给定数组中可能的两个最小子集的长度总和，总和至少为 K

给定一个由N 个整数和一个整数K组成的数组arr[] ，任务是找到两个最小的唯一子集的长度之和，其元素之和至少为 K 。

例子：

Input: arr[] = {2, 4, 5, 6, 7, 8}, K = 16
Output: 6
Explanation:
The subsets {2, 6, 8} and {4, 5, 7} are the two smallest subsets with sum K(= 16).
Therefore, the sum of the lengths of both these subsets = 3 + 3 = 6.

Input: arr[] = {14, 3, 7, 8, 9, 7, 12, 15, 10, 6}, K = 40
Output: 8

方法：根据以下观察可以解决给定的问题：

• 对数组进行排序将问题简化为在索引[i, N]范围内选择总和至少为 K的子数组，然后检查索引[i, N]范围内剩余数组元素的总和是否为K与否。
• 为了实现上述想法，使用二维数组，例如dp[][] ，使得dp[i][j]存储在索引[i, N]范围内的子集的最小总和，其值至少为j.那么过渡状态类似于 0/1 Knapsack 可以定义为：
  • 如果arr[i]的值大于j ，则将dp[i][j]更新为arr[i] 。
  • 否则，将dp[i][j]更新为dp[i + 1][j]和(dp[i + 1][j – arr[i]] + arr[i])的最小值。

请按照以下步骤解决问题：

• 按升序对数组进行排序。
• 初始化一个数组，比如suffix[]，并将数组arr[]的后缀和存储在其中。
• 初始化一个二维数组，比如dp[][]，这样dp[i][j]存储在索引[i, N]范围内具有至少 j值的子集的最小总和。
• 将dp[N][0]初始化为0并将所有其他状态初始化为INT_MAX 。
• 以相反的顺序遍历数组arr[i]并执行以下步骤：
  • 以相反的顺序迭代索引[0, K]的范围并执行以下操作：
    • 如果arr[i]的值至少为j ，则将dp[i][j]的值更新为arr[i] ，因为当前状态的总和至少为 j 。现在，继续迭代。
    • 如果下一个状态的值，即dp[i + 1][j – arr[i]]是INT_MAX ，则将dp[i][j]更新为INT_MAX 。
    • 否则，将dp[i][j]更新为dp[i + 1][j]和(dp[i + 1][j – arr[i]] + arr[i])的最小值以存储总和至少为 j 的所有值。
• 现在，以相反的顺序遍历数组suffix[] ，如果(suffix[i] – dp[i][K]) 的值至少为 K ，则打印(N – i)作为数组大小的总和两个最小的子集形成并跳出循环。
• 否则，打印“-1” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
const int MAX = 1e9;
 
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
int MinimumLength(int A[], int N, int K)
{
    // Sort the array in ascending order
    sort(A, A + N);
 
    // Stores suffix sum of the array
    int suffix[N + 1] = { 0 };
 
    // Update the suffix sum array
    for (int i = N - 1; i >= 0; i--)
        suffix[i] = suffix[i + 1] + A[i];
 
    // Stores all dp-states
    int dp[N + 1][K + 1];
 
    // Initialize all dp-states
    // with a maximum possible value
    for (int i = 0; i <= N; i++)
        for (int j = 0; j <= K; j++)
            dp[i][j] = MAX;
 
    // Base Case
    dp[N][0] = 0;
 
    // Traverse the array arr[]
    for (int i = N - 1; i >= 0; i--) {
 
        // Iterate over the range [0, K]
        for (int j = K; j >= 0; j--) {
 
            // If A[i] is equal to at
            // least the required sum
            // j for the current state
            if (j <= A[i]) {
                dp[i][j] = A[i];
                continue;
            }
 
            // If the next possible
            // state doesn't exist
            if (dp[i + 1][j - A[i]] == MAX)
                dp[i][j] = MAX;
 
            // Otherwise, update the current
            // state to the minimum of the
            // next state and state including
            // the current element A[i]
            else
                dp[i][j] = min(dp[i + 1][j],
                               dp[i + 1][j - A[i]] + A[i]);
        }
    }
 
    // Traverse the suffix sum array
    for (int i = N - 1; i >= 0; i--) {
 
        // If suffix[i] - dp[i][K] >= K
        if (suffix[i] - dp[i][K] >= K) {
 
            // Sum of lengths of the two
            // smallest subsets is obtained
            return N - i;
        }
    }
 
    // Return -1, if there doesn't
    // exist any subset of sum >= K
    return -1;
}
 
// Driver Code
int main()
{
    int arr[] = { 7, 4, 5, 6, 8 };
    int K = 13;
    int N = sizeof(arr) / sizeof(arr[0]);
 
    cout << MinimumLength(arr, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
 
class GFG{
 
static int MAX = (int)(1e9);
 
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
static int MinimumLength(int A[], int N, int K)
{
     
    // Sort the array in ascending order
    Arrays.sort(A);
     
    // Stores suffix sum of the array
    int suffix[] = new int[N + 1];
 
    // Update the suffix sum array
    for(int i = N - 1; i >= 0; i--)
        suffix[i] = suffix[i + 1] + A[i];
 
    // Stores all dp-states
    int dp[][] = new int[N + 1][K + 1];
 
    // Initialize all dp-states
    // with a maximum possible value
    for(int i = 0; i <= N; i++)
        for(int j = 0; j <= K; j++)
            dp[i][j] = MAX;
 
    // Base Case
    dp[N][0] = 0;
 
    // Traverse the array arr[]
    for(int i = N - 1; i >= 0; i--)
    {
         
        // Iterate over the range [0, K]
        for(int j = K; j >= 0; j--)
        {
             
            // If A[i] is equal to at
            // least the required sum
            // j for the current state
            if (j <= A[i])
            {
                dp[i][j] = A[i];
                continue;
            }
 
            // If the next possible
            // state doesn't exist
            if (dp[i + 1][j - A[i]] == MAX)
                dp[i][j] = MAX;
 
            // Otherwise, update the current
            // state to the minimum of the
            // next state and state including
            // the current element A[i]
            else
                dp[i][j] = Math.min(dp[i + 1][j],
                                    dp[i + 1][j - A[i]]
                                         + A[i]);
        }
    }
 
    // Traverse the suffix sum array
    for(int i = N - 1; i >= 0; i--)
    {
         
        // If suffix[i] - dp[i][K] >= K
        if (suffix[i] - dp[i][K] >= K)
        {
             
            // Sum of lengths of the two
            // smallest subsets is obtained
            return N - i;
        }
    }
 
    // Return -1, if there doesn't
    // exist any subset of sum >= K
    return -1;
}
 
// Driver Code
public static void main(String[] args)
{
    int arr[] = { 7, 4, 5, 6, 8 };
    int K = 13;
    int N = arr.length;
 
    System.out.println(MinimumLength(arr, N, K));
}
}
 
// This code is contributed by Kingash

Python3

# Python3 program for the above approach
MAX = 1e9
 
# Function to calculate sum of lengths
# of two smallest subsets with sum >= K
def MinimumLength(A, N, K):
     
    # Sort the array in ascending order
    A.sort()
 
    # Stores suffix sum of the array
    suffix = [0] * (N + 1)
 
    # Update the suffix sum array
    for i in range(N - 1, -1, -1):
        suffix[i] = suffix[i + 1] + A[i]
 
    # Stores all dp-states
    dp = [[0] * (K + 1)] * (N + 1)
 
    # Initialize all dp-states
    # with a maximum possible value
    for i in range(N + 1):
        for j in range(K + 1):
            dp[i][j] = MAX
 
    # Base Case
    dp[N][0] = 0
 
    # Traverse the array arr[]
    for i in range(N - 1, -1, -1):
 
        # Iterate over the range [0, K]
        for j in range(K, -1, -1):
             
            # If A[i] is equal to at
            # least the required sum
            # j for the current state
            if (j <= A[i]) :
                dp[i][j] = A[i]
                continue
             
            # If the next possible
            # state doesn't exist
            if (dp[i + 1][j - A[i]] == MAX):
                dp[i][j] = MAX
 
            # Otherwise, update the current
            # state to the minimum of the
            # next state and state including
            # the current element A[i]
            else :
                dp[i][j] = min(dp[i + 1][j],
                               dp[i + 1][j - A[i]] + A[i])
         
    # Traverse the suffix sum array
    for i in range(N - 1, -1, -1):
 
        # If suffix[i] - dp[i][K] >= K
        if (suffix[i] - dp[i][K] >= K):
 
            # Sum of lengths of the two
            # smallest subsets is obtained
            return N - i
         
    # Return -1, if there doesn't
    # exist any subset of sum >= K
    return -1
 
# Driver Code
arr = [ 7, 4, 5, 6, 8 ]
K = 13
N = len(arr)
 
print(MinimumLength(arr, N, K))
 
# This code is contributed by splevel62

C#

// C# program for the above approach
using System;
 
class GFG{
 
static int MAX = (int)(1e9);
 
