Dados dos enteros N y X . La tarea es imprimir la diferencia absoluta entre los primeros X y los últimos X dígitos en N . Teniendo en cuenta que el número de dígitos es al menos 2*x.
Ejemplos:
Input: N = 21546, X = 2 Output: 25 The first two digit in 21546 is 21. The last two digit in 21546 is 46. The absolute difference of 21 and 46 is 25. Input: N = 351684617, X = 3 Output: 266
Enfoque sencillo:
- Almacene los últimos x dígitos del número en last.
- Encuentra el número de dígitos en el número.
- Elimina todos los dígitos excepto la primera x.
- Almacene los primeros x enteros del número en first.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the
// number of digits in the integer
long long digitsCount(long long n)
{
int len = 0;
while (n > 0) {
len++;
n /= 10;
}
return len;
}
// Function to find the absolute difference
long long absoluteFirstLast(long long n, int x)
{
// Store the last x digits in last
int i = 0, mod = 1;
while (i < x) {
mod *= 10;
i++;
}
int last = n % mod;
// Count the no. of digits in N
long long len = digitsCount(n);
// Remove the digits except the first x
while (len != x) {
n /= 10;
len--;
}
// Store the first x digits in first
int first = n;
// Return the absolute difference between
// the first and last
return abs(first - last);
}
// Driver code
int main()
{
long long n = 21546, x = 2;
cout << absoluteFirstLast(n, x);
return 0;
}
Java
// Java implementation of the above approach
import java.util.*;
class GFG
{
// Function to find the
// number of digits in the integer
static int digitsCount(int n)
{
int len = 0;
while (n > 0)
{
len++;
n /= 10;
}
return len;
}
// Function to find the absolute difference
static int absoluteFirstLast(int n, int x)
{
// Store the last x digits in last
int i = 0, mod = 1;
while (i < x)
{
mod *= 10;
i++;
}
int last = n % mod;
// Count the no. of digits in N
int len = digitsCount(n);
// Remove the digits except the first x
while (len != x)
{
n /= 10;
len--;
}
// Store the first x digits in first
int first = n;
// Return the absolute difference between
// the first and last
return Math.abs(first - last);
}
// Driver code
public static void main(String args[])
{
int n = 21546, x = 2;
System.out.println(absoluteFirstLast(n, x));
}
}
// This code is contributed by
// Surendra_Gangwar
Python3
# Python3 implementation of the above approach # Function to find the # number of digits in the integer def digitsCount(n) : length = 0; while (n > 0) : length += 1; n //= 10; return length; # Function to find the absolute difference def absoluteFirstLast(n, x) : # Store the last x digits in last i = 0 ; mod = 1; while (i < x) : mod *= 10; i += 1; last = n % mod; # Count the no. of digits in N length = digitsCount(n); # Remove the digits except the first x while (length != x) : n //= 10; length -= 1; # Store the first x digits in first first = n; # Return the absolute difference between # the first and last return abs(first - last); # Driver code if __name__ == "__main__" : n = 21546 ; x = 2; print(absoluteFirstLast(n, x)); # This code is contributed by Ryuga
C#
// C# implementation of the approach
using System;
class GFG
{
// Function to find the
// number of digits in the integer
static int digitsCount(int n)
{
int len = 0;
while (n > 0)
{
len++;
n /= 10;
}
return len;
}
// Function to find the absolute difference
static int absoluteFirstLast(int n, int x)
{
// Store the last x digits in last
int i = 0, mod = 1;
while (i < x)
{
mod *= 10;
i++;
}
int last = n % mod;
// Count the no. of digits in N
int len = digitsCount(n);
// Remove the digits except the first x
while (len != x)
{
n /= 10;
len--;
}
// Store the first x digits in first
int first = n;
// Return the absolute difference between
// the first and last
return Math.Abs(first - last);
}
// Driver code
public static void Main(String []args)
{
int n = 21546, x = 2;
Console.Write(absoluteFirstLast(n, x));
}
}
// This code has been contributed by 29AjayKumar
PHP
<?php
// PHP implementation of the above approach
// Function to find the number of
// digits in the integer
function digitsCount($n)
{
$len = 0;
while ($n > 0)
{
$len++;
$n = (int)($n / 10);
}
return $len;
}
// Function to find the absolute difference
function absoluteFirstLast($n, $x)
{
// Store the last x digits in last
$i = 0;
$mod = 1;
while ($i < $x)
{
$mod *= 10;
$i++;
}
$last = $n % $mod;
// Count the no. of digits in N
$len = digitsCount($n);
// Remove the digits except the first x
while ($len != $x)
{
$n = (int)($n / 10);
$len--;
}
// Store the first x digits in first
$first = $n;
// Return the absolute difference
// between the first and last
return abs($first - $last);
}
// Driver code
$n = 21546;
$x = 2;
echo absoluteFirstLast($n, $x);
// This code is contributed by mits
?>
Javascript
<script>
// Javascript implementation of the above approach
// Function to find the
// number of digits in the integer
function digitsCount(n)
{
let len = 0;
while (n > 0)
{
len++;
n = Math.floor(n / 10);
}
return len;
}
// Function to find the absolute difference
function absoluteFirstLast(n, x)
{
// Store the last x digits in last
let i = 0, mod = 1;
while (i < x)
{
mod *= 10;
i++;
}
let last = n % mod;
// Count the no. of digits in N
let len = digitsCount(n);
// Remove the digits except the first x
while (len != x)
{
n = Math.floor(n / 10);
len--;
}
// Store the first x digits in first
let first = n;
// Return the absolute difference between
// the first and last
return Math.abs(first - last);
}
// driver program
let n = 21546, x = 2;
document.write(absoluteFirstLast(n, x));
</script>
Producción:
25
Complejidad de tiempo: O(X+k) donde X es el número entero dado y k es el número de dígitos en n.
Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.