Dado un número num y un número n, la tarea es borrar este n-ésimo dígito del número num, desde el principio y desde el final.
Ejemplos:
Entrada: num = 1234, n = 3
Salida: num_after_deleting_from_starting = 124, num_after_deleting_from_end = 134
Entrada: num = 4516312, n = 2
Salida: num_after_deleting_from_starting = 416312, num_after_deleting_from_end = 451632
Acercarse:
- Para eliminar el enésimo dígito desde el inicio:
- Obtenga el número y el dígito n que se eliminará.
- Contar el número de dígitos
- Repita el número de dígitos del tiempo contándolo con una variable i.
- Si i es igual a (número de dígitos – n), salte, de lo contrario, agregue el i-ésimo dígito como [nuevo_número = (nuevo_número * 10) + i-ésimo dígito].
- Para eliminar el enésimo dígito del final:
- Obtenga el número y el dígito n que se eliminará.
- Repita el número de dígitos del tiempo contándolo con una variable i.
- Si i es igual a (n), salte, de lo contrario, agregue el dígito i como [nuevo_número = (nuevo_número * 10) + i-ésimo dígito].
Implementación:
C++
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;
// Function to delete nth digit
// from starting
int deleteFromStart(int num, int n)
{
// Get the number of digits
int d = log10(num) + 1;
// Declare a variable
// to form the reverse resultant number
int rev_new_num = 0;
// Loop with the number
for (int i = 0; num != 0; i++) {
int digit = num % 10;
num = num / 10;
if (i == (d - n)) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
int new_num = 0;
// Loop with the number
for (int i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = rev_new_num / 10;
}
// Return the resultant number
return new_num;
}
// Function to delete nth digit
// from ending
int deleteFromEnd(int num, int n)
{
// Declare a variable
// to form the reverse resultant number
int rev_new_num = 0;
// Loop with the number
for (int i = 1; num != 0; i++) {
int digit = num % 10;
num = num / 10;
if (i == n) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
int new_num = 0;
// Loop with the number
for (int i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = rev_new_num / 10;
}
// Return the resultant number
return new_num;
}
// Driver code
int main()
{
// Get the number
int num = 1234;
cout << "Number: " << num << endl;
// Get the digit number to be deleted
int n = 3;
cout << "Digit to be deleted: " << n << endl;
// Remove the nth digit from starting
cout << "Number after " << n
<< " digit deleted from starting: "
<< deleteFromStart(num, n) << endl;
// Remove the nth digit from ending
cout << "Number after " << n
<< " digit deleted from ending: "
<< deleteFromEnd(num, n) << endl;
return 0;
}
Java
// Java implementation of above approach
class GFG
{
// Function to delete nth digit
// from starting
static int deleteFromStart(int num, int n)
{
// Get the number of digits
int d = (int)Math.log10(num) + 1;
// Declare a variable
// to form the reverse resultant number
int rev_new_num = 0;
// Loop with the number
for (int i = 0; num != 0; i++) {
int digit = num % 10;
num = num / 10;
if (i == (d - n)) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
int new_num = 0;
// Loop with the number
for (int i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = rev_new_num / 10;
}
// Return the resultant number
return new_num;
}
// Function to delete nth digit
// from ending
static int deleteFromEnd(int num, int n)
{
// Declare a variable
// to form the reverse resultant number
int rev_new_num = 0;
// Loop with the number
for (int i = 1; num != 0; i++) {
int digit = num % 10;
num = num / 10;
if (i == n) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
int new_num = 0;
// Loop with the number
for (int i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = rev_new_num / 10;
}
// Return the resultant number
return new_num;
}
// Driver code
public static void main(String []args)
{
// Get the number
int num = 1234;
System.out.println("Number: " + num );
// Get the digit number to be deleted
int n = 3;
System.out.println("Digit to be deleted: " + n );
// Remove the nth digit from starting
System.out.println("Number after " + n
+ " digit deleted from starting: "
+ deleteFromStart(num, n));
// Remove the nth digit from ending
System.out.println( "Number after " + n
+ " digit deleted from ending: "
+ deleteFromEnd(num, n));
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of above approach
# Function to delete nth digit
# from starting
import math;
def deleteFromStart(num, n):
# Get the number of digits
d = (math.log10(num) + 1);
# Declare a variable to form
# the reverse resultant number
rev_new_num = 0;
# Loop with the number
i = 0;
while (num != 0):
digit = num % 10;
num = int(num / 10);
if (i != (int(d) - n)):
rev_new_num = ((rev_new_num * 10) +
digit);
i += 1;