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
static int MinimumLength(int[] A, int N, int K)
{
     
    // Sort the array in ascending order
    Array.Sort(A);
 
    // Stores suffix sum of the array
    int[] suffix = new int[N + 1];
 
    // Update the suffix sum array
    for(int i = N - 1; i >= 0; i--)
        suffix[i] = suffix[i + 1] + A[i];
 
    // Stores all dp-states
    int[,] dp = new int[N + 1, K + 1];
 
    // Initialize all dp-states
    // with a maximum possible value
    for(int i = 0; i <= N; i++)
        for(int j = 0; j <= K; j++)
            dp[i, j] = MAX;
 
    // Base Case
    dp[N, 0] = 0;
 
    // Traverse the array arr[]
    for(int i = N - 1; i >= 0; i--)
    {
         
        // Iterate over the range [0, K]
        for(int j = K; j >= 0; j--)
        {
             
            // If A[i] is equal to at
            // least the required sum
            // j for the current state
            if (j <= A[i])
            {
                dp[i, j] = A[i];
                continue;
            }
 
            // If the next possible
            // state doesn't exist
            if (dp[i + 1, j - A[i]] == MAX)
                dp[i, j] = MAX;
 
            // Otherwise, update the current
            // state to the minimum of the
            // next state and state including
            // the current element A[i]
            else
                dp[i, j] = Math.Min(dp[i + 1, j],
                                    dp[i + 1, j - A[i]]
                                         + A[i]);
        }
    }
 
    // Traverse the suffix sum array
    for(int i = N - 1; i >= 0; i--)
    {
         
        // If suffix[i] - dp[i][K] >= K
        if (suffix[i] - dp[i, K] >= K)
        {
             
            // Sum of lengths of the two
            // smallest subsets is obtained
            return N - i;
        }
    }
 
    // Return -1, if there doesn't
    // exist any subset of sum >= K
    return -1;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[] arr = { 7, 4, 5, 6, 8 };
    int K = 13;
    int N = arr.Length;
 
    Console.WriteLine(MinimumLength(arr, N, K));
}
}
 
// This code is contributed by ukasp

Javascript

 
// javascript program for the above approach
var max1 = 1000000000;
 
// Function to calculate sum of lengths
// of two smallest subsets with sum >= K
function MinimumLength(A, N, K)
{
0
    // Sort the array in ascending order
    A.sort();
 
    // Stores suffix sum of the array
    var suffix = Array(N + 1).fill(0);
     
    var i;
    // Update the suffix sum array
    for (i = N - 1; i >= 0; i--)
        suffix[i] = suffix[i + 1] + A[i];
 
    // Stores all dp-states
    var dp = new Array(N + 1);
    for (i = 0; i < N+1; i++)
       dp[i] = new Array(K + 1);
 
    // Initialize all dp-states
    // with a max1imum possible value
    var j;
    for (i = 0; i <= N; i++) {
        for (j = 0; j <= K; j++){
            dp[i][j] = max1;
        }
    };
 
    // Base Case
    dp[N][0] = 0;
 
    // Traverse the array arr[]
    for (i = N - 1; i >= 0; i--) {
 
        // Iterate over the range [0, K]
        for (j = K; j >= 0; j--) {
 
            // If A[i] is equal to at
            // least the required sum
            // j for the current state
            if (j <= A[i]) {
                dp[i][j] = A[i];
                continue;
            }
 
            // If the next possible
            // state doesn't exist
            if (dp[i + 1][j - A[i]] == max1)
                dp[i][j] = max1;
 
            // Otherwise, update the current
            // state to the minimum of the
            // next state and state including
            // the current element A[i]
            else
                dp[i][j] = Math.min(dp[i + 1][j],
                               dp[i + 1][j - A[i]] + A[i]);
        }
    }
 
    // Traverse the suffix sum array
    for (i = N - 1; i >= 0; i--) {
 
        // If suffix[i] - dp[i][K] >= K
        if (suffix[i] - dp[i][K] >= K) {
 
            // Sum of lengths of the two
            // smallest subsets is obtained
            return N - i;
        }
    }
 
    // Return -1, if there doesn't
    // exist any subset of sum >= K
    return -1;
}
 
// Driver Code
 
    var arr = [7, 4, 5, 6, 8];
    var K = 13;
    var N = arr.length;
 
    document.write(MinimumLength(arr, N, K));
 
// This code is contributed by SURENDRA_GANGWAR.

输出：

4

时间复杂度： O(N * K)
辅助空间： O(N * K)

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