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📜  由数组的数字组成的两个数字的最小和

📅  最后修改于: 2021-10-28 02:02:53             🧑  作者: Mango

给定一个数字数组(值从 0 到 9),找到由该数组的数字组成的两个数字的最小可能总和。必须使用给定数组的所有数字来形成两个数字。

例子:

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 

由于我们要最小化要形成的两个数字之和,我们必须将所有数字分成两半,并为它们分配半个数字。我们还需要确保前导数字更小。

我们用给定数组的元素构建一个最小堆,这需要 O(n) 最坏的时间。现在我们通过从优先级队列轮询来检索数组的最小值(一次 2 个)并将这两个最小值附加到我们的数字中,直到堆变空,即数组的所有元素都被耗尽。我们返回两个组成数字的总和,这是我们需要的答案。整体复杂度为 O(nlogn),因为 push() 操作需要 O(logn) 并且重复 n 次。

C++
// C++ program to find minimum sum of two numbers
// formed from all digits in a given array.
#include
using namespace std;
 
// Returns sum of two numbers formed
// from all digits in a[]
int minSum(int arr[], int n)
{
    // min Heap
    priority_queue , greater > pq;
 
    // to store the 2 numbers formed by array elements to
    // minimize the required sum
    string num1, num2;
 
    // Adding elements in Priority Queue
    for(int i=0; i


Java
// Java program to find minimum sum of two numbers
// formed from all digits in a given array.
import java.util.PriorityQueue;
 
class MinSum
{
    // Returns sum of two numbers formed
    // from all digits in a[]
    public static long solve(int[] a)
    {
        // min Heap
        PriorityQueue pq = new PriorityQueue();
 
        // to store the 2 numbers formed by array elements to
        // minimize the required sum
        StringBuilder num1 = new StringBuilder();
        StringBuilder num2 = new StringBuilder();
 
        // Adding elements in Priority Queue
        for (int x : a)
            pq.add(x);
 
        // checking if the priority queue is non empty
        while (!pq.isEmpty())
        {
            num1.append(pq.poll()+ "");
            if (!pq.isEmpty())
                num2.append(pq.poll()+ "");
        }
 
        // the required sum calculated
        long sum = Long.parseLong(num1.toString()) +
                   Long.parseLong(num2.toString());
 
        return sum;
    }
 
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = {6, 8, 4, 5, 2, 3};
        System.out.println("The required sum is "+ solve(arr));
    }
}


Python3
# Python3 program to find minimum
# sum of two numbers formed from
# all digits in a given array.
from queue import PriorityQueue
 
# Returns sum of two numbers formed
# from all digits in a[]
def solve(a):
     
    # min Heap
    pq = PriorityQueue()
     
    # To store the 2 numbers
    # formed by array elements to
    # minimize the required sum
    num1 = ""
    num2 = ""
 
    # Adding elements in
    # Priority Queue
    for x in a:
        pq.put(x)
 
    # Checking if the priority
    # queue is non empty
    while not pq.empty():
        num1 += str(pq.get())
        if not pq.empty():
            num2 += str(pq.get())   
 
    # The required sum calculated
    sum = int(num1) + int(num2)
     
    return sum
     
# Driver code
if __name__=="__main__":
     
    arr = [ 6, 8, 4, 5, 2, 3 ]
    print("The required sum is ", solve(arr))
 
# This code is contributed by rutvik_56


C#
// C# program to find minimum sum of two numbers
// formed from all digits in a given array.
using System;
using System.Collections.Generic;
class GFG
{
     
    // Returns sum of two numbers formed
    // from all digits in a[]
    public static long solve(int[] a)
    {
       
        // min Heap
        List pq = new List();
  
        // to store the 2 numbers formed by array elements to
        // minimize the required sum
        string num1 = "";
        string num2 = "";
  
        // Adding elements in Priority Queue
        foreach(int x in a)
            pq.Add(x);
             
        pq.Sort();
  
        // checking if the priority queue is non empty
        while (pq.Count > 0)
        {
            num1 = num1 + pq[0];
            pq.RemoveAt(0);
            if (pq.Count > 0)
            {
                num2 = num2 + pq[0];
                pq.RemoveAt(0);
            }
        }
  
        // the required sum calculated
        int sum = Int32.Parse(num1) + Int32.Parse(num2);
  
        return sum;
    }
     
  // Driver code
  static void Main()
  {
    int[] arr = {6, 8, 4, 5, 2, 3};
    Console.WriteLine("The required sum is "+ solve(arr));
  }
}
 
// This code is contributed by divyesh072019.


Javascript


C++
// C++ program to find minimum sum of two numbers
// formed from all digits in a given array.
#include
using namespace std;
 
// Returns sum of two numbers formed
// from all digits in a[]
int minSum(int a[], int n){
     
    // sort the elements
    sort(a,a+n);
     
    int num1 = 0;
    int num2 = 0;
    for(int i = 0;i


Java
import java.util.Arrays;
//Java program to find minimum sum of two numbers
//formed from all digits in a given array.
public class AQRQ {
 
    //Returns sum of two numbers formed
    //from all digits in a[]
    static int minSum(int a[], int n){
       
     // sort the elements
     Arrays.sort(a);
       
     int num1 = 0;
     int num2 = 0;
     for(int i = 0;i


Python3
# Python 3 program to find minimum
# sum of two numbers formed
# from all digits in an given array
 
# Returns sum of two numbers formed
# from all digits in a[]
def minSum(a, n):
     
    # sorted the elements
    a = sorted(a)
    num1, num2 = 0, 0
     
    for i in range(n):
        if i % 2 == 0:
            num1 = num1 * 10 + a[i]
        else:
            num2 = num2 * 10 + a[i]
     
    return num2 + num1    
     
# Driver code
arr = [5, 3, 0, 7, 4]
n = len(arr)
print("The required sum is",
             minSum(arr, n))
     
# This code is contributed
# by Mohit kumar 29


C#
// C# a program to find minimum sum of two numbers
//formed from all digits in a given array.
 
using System;
 
public class GFG{
     
    //Returns sum of two numbers formed
    //from all digits in a[]
    static int minSum(int []a, int n){
     
    // sort the elements
    Array.Sort(a);
     
    int num1 = 0;
    int num2 = 0;
    for(int i = 0;i


PHP


Javascript


输出:

The required sum is 604

另一种方法:
我们也可以像这样遵循另一种方法,因为我们需要两个数字,使得它们的总和最小,那么我们也需要两个最小数字。如果我们按升序排列我们的数组,那么我们可以用两个数字组成最小的数字,
例如,2 3 4 5 6 8,现在我们可以得到两个从 2 和 3 开始的数字。第一部分现在完成了。继续前进,我们必须形成这样的形式,即它们将包含小数字,即从数组中选择数字来扩展您的两个数字。
即 246、358。现在如果我们看到分析这个,那么我们可以为 num1 选择偶数索引数,为 num2 选择奇数。

下面是实现:

C++

// C++ program to find minimum sum of two numbers
// formed from all digits in a given array.
#include
using namespace std;
 
// Returns sum of two numbers formed
// from all digits in a[]
int minSum(int a[], int n){
     
    // sort the elements
    sort(a,a+n);
     
    int num1 = 0;
    int num2 = 0;
    for(int i = 0;i

Java

import java.util.Arrays;
//Java program to find minimum sum of two numbers
//formed from all digits in a given array.
public class AQRQ {
 
    //Returns sum of two numbers formed
    //from all digits in a[]
    static int minSum(int a[], int n){
       
     // sort the elements
     Arrays.sort(a);
       
     int num1 = 0;
     int num2 = 0;
     for(int i = 0;i

蟒蛇3

# Python 3 program to find minimum
# sum of two numbers formed
# from all digits in an given array
 
# Returns sum of two numbers formed
# from all digits in a[]
def minSum(a, n):
     
    # sorted the elements
    a = sorted(a)
    num1, num2 = 0, 0
     
    for i in range(n):
        if i % 2 == 0:
            num1 = num1 * 10 + a[i]
        else:
            num2 = num2 * 10 + a[i]
     
    return num2 + num1    
     
# Driver code
arr = [5, 3, 0, 7, 4]
n = len(arr)
print("The required sum is",
             minSum(arr, n))
     
# This code is contributed
# by Mohit kumar 29

C#

// C# a program to find minimum sum of two numbers
//formed from all digits in a given array.
 
using System;
 
public class GFG{
     
    //Returns sum of two numbers formed
    //from all digits in a[]
    static int minSum(int []a, int n){
     
    // sort the elements
    Array.Sort(a);
     
    int num1 = 0;
    int num2 = 0;
    for(int i = 0;i

PHP


Javascript


输出:

The required sum is 82

时间复杂度: O(nLogN)