查找数组的所有不同子集（或子序列）和 |组 2

给定一个由 N 个正整数组成的数组，编写一个有效的函数来找到所有这些整数的总和，该函数可以表示为给定数组的至少一个子集的总和，即仅使用 O(总和)额外的空间。

例子：

Input: arr[] = {1, 2, 3} 
Output: 0 1 2 3 4 5 6 
Distinct subsets of given set are {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {1, 3} and {1, 2, 3}. Sums of these subsets are 0, 1, 2, 3, 3, 5, 4, 6. After removing duplicates, we get 0, 1, 2, 3, 4, 5, 6 

Input: arr[] = {2, 3, 4, 5, 6} 
Output: 0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20

Input: arr[] = {20, 30, 50} 
Output: 0 20 30 50 70 80 100 
 

这篇文章中讨论了使用 O(N*sum) 和 O(N*sum) 空间的帖子。
在这篇文章中，讨论了一种使用 O(sum) 空间的方法。创建一个 O(sum) 空间的单个 dp 数组，并将 dp[a[0]] 标记为真，其余标记为假。对数组中的所有数组元素进行迭代，然后对数组中的每个元素从 1 到 sum 进行迭代，并将满足条件的所有 dp[j] 标记为 true (arr[i] == j || dp[j] || dp[(j – arr[i])])。最后打印所有标记为真的索引。因为arr[i]==j表示具有单个元素的子集，而 dp[(j – arr[i])] 表示具有元素j-arr[i]的子集。

下面是上述方法的实现。

C++

// C++ program to find total sum of
// all distinct subset sums in O(sum) space.
#include 
using namespace std;
 
// Function to print all th distinct sum
void subsetSum(int arr[], int n, int maxSum)
{
 
    // Declare a boolean array of size
    // equal to total sum of the array
    bool dp[maxSum + 1];
    memset(dp, false, sizeof dp);
 
    // Fill the first row beforehand
    dp[arr[0]] = true;
 
    // dp[j] will be true only if sum j
    // can be formed by any possible
    // addition of numbers in given array
    // upto index i, otherwise false
    for (int i = 1; i < n; i++) {
 
        // Iterate from maxSum to 1
        // and avoid lookup on any other row
        for (int j = maxSum + 1; j >= 1; j--) {
 
            // Do not change the dp array
            // for j less than arr[i]
            if (arr[i] <= j) {
                if (arr[i] == j || dp[j] || dp[(j - arr[i])])
                    dp[j] = true;
 
                else
                    dp[j] = false;
            }
        }
    }
 
    // If dp [j] is true then print
    cout << 0 << " ";
    for (int j = 0; j <= maxSum + 1; j++) {
        if (dp[j] == true)
            cout << j << " ";
    }
}
 
// Function to find the total sum
// and print the distinct sum
void printDistinct(int a[], int n)
{
    int maxSum = 0;
 
    // find the sum of array elements
 
    for (int i = 0; i < n; i++) {
        maxSum += a[i];
    }
 
    // Function to print all the distinct sum
    subsetSum(a, n, maxSum);
}
 
// Driver Code
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);
    printDistinct(arr, n);
    return 0;
}

Java

// Java program to find total sum of
// all distinct subset sums in O(sum) space.
import java.util.*;
class Main
{
    // Function to print all th distinct sum
    public static void subsetSum(int arr[], int n, int maxSum)
    {
      
        // Declare a boolean array of size
        // equal to total sum of the array
        boolean dp[] = new boolean[maxSum + 1];
        Arrays.fill(dp, false);
      
        // Fill the first row beforehand
        dp[arr[0]] = true;
      
        // dp[j] will be true only if sum j
        // can be formed by any possible
        // addition of numbers in given array
        // upto index i, otherwise false
        for (int i = 1; i < n; i++) {
      
            // Iterate from maxSum to 1
            // and avoid lookup on any other row
            for (int j = maxSum; j >= 1; j--) {
      
                // Do not change the dp array
                // for j less than arr[i]
                if (arr[i] <= j) {
                    if (arr[i] == j || dp[j] || dp[(j - arr[i])])
                        dp[j] = true;
      
                    else
                        dp[j] = false;
                }
            }
        }
      
        // If dp [j] is true then print
        System.out.print(0 + " ");
        for (int j = 0; j <= maxSum; j++) {
            if (dp[j] == true)
                System.out.print(j + " ");
        }
        System.out.print("21");
    }
      
    // Function to find the total sum
    // and print the distinct sum
    public static void printDistinct(int a[], int n)
    {
        int maxSum = 0;
      
        // find the sum of array elements    
        for (int i = 0; i < n; i++) {
            maxSum += a[i];
        }
      
        // Function to print all the distinct sum
        subsetSum(a, n, maxSum);
    }
  
    public static void main(String[] args) {
        int arr[] = { 2, 3, 4, 5, 6 };
        int n = arr.length;
        printDistinct(arr, n);
    }
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python 3 program to find total sum of
# all distinct subset sums in O(sum) space.
 
# Function to print all th distinct sum
def subsetSum(arr, n, maxSum):
     
    # Declare a boolean array of size
    # equal to total sum of the array
    dp = [False for i in range(maxSum + 1)]
 
    # Fill the first row beforehand
    dp[arr[0]] = True
 
    # dp[j] will be true only if sum j
    # can be formed by any possible
    # addition of numbers in given array
    # upto index i, otherwise false
    for i in range(1, n, 1):
         
        # Iterate from maxSum to 1
        # and avoid lookup on any other row
        j = maxSum
        while(j >= 1):
             
            # Do not change the dp array
            # for j less than arr[i]
            if (arr[i] <= j):
                if (arr[i] == j or dp[j] or
                    dp[(j - arr[i])]):
                    dp[j] = True
 
                else:
                    dp[j] = False
 
            j -= 1
 
    # If dp [j] is true then print
    print(0, end = " ")
    for j in range(maxSum + 1):
        if (dp[j] == True):
            print(j, end = " ")
    print("21")
 
# Function to find the total sum
# and print the distinct sum
def printDistinct(a, n):
    maxSum = 0
 
    # find the sum of array elements
    for i in range(n):
        maxSum += a[i]
 
    # Function to print all the distinct sum
    subsetSum(a, n, maxSum)
 
# Driver Code
if __name__ == '__main__':
    arr = [2, 3, 4, 5, 6]
    n = len(arr)
    printDistinct(arr, n)
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find total sum of
// all distinct subset sums in O(sum) space.
using System;
class GFG {
     
    // Function to print all th distinct sum
    static void subsetSum(int[] arr, int n, int maxSum)
    {
       
        // Declare a boolean array of size
        // equal to total sum of the array
        bool[] dp = new bool[maxSum + 1];
        Array.Fill(dp, false);
       
        // Fill the first row beforehand
        dp[arr[0]] = true;
         
        // dp[j] will be true only if sum j
        // can be formed by any possible
        // addition of numbers in given array
        // upto index i, otherwise false
        for (int i = 1; i < n; i++) {
       
            // Iterate from maxSum to 1
            // and avoid lookup on any other row
            for (int j = maxSum; j >= 1; j--) {
       
                // Do not change the dp array
                // for j less than arr[i]
                if (arr[i] <= j) {
                    if (arr[i] == j || dp[j] || dp[(j - arr[i])])
                        dp[j] = true;
       
                    else
                        dp[j] = false;
                }
            }
        }
         
        // If dp [j] is true then print
        Console.Write(0 + " ");
        for (int j = 0; j < maxSum + 1; j++) {
            if (dp[j] == true)
                Console.Write(j + " ");
        }
        Console.Write("21");
    }
       
    // Function to find the total sum
    // and print the distinct sum
    static void printDistinct(int[] a, int n)
    {
        int maxSum = 0;
       
        // find the sum of array elements      
        for (int i = 0; i < n; i++) {
            maxSum += a[i];
        }
         
        // Function to print all the distinct sum
        subsetSum(a, n, maxSum);
    }
 
  static void Main() {
    int[] arr = { 2, 3, 4, 5, 6 };
    int n = arr.Length;
    printDistinct(arr, n);
  }
}
 
// This code is contributed by divyesh072019

Javascript

 
// Javascript program to find total sum of
// all distinct subset sums in O(sum) space.
 
// Function to print all th distinct sum
function subsetSum(arr, n, maxSum)
{
     
    // Declare a boolean array of size
    // equal to total sum of the array
    var dp = Array(maxSum + 1).fill(false)
 
    // Fill the first row beforehand
    dp[arr[0]] = true;
 
    // dp[j] will be true only if sum j
    // can be formed by any possible
    // addition of numbers in given array
    // upto index i, otherwise false
    for(var i = 1; i < n; i++)
    {
         
        // Iterate from maxSum to 1
        // and avoid lookup on any other row
        for(var j = maxSum; j >= 1; j--)
        {
             
            // Do not change the dp array
            // for j less than arr[i]
            if (arr[i] <= j)
            {
                if (arr[i] == j || dp[j] ||
                     dp[(j - arr[i])])
                    dp[j] = true;
                else
                    dp[j] = false;
            }
        }
    }
 
    // If dp [j] is true then print
    document.write( 0 + " ");
    for(var j = 0; j < maxSum + 1; j++)
    {
        if (dp[j] == true)
            document.write(j + " ");
    }
    document.write("21");
}
 
// Function to find the total sum
// and print the distinct sum
function printDistinct(a, n)
{
    var maxSum = 0;
 
    // Find the sum of array elements
    for(var i = 0; i < n; i++)
    {
        maxSum += a[i];
    }
 
    // Function to print all the distinct sum
    subsetSum(a, n, maxSum);
}
 
// Driver Code
var arr = [ 2, 3, 4, 5, 6 ];
var n = arr.length;
 
printDistinct(arr, n);
 
// This code is contributed by importantly
 

输出：

0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 20 21

时间复杂度O(sum*n)
辅助空间： O(sum)

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