# Declare a variable to form the
# resultant number
new_num = 0;
# Loop with the number
i = 0;
while (rev_new_num != 0):
new_num = ((new_num * 10) +
(rev_new_num % 10));
rev_new_num = int(rev_new_num / 10);
i += 1;
# Return the resultant number
return new_num;
# Function to delete nth digit
# from ending
def deleteFromEnd(num, n):
# Declare a variable to form
# the reverse resultant number
rev_new_num = 0;
# Loop with the number
i = 1;
while (num != 0):
digit = num % 10;
num = int(num / 10);
if (i != n):
rev_new_num = ((rev_new_num * 10) +
digit);
i += 1;
# Declare a variable
# to form the resultant number
new_num = 0;
# Loop with the number
i = 0;
while (rev_new_num != 0):
new_num = ((new_num * 10) +
(rev_new_num % 10));
rev_new_num = int(rev_new_num / 10);
i += 1;
# Return the resultant number
return new_num;
# Driver code
# Get the number
num = 1234;
print("Number:", num);
# Get the digit number to be deleted
n = 3;
print("Digit to be deleted:", n);
# Remove the nth digit from starting
print("Number after", n,
"digit deleted from starting:",
deleteFromStart(num, n));
# Remove the nth digit from ending
print("Number after", n,
"digit deleted from ending:",
deleteFromEnd(num, n));
# This code is contributed by chandan_jnu
C#
// C# implementation of the above approach
using System;
class GFG
{
// Function to delete nth digit
// from starting
static int deleteFromStart(int num, int n)
{
// Get the number of digits
int d = (int)Math.Log10(num) + 1;
// Declare a variable
// to form the reverse resultant number
int rev_new_num = 0;
// Loop with the number
for (int i = 0; num != 0; i++) {
int digit = num % 10;
num = num / 10;
if (i == (d - n)) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
int new_num = 0;
// Loop with the number
for (int i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = rev_new_num / 10;
}
// Return the resultant number
return new_num;
}
// Function to delete nth digit
// from ending
static int deleteFromEnd(int num, int n)
{
// Declare a variable
// to form the reverse resultant number
int rev_new_num = 0;
// Loop with the number
for (int i = 1; num != 0; i++) {
int digit = num % 10;
num = num / 10;
if (i == n) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
int new_num = 0;
// Loop with the number
for (int i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = rev_new_num / 10;
}
// Return the resultant number
return new_num;
}
// Driver code
public static void Main()
{
// Get the number
int num = 1234;
Console.WriteLine("Number: " + num );
// Get the digit number to be deleted
int n = 3;
Console.WriteLine("Digit to be deleted: " + n );
// Remove the nth digit from starting
Console.WriteLine("Number after " + n
+ " digit deleted from starting: "
+ deleteFromStart(num, n));
// Remove the nth digit from ending
Console.WriteLine( "Number after " + n
+ " digit deleted from ending: "
+ deleteFromEnd(num, n));
}
}
// This code is contributed by ihritik
PHP
<?php
//PHP implementation of above approach
// Function to delete nth digit
// from starting
function deleteFromStart($num, $n)
{
// Get the number of digits
$d = (log10($num) + 1);
// Declare a variable
// to form the reverse resultant number
$rev_new_num = 0;
// Loop with the number
for ($i = 0; $num != 0; $i++) {
$digit = $num % 10;
$num = (int)$num / 10;
if ($i == ($d - $n)) {
continue;
}
else {
$rev_new_num = ($rev_new_num * 10) + $digit;
}
}
// Declare a variable
// to form the resultant number
$new_num = 0;
// Loop with the number
for ($i = 0; $rev_new_num != 0; $i++) {
$new_num = ($new_num * 10)
+ ($rev_new_num % 10);
$rev_new_num = (int)$rev_new_num / 10;
}
// Return the resultant number
return $new_num;
}
// Function to delete nth digit
// from ending
function deleteFromEnd($num, $n)
{
// Declare a variable
// to form the reverse resultant number
$rev_new_num = 0;
// Loop with the number
for ($i = 1; $num != 0; $i++) {
$digit = $num % 10;
$num = (int)$num / 10;
if ($i == $n) {
continue;
}
else {
$rev_new_num = ($rev_new_num * 10) + $digit;
}
}
// Declare a variable
// to form the resultant number
$new_num = 0;
// Loop with the number
for ($i = 0; $rev_new_num != 0; $i++) {
$new_num = ($new_num * 10)
+ ($rev_new_num % 10);
$rev_new_num = (int)$rev_new_num / 10;
}
// Return the resultant number
return $new_num;
}
// Driver code
// Get the number
$num = 1234;
echo "Number: " , $num ,"\n";
// Get the digit number to be deleted
$n = 3;
echo "Digit to be deleted: " ,$n ,"\n";
// Remove the nth digit from starting
echo "Number after " , $n,
" digit deleted from starting: ",
deleteFromStart($num, $n),"\n";
// Remove the nth digit from ending
echo "Number after " , $n,
" digit deleted from ending: ",
deleteFromEnd($num, $n) ,"\n";
?>
// This code is contributed by jit_t.
Javascript
<script>
// Javascript implementation of
// the above approach
// Function to delete nth digit
// from starting
function deleteFromStart(num, n)
{
// Get the number of digits
let d = parseInt(Math.log10(num), 10) + 1;
// Declare a variable
// to form the reverse resultant number
let rev_new_num = 0;
// Loop with the number
for (let i = 0; num != 0; i++) {
let digit = num % 10;
num = parseInt(num / 10, 10);
if (i == (d - n)) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
let new_num = 0;
// Loop with the number
for (let i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = parseInt(rev_new_num / 10, 10);
}
// Return the resultant number
return new_num;
}
// Function to delete nth digit
// from ending
function deleteFromEnd(num, n)
{
// Declare a variable
// to form the reverse resultant number
let rev_new_num = 0;
// Loop with the number
for (let i = 1; num != 0; i++) {
let digit = num % 10;
num = parseInt(num / 10, 10);
if (i == n) {
continue;
}
else {
rev_new_num = (rev_new_num * 10) + digit;
}
}
// Declare a variable
// to form the resultant number
let new_num = 0;
// Loop with the number
for (let i = 0; rev_new_num != 0; i++) {
new_num = (new_num * 10)
+ (rev_new_num % 10);
rev_new_num = parseInt(rev_new_num / 10, 10);
}
// Return the resultant number
return new_num;
}
// Get the number
let num = 1234;
document.write("Number: " + num + "</br>");
// Get the digit number to be deleted
let n = 3;
document.write("Digit to be deleted: " + n + "</br>");
// Remove the nth digit from starting
document.write("Number after " + n
+ " digit deleted from starting: "
+ deleteFromStart(num, n) + "</br>");
// Remove the nth digit from ending
document.write( "Number after " + n
+ " digit deleted from ending: "
+ deleteFromEnd(num, n));
</script>
Producción:
Number: 1234 Digit to be deleted: 3 Number after 3 digit deleted from starting: 124 Number after 3 digit deleted from ending: 134
Análisis de Complejidad:
Complejidad temporal: O(log(N)).
Espacio Auxiliar: O(1), ya que no se ha ocupado ningún espacio extra.
Otro enfoque: (para convertir el número en string)
C++
// C++ implementation to delete nth digit
// from starting with O(logN) time complexity.
#include<bits/stdc++.h>
using namespace std;
// function to delete nth number from starting
static string fromStart(string inp, int del)
{
string inp1 = inp.substr(0, del - 1);
string inp2 = inp.substr(del, inp.length());
return inp1 + inp2;
}
// function to delete nth number from ending
static string fromEnd(string inp, int del)
{
string inp1 = inp.substr(0, inp.length() - del);
string inp2 = inp.substr(inp.length() - del + 1,
inp.length());
return inp1 + inp2;
}
// Driver Code
int main()
{
int in = 1234;
// type cast input number to string
stringstream ss;
ss << in;
string inp = ss.str();
int del = 3;
cout << "num_after_deleting_from_starting "
<< fromStart(inp, del) << endl;
cout << "num_after_deleting_from_ending "
<< fromEnd(inp, del) << endl;
return 0;
}
// This code is contributed by chandan_jnu
Java
// Java implementation to delete nth digit
// from starting with O(logN) time complexity.
public class DeleteN {
public static void main(String args[]) {
int in = 1234;
// type cast input number to string
String inp = Integer.toString(in);
int del = 3;
System.out.println("num_after_deleting_from_starting " + fromStart(inp, del));
System.out.println("num_after_deleting_from_ending " + fromEnd(inp, del));
}
// function to delete nth number from starting
static String fromStart(String inp, int del) {
try {
String inp1 = inp.substring(0, del - 1);
String inp2 = inp.substring(del, inp.length());
return inp1 + inp2;
}
catch (Exception e) {
return "Check Input";
}
}
// function to delete nth number from ending
static String fromEnd(String inp, int del) {
try {
String inp1 = inp.substring(0, inp.length() - del);
String inp2 = inp.substring(inp.length() - del + 1, inp.length());
return inp1 + inp2;
}
catch (Exception e) {
return "Check Input";
}
}
}
Python3
# Python3 implementation to delete nth digit
# from starting with O(logN) time complexity.
# function to del1ete nth number
# from starting
def fromStart(inp, del11):
inp1 = inp[0:del1 - 1];
inp2 = inp[del1:len(inp)];
return inp1 + inp2;
# function to delete nth number
# from ending
def fromEnd(inp, del1):
inp1 = inp[0:len(inp) - del1];
inp2 = inp[len(inp) - del1 + 1:len(inp)];
return inp1 + inp2;
# Driver Code
in1 = 1234;
# type cast input number to string
inp = str(in1);
del1 = 3;
print("num_after_deleting_from_starting",
fromStart(inp, del1));
print("num_after_deleting_from_ending",
fromEnd(inp, del1));
# This code is contributed by chandan_jnu
C#
// C# implementation to delete nth digit
// from starting with O(logN) time complexity.
using System ;
public class DeleteN {
public static void Main() {
int num = 1234;
// type cast input number to string
string inp = Convert.ToString(num) ;
int del = 3;
Console.WriteLine("num_after_deleting_from_starting "
+ fromStart(inp, del));
Console.WriteLine("num_after_deleting_from_ending "
+ fromEnd(inp, del));
}
// function to delete nth number from starting
static String fromStart(string inp, int del) {
try {
string inp1 = inp.Substring(0, del - 1);
string inp2 = inp.Substring(del, inp.Length - del);
return inp1 + inp2;
}
catch (Exception ) {
return "Check Input";
}
}
// function to delete nth number from ending
static String fromEnd(string inp, int del) {
try {
string inp1 = inp.Substring(0, inp.Length - del);
string inp2 = inp.Substring(inp.Length - del + 1, del - 1);
return inp1 + inp2;
}
catch (Exception e) {
Console.WriteLine(e) ;
return "Check Input";
}
}
}
// This code is contributed by Ryuga
PHP
<?php
// PHP implementation to delete nth digit
// from starting with O(logN) time complexity.
// function to delete nth number from starting
function fromStart($inp, $del)
{
$inp1 = substr($inp, 0, $del - 1);
$inp2 = substr($inp, $del, strlen($inp));
return $inp1.$inp2;
}
// function to delete nth number from ending
function fromEnd($inp, $del)
{
$inp1 = substr($inp, 0, strlen($inp) - $del);
$inp2 = substr($inp, strlen($inp) - $del + 1,
strlen($inp));
return $inp1.$inp2;
}
// Driver Code
$in = 1234;
// type cast input number to string
$inp = strval($in);
$del = 3;
print("num_after_deleting_from_starting " .
fromStart($inp, $del) . "\n");
print("num_after_deleting_from_ending " .
fromEnd($inp, $del));
// This code is contributed by chandan_jnu
?>
Javascript
<script>
// Javascript implementation to delete nth digit
// from starting with O(logN) time complexity.
// function to delete nth number from starting
function fromStart(inp, del) {
let inp1 = inp.substring(0, del - 1);
let inp2 = inp.substring(del, inp.length);
return inp1 + inp2;
}
function fromEnd(inp, del) {
let inp1 = inp.substring(0, inp.length - del);
let inp2 = inp.substring(inp.length - del + 1, inp.length - del + 1 + inp.length);
return inp1 + inp2;
}
let In = 1234;
// type cast input number to string
let inp = In.toString();
let del = 3;
document.write("num_after_deleting_from_starting " + fromStart(inp, del) + "</br>");
document.write("num_after_deleting_from_ending " + fromEnd(inp, del) + "</br>");
// This code is contributed by mukesh07.
</script>
Producción:
num_after_deleting_from_starting 124 num_after_deleting_from_ending 134
Complejidad del tiempo: O(log(N)).
Espacio Auxiliar: O(log(N)).
Publicación traducida automáticamente
Artículo escrito por Mostafijur Rahaman y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA