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📜  给定数组中两个最小子集的长度总和至少为K

📅  最后修改于: 2021-05-17 23:49:29             🧑  作者: Mango

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唯一子集的长度之和,其元素之和至少为K。

例子:

方法:可以根据以下观察结果解决给定问题:

  • 对数组进行排序可以解决以下问题:选择索引[i,N]的范围内总和至少为K的子数组,然后检查索引[i,N]范围内的其余数组元素的总和是否为是否为K。
  • 为了实现上述想法,使用二维数组dp [] [] ,使得dp [i] [j]存储索引值至少[i,N]的子集的最小和。 j。那么过渡状态类似于0/1背包,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](dp [i +1] [j – arr [i]] + arr [i])的最小值。

请按照以下步骤解决问题:

  • 以升序对数组进行排序。
  • 初始化一个数组,例如suffix [],并在其中存储数组arr []的后缀和。
  • 初始化一个二维数组,例如dp [] [],以使dp [i] [j]存储索引值至少为j的索引[i,N]范围内子集的最小和。
  • 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:
    • 以相反的顺序遍历索引[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]]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](dp [i +1] [j – arr [i]] + arr [i])的最小值,以存储所有值的总和至少为j
  • 现在,以相反的顺序遍历数组suffix [] ,如果(suffix [i] – dp [i] [K])的至少为K ,则打印(N – i)作为形成了两个最小的子集并打破了循环。
  • 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];
 
    // Intialize 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];
 
    // Intialize 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)
 
    # Intialize 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];
 
    // Intialize 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


输出:
4

时间复杂度: O(N * K)
辅助空间: O(N * K)