Dada una array arr[] de dígitos y un entero K , la tarea es encontrar num(arr) + K donde num(arr) es el número formado al concatenar todos los dígitos de la array.
Ejemplos:
Entrada: arr[] = {2, 7, 4}, K = 181
Salida: 455
274 + 181 = 455Entrada: arr[] = {6}, K = 815
Salida: 821
6 + 815 = 821
Enfoque: comience a recorrer la array de dígitos desde el final y comience a agregar los dígitos de K uno por uno al dígito actual, si se genera algún acarreo, agréguelo a la siguiente adición de dígitos. Después de que se hayan agregado todos los dígitos de K, comience a recorrer la array de dígitos desde la izquierda y comience a imprimir los elementos que, al concatenar, darán el número resultante.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the vector containing the answer
vector<int> addToArrayForm(vector<int>& A, int K)
{
// Vector v is to store each digits sum
// and vector ans is to store the answer
vector<int> v, ans;
// No carry in the beginning
int rem = 0;
int i = 0;
// Start loop from the end
// and take element one by one
for (i = A.size() - 1; i >= 0; i--) {
// Array index and last digit of number
int my = A[i] + K % 10 + rem;
if (my > 9) {
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.push_back(my % 10);
}
else {
v.push_back(my);
rem = 0;
}
K = K / 10;
}
// K value is greater then 0
while (K > 0) {
// Push digits of K one by one in the array
int my = K % 10 + rem;
v.push_back(my % 10);
// Also maintain carry with summation
if (my / 10 > 0)
rem = 1;
else
rem = 0;
K = K / 10;
}
if (rem > 0)
v.push_back(rem);
// Reverse the elements of vector v
// and store it in vector ans
for (int i = v.size() - 1; i >= 0; i--)
ans.push_back(v[i]);
return ans;
}
// Driver code
int main()
{
vector<int> A{ 2, 7, 4 };
int K = 181;
vector<int> ans = addToArrayForm(A, K);
// Print the answer
for (int i = 0; i < ans.size(); i++)
cout << ans[i];
return 0;
}
Java
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the vector containing the answer
static ArrayList<Integer> addToArrayForm(ArrayList<Integer> A, int K)
{
// ArrayList v is to store each digits sum
// and ArrayList ans is to store the answer
ArrayList<Integer> v = new ArrayList<Integer>();
ArrayList<Integer> ans = new ArrayList<Integer>();
// No carry in the beginning
int rem = 0;
int i = 0;
// Start loop from the end
// and take element one by one
for (i = A.size() - 1; i >= 0; i--)
{
// Array index and last digit of number
int my = A.get(i) + K % 10 + rem;
if (my > 9)
{
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.add(my % 10);
}
else
{
v.add(my);
rem = 0;
}
K = K / 10;
}
// K value is greater then 0
while (K > 0)
{
// Push digits of K one by one in the array
int my = K % 10 + rem;
v.add(my % 10);
// Also maintain carry with summation
if (my / 10 > 0)
rem = 1;
else
rem = 0;
K = K / 10;
}
if (rem > 0)
v.add(rem);
// Reverse the elements of vector v
// and store it in vector ans
for (int j = v.size() - 1; j >= 0; j--)
ans.add(v.get(j));
return ans;
}
// Driver code
public static void main (String[] args)
{
ArrayList<Integer> A = new ArrayList<Integer>();
A.add(2);
A.add(7);
A.add(4);
int K = 181;
ArrayList<Integer> ans = addToArrayForm(A, K);
// Print the answer
for (int i = 0; i < ans.size(); i++)
System.out.print(ans.get(i));
}
}
// This code is contributed by ihritik
Python3
# Python3 implementation of the approach # Function to return the vector # containing the answer def addToArrayForm(A, K): # Vector v is to store each digits sum # and vector ans is to store the answer v, ans = [], [] # No carry in the beginning rem, i = 0, 0 # Start loop from the end # and take element one by one for i in range(len(A) - 1, -1, -1): # Array index and last digit of number my = A[i] + (K % 10) + rem if my > 9: # Maintain carry of summation rem = 1 # Push the digit value into the array v.append(my % 10) else: v.append(my) rem = 0 K = K // 10 # K value is greater then 0 while K > 0: # Push digits of K one by one in the array my = (K % 10) + rem v.append(my % 10) # Also maintain carry with summation if my // 10 > 0: rem = 1 else: rem = 0 K = K // 10 if rem > 0: v.append(rem) # Reverse the elements of vector v # and store it in vector ans for i in range(len(v) - 1, -1, -1): ans.append(v[i]) return ans # Driver code if __name__ == "__main__": A = [2, 7, 4] K = 181 ans = addToArrayForm(A, K) # Print the answer for i in range(0, len(ans)): print(ans[i], end = "") # This code is contributed by Rituraj Jain
C#
// C# implementation of the approach
using System;
using System.Collections;
class GFG
{
// Function to return the vector containing the answer
static ArrayList addToArrayForm(ArrayList A, int K)
{
// ArrayList v is to store each digits sum
// and ArrayList ans is to store the answer
ArrayList v = new ArrayList();
ArrayList ans = new ArrayList();
// No carry in the beginning
int rem = 0;
int i = 0;
// Start loop from the end
// and take element one by one
for (i = A.Count - 1; i >= 0; i--)
{
// Array index and last digit of number
int my = (int)A[i] + K % 10 + rem;
if (my > 9)
{
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.Add(my % 10);
}
else
{
v.Add(my);
rem = 0;
}
K = K / 10;
}
// K value is greater then 0
while (K > 0)
{
// Push digits of K one by one in the array
int my = K % 10 + rem;
v.Add(my % 10);
// Also maintain carry with summation
if (my / 10 > 0)
rem = 1;
else
rem = 0;
K = K / 10;
}
if (rem > 0)
v.Add(rem);
// Reverse the elements of vector v
// and store it in vector ans
for (int j = v.Count - 1; j >= 0; j--)
ans.Add((int)v[j]);
return ans;
}
// Driver code
static void Main()
{
ArrayList A = new ArrayList();
A.Add(2);
A.Add(7);
A.Add(4);
int K = 181;
ArrayList ans = addToArrayForm(A, K);
// Print the answer
for (int i = 0; i < ans.Count; i++)
Console.Write((int)ans[i]);
}
}
// This code is contributed by mits
PHP
<?php
// Php implementation of the approach
// Function to return the vector containing the answer
function addToArrayForm($A, $K)
{
// Vector v is to store each digits sum
// and vector ans is to store the answer
$v = array();
$ans = array();
// No carry in the beginning
$rem = 0;
$i = 0;
// Start loop from the end
// and take element one by one
for ($i = count($A) - 1; $i >= 0; $i--)
{
// Array index and last digit of number
$my = $A[$i] + $K % 10 + $rem;
if ($my > 9)
{
// Maintain carry of summation
$rem = 1;
// Push the digit value into the array
array_push($v,$my % 10);
}
else
{
array_push($v,$my);
$rem = 0;
}
$K = floor($K / 10);
}
// K value is greater then 0
while ($K > 0)
{
// Push digits of K one by one in the array
$my = $K % 10 + $rem;
array_push($v,$my % 10);
// Also maintain carry with summation
if ($my / 10 > 0)
$rem = 1;
else
$rem = 0;
$K = floor($K / 10);
}
if ($rem > 0)
array_push($v,$rem);
// Reverse the elements of vector v
// and store it in vector ans
for ($i = count($v) - 1; $i >= 0; $i--)
array_push($ans,$v[$i]);
return $ans;
}
// Driver code
$A = array( 2, 7, 4 );
$K = 181;
$ans = addToArrayForm($A, $K);
// Print the answer
for ($i = 0; $i < count($ans); $i++)
echo $ans[$i];
// This code is contributed by Ryuga
?>
Javascript
<script>
// Javascript implementation of the approach
// Function to return the vector
// containing the answer
function addToArrayForm(A, K)
{
// ArrayList v is to store each digits sum
// and ArrayList ans is to store the answer
let v = [];
let ans = [];
// No carry in the beginning
let rem = 0;
let i = 0;
// Start loop from the end
// and take element one by one
for(i = A.length - 1; i >= 0; i--)
{
// Array index and last digit of number
let my = A[i] + K % 10 + rem;
if (my > 9)
{
// Maintain carry of summation
rem = 1;
// Push the digit value into the array
v.push(my % 10);
}
else
{
v.push(my);
rem = 0;
}
K = parseInt(K / 10, 10);
}
// K value is greater then 0
while (K > 0)
{
// Push digits of K one by one in the array
let my = K % 10 + rem;
v.push(my % 10);
// Also maintain carry with summation
if (parseInt(my / 10, 10) > 0)
rem = 1;
else
rem = 0;
K = parseInt(K / 10, 10);
}
if (rem > 0)
v.push(rem);
// Reverse the elements of vector v
// and store it in vector ans
for(let j = v.length - 1; j >= 0; j--)
ans.push(v[j]);
return ans;
}
// Driver code
let A = [];
A.push(2);
A.push(7);
A.push(4);
let K = 181;
let ans = addToArrayForm(A, K);
// Print the answer
for(let i = 0; i < ans.length; i++)
document.write(ans[i]);
// This code is contributed by divyeshrabadiya07
</script>
Producción:
455
Complejidad temporal: O(n + log K)
Espacio Auxiliar: O(n + log K)
Publicación traducida automáticamente
Artículo escrito por Sumit bangar y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA