Dada una array de dígitos (los valores son del 0 al 9), encuentre la suma mínima posible de dos números formados a partir de los dígitos de la array. Todos los dígitos de la array dada deben usarse para formar los dos números.
Ejemplos:
Input: [6, 8, 4, 5, 2, 3] Output: 604 The minimum sum is formed by numbers 358 and 246 Input: [5, 3, 0, 7, 4] Output: 82 The minimum sum is formed by numbers 35 and 047
Se formará un número mínimo a partir de un conjunto de dígitos cuando el dígito más pequeño aparezca en la posición más significativa y el siguiente dígito más pequeño aparezca en la siguiente posición más significativa y así sucesivamente. La idea es
ordenar la array en orden creciente y construir dos números alternando la selección dígitos de la array. Entonces, el primer número está formado por dígitos presentes en posiciones impares en la array y el segundo número está formado por dígitos de posiciones pares en la array. Finalmente, devolvemos la suma del primer y segundo número.
A continuación se muestra la implementación de la idea anterior.
C++
// C++ program to find minimum sum of two numbers
// formed from digits of the array.
#include <bits/stdc++.h>
using namespace std;
// Function to find and return minimum sum of
// two numbers formed from digits of the array.
int solve(int arr[], int n)
{
// sort the array
sort(arr, arr + n);
// let two numbers be a and b
int a = 0, b = 0;
for (int i = 0; i < n; i++)
{
// fill a and b with every alternate digit
// of input array
if (i & 1)
a = a*10 + arr[i];
else
b = b*10 + arr[i];
}
// return the sum
return a + b;
}
// Driver code
int main()
{
int arr[] = {6, 8, 4, 5, 2, 3};
int n = sizeof(arr)/sizeof(arr[0]);
cout << "Sum is " << solve(arr, n);
return 0;
}
Java
// Java program to find minimum sum of two numbers
// formed from digits of the array.
import java.util.Arrays;
class GFG {
// Function to find and return minimum sum of
// two numbers formed from digits of the array.
static int solve(int arr[], int n)
{
// sort the array
Arrays.sort(arr);
// let two numbers be a and b
int a = 0, b = 0;
for (int i = 0; i < n; i++)
{
// fill a and b with every alternate
// digit of input array
if (i % 2 != 0)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
}
//driver code
public static void main (String[] args)
{
int arr[] = {6, 8, 4, 5, 2, 3};
int n = arr.length;
System.out.print("Sum is "
+ solve(arr, n));
}
}
//This code is contributed by Anant Agarwal.
Python3
# Python3 program to find minimum sum of two
# numbers formed from digits of the array.
# Function to find and return minimum sum of
# two numbers formed from digits of the array.
def solve(arr, n):
# sort the array
arr.sort()
# let two numbers be a and b
a = 0; b = 0
for i in range(n):
# Fill a and b with every alternate
# digit of input array
if (i % 2 != 0):
a = a * 10 + arr[i]
else:
b = b * 10 + arr[i]
# return the sum
return a + b
# Driver code
arr = [6, 8, 4, 5, 2, 3]
n = len(arr)
print("Sum is ", solve(arr, n))
# This code is contributed by Anant Agarwal.
C#
// C# program to find minimum
// sum of two numbers formed
// from digits of the array.
using System;
class GFG
{
// Function to find and return
// minimum sum of two numbers
// formed from digits of the array.
static int solve(int []arr, int n)
{
// sort the array
Array.Sort(arr);
// let two numbers be a and b
int a = 0, b = 0;
for (int i = 0; i < n; i++)
{
// fill a and b with every alternate digit
// of input array
if (i % 2 != 0)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
}
// Driver code
public static void Main ()
{
int []arr = {6, 8, 4, 5, 2, 3};
int n = arr.Length;
Console.WriteLine("Sum is " + solve(arr, n));
}
}
// This code is contributed by Anant Agarwal.
PHP
<?php
// PHP program to find minimum
// sum of two numbers formed
// from digits of the array.
// Function to find and return
// minimum sum of two numbers
// formed from digits of the array.
function solve($arr, $n)
{
// sort the array
sort($arr); sort($arr , $n);
// let two numbers be a and b
$a = 0; $b = 0;
for ($i = 0; $i < $n; $i++)
{
// fill a and b with every
// alternate digit of input array
if ($i & 1)
$a = $a * 10 + $arr[$i];
else
$b = $b * 10 + $arr[$i];
}
// return the sum
return $a + $b;
}
// Driver code
$arr = array(6, 8, 4, 5, 2, 3);
$n = sizeof($arr);
echo "Sum is " , solve($arr, $n);
// This code is contributed by nitin mittal.
?>
Javascript
<script>
// Javascript program to find minimum sum of two numbers
// formed from digits of the array.
// Function to find and return minimum sum of
// two numbers formed from digits of the array.
function solve(arr, n)
{
// sort the array
arr.sort();
// let two numbers be a and b
let a = 0, b = 0;
for (let i = 0; i < n; i++)
{
// fill a and b with every alternate
// digit of input array
if (i % 2 != 0)
a = a * 10 + arr[i];
else
b = b * 10 + arr[i];
}
// return the sum
return a + b;
}
// Driver Code
let arr = [6, 8, 4, 5, 2, 3];
let n = arr.length;
document.write("Sum is "
+ solve(arr, n));
</script>
Producción :
Sum is 604
Método 2 (para números grandes)
Cuando tenemos que lidiar con números muy grandes (como en la sección PRÁCTICA de esta pregunta), el enfoque anterior no funcionará. La idea básica de abordar la pregunta es la misma que la anterior, pero en lugar de usar números, usaremos strings para manejar la suma.
Para sumar dos números dados en forma de string, puede consultar this .
C++
#include <bits/stdc++.h>
using namespace std;
string solve(int arr[], int n)
{
// code here
// sorting of array O(nlogn)
sort(arr, arr + n);
// Two String for storing our two minimum numbers
string a = "", b = "";
// string string alternatively
for (int i = 0; i < n; i += 2)
{
a += (arr[i] + '0');
}
for (int i = 1; i < n; i += 2)
{
b += (arr[i] + '0');
}
int j = a.length() - 1;
int k = b.length() - 1;
// as initial carry is zero
int carry = 0;
string ans = "";
while (j >= 0 && k >= 0)
{
int sum = 0;
sum += (a[j] - '0') + (b[k] - '0') + carry;
ans += to_string(sum % 10);
carry = sum / 10;
j--;
k--;
}
// if string b is over and string a is left
// here we dont need to put here while condition
// as it would run at max one time. Because the difference
// between both the strings could be at max 1.
while (j >= 0)
{
int sum = 0;
sum += (a[j] - '0') + carry;
ans += to_string(sum % 10);
carry = sum / 10;
j--;
}
// if string a is over and string b is left
while (k >= 0)
{
int sum = 0;
sum += (b[k] - '0') + carry;
ans += to_string(sum % 10);
carry = sum / 10;
k--;
}
// if carry is left
if (carry)
{
ans += to_string(carry);
}
// to remove leading zeroes as they will be ahead of our sum
while (!ans.empty() and ans.back() == '0')
ans.pop_back();
// reverse our final string because we were storing sum from left to right
reverse(ans.begin(), ans.end());
return ans;
}
// Driver Code Starts.
int main()
{
int arr[] = {6, 8, 4, 5, 2, 3};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Sum is " << solve(arr, n);
return 0;
} // Driver Code Ends
Python3
# Python code for the approach
def solve(arr, n):
# sorting of array O(nlogn)
arr.sort()
# Two String for storing our two minimum numbers
a,b = "",""
# string string alternatively
for i in range(0,n,2):
a += str(arr[i])
for i in range(1,n,2):
b += str(arr[i])
j = len(a) - 1
k = len(b) - 1
# as initial carry is zero
carry = 0
ans = ""
while (j >= 0 and k >= 0):
sum = 0
sum += (ord(a[j]) - ord('0') + ord(b[k]) - ord('0')) + carry
ans += str(sum % 10)
carry = sum // 10
j -= 1
k -= 1
# if string b is over and string a is left
# here we dont need to put here while condition
# as it would run at max one time. Because the difference
# between both the strings could be at max 1.
while (j >= 0):
sum = 0
sum += (a[j] - '0') + carry
ans += (sum % 10).toString()
carry = sum // 10
j -= 1
# if string a is over and string b is left
while (k >= 0):
sum = 0
sum += ord(b[k]) - ord('0') + carry
ans += str(sum % 10)
carry = (sum // 10)
k -= 1
# if carry is left
if (carry):
ans += str(carry)
# to remove leading zeroes as they will be ahead of our sum
while (len(ans) and ans[len(ans) - 1] == '0'):
ans.pop()
# reverse our final string because we were storing sum from left to right
ans = ans[::-1]
return ans
# Driver Code
arr = [6, 8, 4, 5, 2, 3]
n = len(arr)
print("Sum is " + solve(arr, n))
# This code is contributed by shinjanpatra
Javascript
<script>
function solve(arr, n)
{
// sorting of array O(nlogn)
arr.sort();
// Two String for storing our two minimum numbers
let a = "", b = "";
// string string alternatively
for (let i = 0; i < n; i += 2)
{
a += arr[i];
}
for (let i = 1; i < n; i += 2)
{
b += arr[i];
}
let j = a.length - 1;
let k = b.length - 1;
// as initial carry is zero
let carry = 0;
let ans = "";
while (j >= 0 && k >= 0)
{
let sum = 0;
sum += (a.charCodeAt(j) - '0'.charCodeAt(0)) + (b.charCodeAt(k) - '0'.charCodeAt(0)) + carry;
ans += (sum % 10).toString();
carry = Math.floor(sum / 10);
j--;
k--;
}
// if string b is over and string a is left
// here we dont need to put here while condition
// as it would run at max one time. Because the difference
// between both the strings could be at max 1.
while (j >= 0)
{
let sum = 0;
sum += (a[j] - '0') + carry;
ans += (sum % 10).toString();
carry = Math.floor(sum / 10);
j--;
}
// if string a is over and string b is left
while (k >= 0)
{
let sum = 0;
sum += (b.charCodeAt(k) - '0'.charCodeAt(0)) + carry;
ans += (sum % 10).toString();
carry = Math.floor(sum / 10);
k--;
}
// if carry is left
if (carry)
{
ans += carry.toString();
}
// to remove leading zeroes as they will be ahead of our sum
while (ans.length && ans[ans.length-1] == '0')
ans.pop();
// reverse our final string because we were storing sum from left to right
ans = ans.split('').reverse().join('');
return ans;
}
// Driver Code Starts.
let arr = [6, 8, 4, 5, 2, 3];
let n = arr.length;
document.write("Sum is " + solve(arr, n));
// This code is contributed by shinjanpatra
</script>
Producción
Sum is 604
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA