获得所需总和K所需的最小减少量的计数

给定整数N对和整数K，任务是找到降低的最小数目需要使得每对的第一元素的总和为≤ķ。每次减少涉及减小的一对的第一值到它的第二值。如果它是不可能使之和≤K，打印-1。

Examples: 
Input: N = 5, K = 32 
10 6 
6 4 
8 5 
9 8 
5 2 
Output: 2 
Explanation: 
Total Sum = 10 + 6 + 8 + 9 + 5 = 38 > K 
Reducing 10 – > 6 and 8 – > 5 reduces the sum to 31( 6 + 6 + 5 + 9 + 5) which is less than K.

Input: N = 4, K = 25 
10 5 
20 9 
12 10 
4 2 
Output: -1 

方法：
请按照以下步骤解决问题：

1. 计算每对第一个元素的总和。如果总和已经≤K，则打印0。
2. 根据它们之间的差异对给定的对进行排序。
3. 计算需要以非递增顺序添加的对对的差异数，以使总和小于K。
4. 如果遍历所有对之后总和超过K，则打印-1。否则，打印计数。

下面是上述方法的实现：

C++

// C++ Program to find the count of
// minimum reductions required to
// get the required sum K
#include 
using namespace std;
 
// Function to return the count
// of minimum reductions
int countReductions(
    vector >& v,
    int K)
{
 
    int sum = 0;
    for (auto i : v) {
        sum += i.first;
    }
 
    // If the sum is already
    // less than K
    if (sum <= K) {
        return 0;
    }
 
    // Sort in non-increasing
    // order of difference
    sort(v.begin(), v.end(),
         [&](
             pair a,
             pair b) {
             return (a.first - a.second)
                    > (b.first - b.second);
         });
 
    int i = 0;
    while (sum > K && i < v.size()) {
        sum -= (v[i].first
                - v[i].second);
        i++;
    }
 
    if (sum <= K)
        return i;
 
    return -1;
}
 
// Driver Code
int main()
{
    int N = 4, K = 25;
 
    vector > v(N);
    v[0] = { 10, 5 };
    v[1] = { 20, 9 };
    v[2] = { 12, 10 };
    v[3] = { 4, 2 };
 
    // Function Call
    cout << countReductions(v, K)
         << endl;
    return 0;
}

Java

// Java program to find the count of
// minimum reductions required to
// get the required sum K
import java.util.*;
 
class GFG{
     
// Function to return the count
// of minimum reductions
static int countReductions(ArrayList v,
                           int K)
{
    int sum = 0;
    for(int[] i : v)
    {
        sum += i[0];
    }
 
    // If the sum is already
    // less than K
    if (sum <= K)
    {
        return 0;
    }
 
    // Sort in non-increasing
    // order of difference
    Collections.sort(v, (a, b) -> Math.abs(b[0] - b[1]) -
                                  Math.abs(a[0] - a[1]));
 
    int i = 0;
    while (sum > K && i < v.size())
    {
        sum -= (v.get(i)[0] - v.get(i)[1]);
        i++;
    }
 
    if (sum <= K)
        return i;
 
    return -1;
}
 
// Driver code
public static void main(String[] args)
{
    int N = 4, K = 25;
 
    ArrayList v = new ArrayList<>();
 
    v.add(new int[] { 10, 5 });
    v.add(new int[] { 20, 9 });
    v.add(new int[] { 12, 10 });
    v.add(new int[] { 4, 2 });
 
    // Function Call
    System.out.println(countReductions(v, K));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 program to find the count of
# minimum reductions required to
# get the required sum K
from typing import Any, List
 
# Function to return the count
# of minimum reductions
def countReductions(v: List[Any], K: int) -> int:
 
    sum = 0
     
    for i in v:
        sum += i[0]
 
    # If the sum is already
    # less than K
    if (sum <= K):
        return 0
 
    # Sort in non-increasing
    # order of difference
    v.sort(key = lambda a : a[0] - a[1])
 
    i = 0
     
    while (sum > K and i < len(v)):
        sum -= (v[i][0] - v[i][1])
        i += 1
 
    if (sum <= K):
        return i
 
    return -1
 
# Driver Code
if __name__ == "__main__":
 
    N = 4
    K = 25
 
    v = [[0, 0] for _ in range(N)]
    v[0] = [10, 5]
    v[1] = [20, 9]
    v[2] = [12, 10]
    v[3] = [4, 2]
 
    # Function Call
    print(countReductions(v, K))
 
# This code is contributed by sanjeev2552

输出：

-1

时间复杂度： O(NlogN)
辅助空间： O(1)