由 O(n) 中数组的数字组成的两个数字的最小和
给定一个数字数组（值从 0 到 9），找到由数组的数字组成的两个数字的最小可能和。必须使用给定数组的所有数字来形成这两个数字。
例子：
Input: arr[] = {6, 8, 4, 5, 2, 3}
Output: 604
246 + 358 = 604
Input: arr[] = {5, 3, 0, 7, 4}
Output: 82
方法：当最小的数字出现在最高有效位置时，将由一组数字形成最小数字，然后最小的数字出现在下一个最高有效位置，依此类推……
这个想法是通过从数组中交替选择数字来构建两个数字（假设它按升序排序）。因此，第一个数字由数组中奇数位置的数字组成，第二个数字由数组中偶数位置的数字组成。最后，我们返回第一个和第二个数字的总和。为了降低时间复杂度，可以使用数字频率数组在 O(n) 中对数组进行排序，因为原始数组的每个元素都是单个数字，即最多可以有 10 个不同的元素。
下面是上述方法的实现：
C++
// C++ implementation of above approach
#include
using namespace std;
// Function to return the required minimum sum
int minSum(vector arr, int n)
{
// Array to store the
// frequency of each digit
int MAX = 10;
int *freq = new int[MAX];
for (int i = 0; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
int k = 0;
for (int i = 0; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for (int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
// Generating numbers alternatively
for (int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
}
// Driver code
int main(void)
{
vectorarr = { 6, 8, 4, 5, 2, 3 };
int n = arr.size();
cout << minSum(arr, n);
}
// This code is contributed by ankush_953 Java
// Java implementation of above approach
public class GFG {
public static final int MAX = 10;
// Function to return the required minimum sum
static int minSum(int arr[], int n)
{
// Array to store the
// frequency of each digit
int freq[] = new int[MAX];
for (int i = 0; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
int k = 0;
for (int i = 0; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for (int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
// Generating numbers alternatively
for (int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 6, 8, 4, 5, 2, 3 };
int n = arr.length;
System.out.print(minSum(arr, n));
}
}Python3
# Python implementation of above approach
# Function to return the required minimum sum
def minSum(arr, n):
# Array to store the
# frequency of each digit
MAX = 10
freq = [0]*MAX
for i in range(n):
# Store count of every digit
freq[arr[i]] += 1
# Update arr[] such that it is
# sorted in ascending
k = 0
for i in range(MAX):
# Adding elements in arr[]
# in sorted order
for j in range(0,freq[i]):
arr[k] = i
k += 1
num1 = 0
num2 = 0
# Generating numbers alternatively
for i in range(n):
if i % 2 == 0:
num1 = num1 * MAX + arr[i]
else:
num2 = num2 * MAX + arr[i]
# Return the minimum possible sum
return num1 + num2
# Driver code
arr = [ 6, 8, 4, 5, 2, 3 ]
n = len(arr);
print(minSum(arr, n))
#This code is contributed by ankush_953C#
// C# implementation of above approach
using System;
class GFG {
public static int MAX = 10;
// Function to return the required minimum sum
static int minSum(int[] arr, int n)
{
// Array to store the
// frequency of each digit
int[] freq = new int[MAX];
for (int i = 0; i < n; i++) {
// Store count of every digit
freq[arr[i]]++;
}
// Update arr[] such that it is
// sorted in ascending
int k = 0;
for (int i = 0; i < MAX; i++) {
// Adding elements in arr[]
// in sorted order
for (int j = 0; j < freq[i]; j++) {
arr[k++] = i;
}
}
int num1 = 0;
int num2 = 0;
// Generating numbers alternatively
for (int i = 0; i < n; i++) {
if (i % 2 == 0)
num1 = num1 * MAX + arr[i];
else
num2 = num2 * MAX + arr[i];
}
// Return the minimum possible sum
return num1 + num2;
}
// Driver code
static public void Main()
{
int[] arr = { 6, 8, 4, 5, 2, 3 };
int n = arr.Length;
Console.WriteLine(minSum(arr, n));
}
}
// This code is contributed by jit_t.Javascript
输出：
604时间复杂度： O(n)