 1));
}
// Function to replace array
// elements with nearest power of K
static void replaceWithNearestPowerOfK(int []arr, int K, int n)
{
// Traverse the array
for (int i = 0; i < n; i++) {
// Convert integer into a string
int num = arr[i];
string strEle = num.ToString();
for (int c = 0; c < strEle.Length; c++) {
// If K is found, then replace
// with the nearest power of K
if ((strEle[c]-'0') == K) {
arr[i] = nearestPow(arr[i], K);
break;
}
}
}
// Print the array
for (int i = 0; i < n; i++)
Console.Write(arr[i]+" ");
}
// Driver Code
public static void Main()
{
// Given array
int []arr = { 432, 953, 232, 333 };
int n = arr.Length;
// Given value of K
int K = 3;
// Function call to replace array
// elements with nearest power of K
replaceWithNearestPowerOfK(arr, K, n);
}
}
// This code is contributed by ipg2016107.
<script>
// Javascript program for the
// above approach
// Function to calculate the
// power of base nearest to x
function nearestPow( x, base)
{
// Stores logX to the base K
let k =
Math.floor(Math.log(x) / Math.log(base));
if (Math.abs(Math.pow(base, k) - x) <
Math.abs(Math.pow(base, (k + 1)) - x))
return Math.pow(base, k);
else
return Math.pow(base, (k + 1));
}
// Function to replace array
// elements with nearest power of K
function replaceWithNearestPowerOfK( arr, K, n)
{
// Traverse the array
for (let i = 0; i < n; i++) {
// Convert integer into a string
let strEle = arr[i].toString();
for (let c = 0; c < strEle.length; c++)
{
// If K is found, then replace
// with the nearest power of K
if ((strEle[c] - '0') == K) {
arr[i] = nearestPow(arr[i], K);
break;
}
}
}
// Print the array
for (let i = 0; i < n; i++)
document.write(arr[i] + " ")
}
// Driver Code
// Given array
let arr = [ 432, 953, 232, 333 ];
let n = arr.length;
// Given value of K
let K = 3;
// Function call to replace array
// elements with nearest power of K
replaceWithNearestPowerOfK(arr, K, n)
// This code is contributed by Hritik
</script>
Output: [243, 729, 243, 243]
Time Complexity: O(N)
Auxiliary Space: O(1)
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