通过将数组拆分为两个子集，最大和的最小子集

给定一个由N 个整数组成的数组arr[] ，任务是打印通过将数组拆分为两个子集而获得的两个子集中较小的一个，从而使较小子集的总和最大化。

例子：

Input: arr[] = {5, 3, 2, 4, 1, 2}
Output: 4 5
Explanation:
Split the array into two subsets as {4, 5} and {1, 2, 2, 3}.
The subset {4, 5} is of minimum length, i.e. 2, having maximum sum = 4 + 5 = 9.

Input: arr[] = {20, 15, 20, 50, 20}
Output: 15 50

方法：给定的问题可以通过使用散列和排序来解决。
请按照以下步骤解决问题：

• 初始化一个 HashMap，比如M ，以存储数组arr[]中每个字符的频率。
• 遍历数组arr[]并增加 HashMap M中每个字符的计数。
• 初始化 2 个变量，例如S和flag ，以存储第一个子集的总和并分别存储答案是否存在。
• 按升序对数组arr[]进行排序。
• 初始化一个 ArrayList，比如ans ，以存储结果子集的元素。
• 以相反的顺序遍历数组arr[]并执行以下步骤：
  • 将当前字符的频率存储在一个变量中，比如F 。
  • 如果(F + ans.size())小于 ( N – (F + ans.size()))然后将元素arr[i]附加到 ArrayList ans F number of 次。
  • 将i的值减少F 。
  • 如果S的值大于数组元素的总和，则将标志标记为真然后中断。
• 完成上述步骤后，如果flag 的值为true ，则打印 ArrayList ans作为结果子集。否则，打印-1 .the

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to split array elements
// into two subsets having sum of
// the smaller subset maximized
static void findSubset(vector arr)
{
   
    // Stores the size of the array
    int N = arr.size();
 
    // Stores the frequency
    // of array elements
    map mp;
 
    // Stores the total
    // sum of the array
    int totSum = 0;
 
    // Stores the sum of
    // the resultant set
    int s = 0;
 
    // Stores if it is possible
    // to split the array that
    // satisfies the conditions
    int flag = 0;
 
    // Stores the elements
    // of the first subseta
    vector ans;
 
    // Traverse the array arr[]
    for (int i = 0;
         i < arr.size(); i++) {
 
        // Increment total sum
        totSum += arr[i];
 
        // Increment count of arr[i]
        mp[arr[i]]=mp[arr[i]]+1;
      } 
 
    // Sort the array arr[]
    sort(arr.begin(),arr.end());
 
    // Stores the index of the
    // last element of the array
    int i = N - 1;
 
    // Traverse the array arr[]
    while (i >= 0) {
 
        // Stores the frequency
        // of arr[i]
        int frq = mp[arr[i]];
 
        // If frq + ans.size() is
        // at most remaining size
        if ((frq + ans.size())
            < (N - (frq + ans.size())))
        {
 
            for (int k = 0; k < frq; k++)
            {
 
                // Append arr[i] to ans
                ans.push_back(arr[i]);
 
                // Decrement totSum by arr[i]
                totSum -= arr[i];
 
                // Increment s by arr[i]
                s += arr[i];
 
                i--;
            }
        }
 
        // Otherwise, decrement i
        // by frq
        else {
            i -= frq;
        }
 
        // If s is greater
        // than totSum
        if (s > totSum) {
 
            // Mark flag 1
            flag = 1;
            break;
        }
    }
 
    // If flag is equal to 1
    if (flag == 1) {
 
        // Print the arrList ans
        for (i = ans.size() - 1;
             i >= 0; i--) {
 
            cout< arr = { 5, 3, 2, 4, 1, 2 };
    findSubset(arr);
}
 
// This code is contributed by mohit kumar 29.

Java

// Java program for above approach
 
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG {
 
    // Function to split array elements
    // into two subsets having sum of
    // the smaller subset maximized
    static void findSubset(int[] arr)
    {
        // Stores the size of the array
        int N = arr.length;
 
        // Stores the frequency
        // of array elements
        Map map
            = new HashMap<>();
 
        // Stores the total
        // sum of the array
        int totSum = 0;
 
        // Stores the sum of
        // the resultant set
        int s = 0;
 
        // Stores if it is possible
        // to split the array that
        // satisfies the conditions
        int flag = 0;
 
        // Stores the elements
        // of the first subset
        ArrayList ans
            = new ArrayList<>();
 
        // Traverse the array arr[]
        for (int i = 0;
             i < arr.length; i++) {
 
            // Increment total sum
            totSum += arr[i];
 
            // Increment count of arr[i]
            map.put(arr[i],
                    map.getOrDefault(
                        arr[i], 0)
                        + 1);
        }
 
        // Sort the array arr[]
        Arrays.sort(arr);
 
        // Stores the index of the
        // last element of the array
        int i = N - 1;
 
        // Traverse the array arr[]
        while (i >= 0) {
 
            // Stores the frequency
            // of arr[i]
            int frq = map.get(arr[i]);
 
            // If frq + ans.size() is
            // at most remaining size
            if ((frq + ans.size())
                < (N - (frq + ans.size()))) {
 
                for (int k = 0; k < frq; k++) {
 
                    // Append arr[i] to ans
                    ans.add(arr[i]);
 
                    // Decrement totSum by arr[i]
                    totSum -= arr[i];
 
                    // Increment s by arr[i]
                    s += arr[i];
 
                    i--;
                }
            }
 
            // Otherwise, decrement i
            // by frq
            else {
                i -= frq;
            }
 
            // If s is greater
            // than totSum
            if (s > totSum) {
 
                // Mark flag 1
                flag = 1;
                break;
            }
        }
 
        // If flag is equal to 1
        if (flag == 1) {
 
            // Print the arrList ans
            for (i = ans.size() - 1;
                 i >= 0; i--) {
 
                System.out.print(
                    ans.get(i) + " ");
            }
        }
 
        // Otherwise, print "-1"
        else {
            System.out.print(-1);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] arr = { 5, 3, 2, 4, 1, 2 };
        findSubset(arr);
    }
}

Python3

# Python 3 program for the above approach
from collections import defaultdict
 
# Function to split array elements
# into two subsets having sum of
# the smaller subset maximized
def findSubset(arr):
 
    # Stores the size of the array
    N = len(arr)
 
    # Stores the frequency
    # of array elements
    mp = defaultdict(int)
 
    # Stores the total
    # sum of the array
    totSum = 0
 
    # Stores the sum of
    # the resultant set
    s = 0
 
    # Stores if it is possible
    # to split the array that
    # satisfies the conditions
    flag = 0
 
    # Stores the elements
    # of the first subseta
    ans = []
 
    # Traverse the array arr[]
    for i in range(len(arr)):
 
        # Increment total sum
        totSum += arr[i]
 
        # Increment count of arr[i]
        mp[arr[i]] = mp[arr[i]]+1
 
    # Sort the array arr[]
    arr.sort()
 
    # Stores the index of the
    # last element of the array
    i = N - 1
 
    # Traverse the array arr[]
    while (i >= 0):
 
        # Stores the frequency
        # of arr[i]
        frq = mp[arr[i]]
 
        # If frq + ans.size() is
        # at most remaining size
        if ((frq + len(ans))
                < (N - (frq + len(ans)))):
 
            for k in range(frq):
 
                # Append arr[i] to ans
                ans.append(arr[i])
 
                # Decrement totSum by arr[i]
                totSum -= arr[i]
 
                # Increment s by arr[i]
                s += arr[i]
                i -= 1
 
        # Otherwise, decrement i
        # by frq
        else:
            i -= frq
 
        # If s is greater
        # than totSum
        if (s > totSum):
 
            # Mark flag 1
            flag = 1
            break
 
    # If flag is equal to 1
    if (flag == 1):
 
        # Print the arrList ans
        for i in range(len(ans) - 1, -1, -1):
 
            print(ans[i], end = " ")
 
    # Otherwise, print "-1"
    else:
        print(-1)
 
# Driver Code
if __name__ == "__main__":
 
    arr = [5, 3, 2, 4, 1, 2]
    findSubset(arr)
 
    # This code is contributed by ukasp.

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
   
// Function to split array elements
// into two subsets having sum of
// the smaller subset maximized
static void findSubset(List arr)
{
   
    // Stores the size of the array
    int N = arr.Count;
    int i;
 
    // Stores the frequency
    // of array elements
    Dictionary mp = new Dictionary();
 
    // Stores the total
    // sum of the array
    int totSum = 0;
 
    // Stores the sum of
    // the resultant set
    int s = 0;
 
    // Stores if it is possible
    // to split the array that
    // satisfies the conditions
    int flag = 0;
 
    // Stores the elements
    // of the first subseta
    List ans = new List();
 
    // Traverse the array arr[]
    for (i = 0;
         i < arr.Count; i++) {
 
        // Increment total sum
        totSum += arr[i];
 
        // Increment count of arr[i]
        if(mp.ContainsKey(arr[i]))
         mp[arr[i]]=mp[arr[i]]+1;
        else
          mp.Add(arr[i],1);
      } 
 
    // Sort the array arr[]
    arr.Sort();
 
    // Stores the index of the
    // last element of the array
    i = N - 1;
 
    // Traverse the array arr[]
    while (i >= 0) {
 
        // Stores the frequency
        // of arr[i]
        int frq = mp[arr[i]];
 
        // If frq + ans.size() is
        // at most remaining size
        if ((frq + ans.Count)
            < (N - (frq + ans.Count)))
        {
 
            for (int k = 0; k < frq; k++)
            {
 
                // Append arr[i] to ans
                ans.Add(arr[i]);
 
                // Decrement totSum by arr[i]
                totSum -= arr[i];
 
                // Increment s by arr[i]
                s += arr[i];
 
                i--;
            }
        }
 
        // Otherwise, decrement i
        // by frq
        else {
            i -= frq;
        }
 
        // If s is greater
        // than totSum
        if (s > totSum) {
 
            // Mark flag 1
            flag = 1;
            break;
        }
    }
 
    // If flag is equal to 1
    if (flag == 1) {
 
        // Print the arrList ans
        for (i = ans.Count - 1;
             i >= 0; i--) {
 
            Console.Write(ans[i]+" ");
        }
    }
 
    // Otherwise, print "-1"
    else {
        Console.Write(-1);
    }
}
 
// Driver Code
public static void Main()
{
    List arr = new List(){ 5, 3, 2, 4, 1, 2 };
    findSubset(arr);
}
 
}
 
// This code is contributed by ipg2016107.

Javascript

 
// Javascript program for the above approach
 
 
// Function to split array elements
// into two subsets having sum of
// the smaller subset maximized
function findSubset(arr)
{
   
    // Stores the size of the array
    var N = arr.length;
 
    // Stores the frequency
    // of array elements
    var mp = new Map();
 
    // Stores the total
    // sum of the array
    var totSum = 0;
 
    // Stores the sum of
    // the resultant set
    var s = 0;
 
    // Stores if it is possible
    // to split the array that
    // satisfies the conditions
    var flag = 0;
 
    // Stores the elements
    // of the first subseta
    var ans = [];
 
    // Traverse the array arr[]
    for (var i = 0;
         i < arr.length; i++) {
 
        // Increment total sum
        totSum += arr[i];
 
        // Increment count of arr[i]
        if(mp.has(arr[i]))
            mp.set(arr[i], mp.get(arr[i])+1)
        else
            mp.set(arr[i], 1);
      } 
 
    // Sort the array arr[]
    arr.sort((a,b)=> a-b)
 
    // Stores the index of the
    // last element of the array
    var i = N - 1;
 
    // Traverse the array arr[]
    while (i >= 0) {
 
        // Stores the frequency
        // of arr[i]
        var frq = mp.get(arr[i]);
 
        // If frq + ans.size() is
        // at most remaining size
        if ((frq + ans.length)
            < (N - (frq + ans.length)))
        {
 
            for (var k = 0; k < frq; k++)
            {
 
                // Append arr[i] to ans
                ans.push(arr[i]);
 
                // Decrement totSum by arr[i]
                totSum -= arr[i];
 
                // Increment s by arr[i]
                s += arr[i];
 
                i--;
            }
        }
 
        // Otherwise, decrement i
        // by frq
        else {
            i -= frq;
        }
 
        // If s is greater
        // than totSum
        if (s > totSum) {
 
            // Mark flag 1
            flag = 1;
            break;
        }
    }
 
    // If flag is equal to 1
    if (flag == 1) {
 
        // Print the arrList ans
        for (i = ans.length - 1;
             i >= 0; i--) {
 
            document.write( ans[i] + " ");
        }
    }
 
    // Otherwise, print "-1"
    else {
        document.write(-1);
    }
}
 
// Driver Code
var arr = [5, 3, 2, 4, 1, 2 ];
findSubset(arr);
 
// This code is contributed by rutvik_56.

输出：

4 5

时间复杂度： O(N*log N)
辅助空间： O(N)

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