总和至少为剩余数组元素的对子集的最小大小

给定两个由N个正整数组成的数组A[]和B[] ，任务是找到元素对(A[i], B[i])的子集的最小大小，使得所有元素的总和子集对至少是不包括在子集中的剩余数组元素A[]的总和，即(A[0] + B[0] + A[1] + B[1] + … + A[K – 1] + B[K – 1] >= A[K + 1] + … + A[N – 1] 。

例子：

Input: A[] = {3, 2, 2}, B[] = {2, 3, 1}
Output: 1
Explanation:
Choose the subset as {(3, 2)}. Now, the sum of subsets is 3 + 2 = 5, which greater than sum of remaining A[] = 3 + 1 = 4.

Input: A[] = {2, 2, 2, 2, 2}, B[] = {1, 1, 1, 1, 1}  
Output: 3

朴素方法：最简单的方法是生成给定元素对的所有可能子集，并打印满足给定标准并具有最小大小的子集的大小。

时间复杂度： O(2 ^N )
辅助空间： O(1)

有效的方法：给定的问题可以通过使用贪心方法来解决，这个想法是使用排序并且可以观察到选择第ⁱ对作为结果子集的一部分，两个子集之间的差异减少了（ 2*S[i] + U[i]) 。请按照以下步骤解决问题：

• 初始化一个数组，比如大小为N的差异 [] ，它存储每对元素的(2*A[i] + B[i])的值。
• 初始化一个变量，比如sum ，它跟踪剩余数组元素A[i]的总和。
• 初始化一个变量，比如K来跟踪结果子集的大小。
• 在[0, N)范围内迭代，并将差异 [i]的值更新为(2*A[i] + B[i])的值，并将sum的值减去A[i]的值。
• 以升序对给定的数组差异 []进行排序。
• 以相反的方式遍历数组difference[]直到sum为负，然后将difference[i]的值添加到sum中，并将K的值增加1 。
• 完成上述步骤后，打印出K的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find the minimum size
// of subset satisfying given criteria
void maximizeApples(vector& U, vector& S, int N)
{
   
    // Stores the value of 2*S[i] + U[i]
    vector x(N);
 
    // Stores the difference between
    // the 2 subsets
    int sum = 0;
 
    for (int i = 0; i < N; i++) {
 
        // Update the value of sum
        sum -= S[i];
        x[i] += 2 * S[i] + U[i];
    }
 
    // Sort the array X[] in an
    // ascending order
    sort(x.begin(), x.end());
    int ans = 0;
    int j = N - 1;
 
    // Traverse the array
    while (sum <= 0) {
 
        // Update the value of sum
        sum += x[j--];
 
        // Increment value of ans
        ans++;
    }
 
    // Print the resultant ans
    cout << ans;
}
 
// Driver Code
int main()
{
    vector A = { 1, 1, 1, 1, 1 };
    vector B = { 2, 2, 2, 2, 2 };
    int N = A.size();
    maximizeApples(A, B, N);
 
    return 0;
}
 
    // This code is contributed by rakeshsahni

Java

// Java program for the above approach
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to find the minimum size
    // of subset satisfying given criteria
    private static void maximizeApples(
        int[] U, int[] S, int N)
    {
        // Stores the value of 2*S[i] + U[i]
        int[] x = new int[N];
 
        // Stores the difference between
        // the 2 subsets
        int sum = 0;
 
        for (int i = 0; i < N; i++) {
 
            // Update the value of sum
            sum -= S[i];
            x[i] += 2 * S[i] + U[i];
        }
 
        // Sort the array X[] in an
        // ascending order
        Arrays.sort(x);
        int ans = 0;
        int j = N - 1;
 
        // Traverse the array
        while (sum <= 0) {
 
            // Update the value of sum
            sum += x[j--];
 
            // Increment value of ans
            ans++;
        }
 
        // Print the resultant ans
        System.out.println(ans);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int[] A = new int[] { 1, 1, 1, 1, 1 };
        int[] B = new int[] { 2, 2, 2, 2, 2 };
        int N = A.length;
        maximizeApples(A, B, N);
    }
}

Python3

# Python program for the above approach
 
# Function to find the minimum size
# of subset satisfying given criteria
def maximizeApples(U, S, N):
 
    # Stores the value of 2*S[i] + U[i]
    x = [0]*N
 
    # Stores the difference between
    # the 2 subsets
    sum = 0
 
    for i in range(N):
 
        # Update the value of sum
        sum -= S[i]
        x[i] += 2 * S[i] + U[i]
 
    # Sort the array X[] in an
    # ascending order
    x.sort()
    ans = 0
    j = N - 1
 
    # Traverse the array
    while sum <= 0:
 
        # Update the value of sum
        sum += x[j]
        j = j - 1
 
        # Increment value of ans
        ans = ans + 1
 
    # Print the resultant ans
    print(ans)
 
# Driver Code
A = [1, 1, 1, 1, 1]
B = [2, 2, 2, 2, 2]
N = len(A)
maximizeApples(A, B, N)
 
# This code is contributed by Potta Lokesh

C#

//  c# program for the above approach
using System.Collections.Generic;
using System;
class GFG
{
 
    // Function to find the minimum size
    // of subset satisfying given criteria
    static void maximizeApples(int[] U, int[] S, int N)
    {
       
        // Stores the value of 2*S[i] + U[i]
        List x = new List();
       
        // Stores the difference between
        // the 2 subsets
        int sum = 0;
 
        for (int i = 0; i < N; i++)
        {
           
            // Update the value of sum
            sum -= S[i];
            x.Add( 2 * S[i] + U[i]);
        }
 
        // Sort the array X[] in an
        // ascending order
        x.Sort();
        int ans = 0;
        int j = N - 1;
 
        // Traverse the array
        while (sum <= 0) {
 
            // Update the value of sum
            sum += x[j--];
 
            // Increment value of ans
            ans++;
        }
 
        // Print the resultant ans
        Console.WriteLine(ans);
    }
 
    // Driver Code
    public static void Main()
    {
        int[] A = new int[] { 1, 1, 1, 1, 1 };
        int[] B = new int[] { 2, 2, 2, 2, 2 };
        int N = A.Length;
        maximizeApples(A, B, N);
    }
}
 
// This code is contributed by amreshkumar3.

Javascript

// Javascript program for the above approach
 
// Function to find the minimum size
// of subset satisfying given criteria
function maximizeApples(U, S, N) {
  // Stores the value of 2*S[i] + U[i]
  let x = new Array(N).fill(0);
 
  // Stores the difference between
  // the 2 subsets
  let sum = 0;
 
  for (let i = 0; i < N; i++) {
    // Update the value of sum
    sum -= S[i];
    x[i] += 2 * S[i] + U[i];
  }
 
  // Sort the array X[] in an
  // ascending order
  x.sort((a, b) => a - b);
 
  let ans = 0;
  let j = N - 1;
 
  // Traverse the array
  while (sum <= 0) {
    // Update the value of sum
    sum += x[j--];
 
    // Increment value of ans
    ans++;
  }
 
  // Print the resultant ans
  document.write(ans);
}
 
// Driver Code
 
let A = [1, 1, 1, 1, 1];
let B = [2, 2, 2, 2, 2];
let N = A.length;
maximizeApples(A, B, N);
 
// This code is contributed by gfgking.

输出：

3

时间复杂度： O(N*log N)
辅助空间： O(N)