Perfect Sum Problem (Print all subsets with given sum)

Given an array of integers and a sum, the task is to print all subsets of given array with sum equal to given sum.

Examples:

Input : arr[] = {2, 3, 5, 6, 8, 10}
        sum = 10
Output : 5 2 3
         2 8
         10

Input : arr[] = {1, 2, 3, 4, 5}
        sum = 10
Output : 4 3 2 1 
         5 3 2 
         5 4 1 

Asked in Tesco

This problem is mainly an extension of Subset Sum Problem. Here we not only need to find if there is a subset with given sum, but also need to print all subsets with given sum.

Like previous post, we build a 2D array dp[][] such that dp[i][j] stores true if sum j is possible with array elements from 0 to i.
After filling dp[][], we recursively traverse it from dp[n-1][sum]. For cell being traversed, we store path before reaching it and consider two possibilities for the element.
1) Element is included in current path.
2) Element is not included in current path.

Whenever sum becomes 0, we stop the recursive calls and print current path.

Below is C++ implementation of above idea.

// C++ program to count all subsets with
// given sum.
#include <bits/stdc++.h>
using namespace std;

// dp[i][j] is going to store true if sum j is
// possible with array elements from 0 to i.
bool** dp;

void display(const vector<int>& v)
{
    for (int i = 0; i < v.size(); ++i)
        printf("%d ", v[i]);
    printf("n");
}

// A recursive function to print all subsets with the
// help of dp[][]. Vector p[] stores current subset.
void printSubsetsRec(int arr[], int i, int sum, vector<int>& p)
{
    // If we reached end and sum is non-zero. We print
    // p[] only if arr[0] is equal to sun OR dp[0][sum]
    // is true.
    if (i == 0 && sum != 0 && dp[0][sum])
    {
        p.push_back(arr[i]);
        display(p);
        return;
    }

    // If sum becomes 0
    if (i == 0 && sum == 0)
    {
        display(p);
        return;
    }

    // If given sum can be achieved after ignoring
    // current element.
    if (dp[i-1][sum])
    {
        // Create a new vector to store path
        vector<int> b = p;
        printSubsetsRec(arr, i-1, sum, b);
    }

    // If given sum can be achieved after considering
    // current element.
    if (sum >= arr[i] && dp[i-1][sum-arr[i]])
    {
        p.push_back(arr[i]);
        printSubsetsRec(arr, i-1, sum-arr[i], p);
    }
}

// Prints all subsets of arr[0..n-1] with sum 0.
void printAllSubsets(int arr[], int n, int sum)
{
    if (n == 0 || sum < 0)
       return;

    // Sum 0 can always be achieved with 0 elements
    dp = new bool*[n];
    for (int i=0; i<n; ++i)
    {
        dp[i] = new bool[sum + 1];
        dp[i][0] = true;
    }

    // Sum arr[0] can be achieved with single element
    if (arr[0] <= sum)
       dp[0][arr[0]] = true;

    // Fill rest of the entries in dp[][]
    for (int i = 1; i < n; ++i)
        for (int j = 0; j < sum + 1; ++j)
            dp[i][j] = (arr[i] <= j) ? dp[i-1][j] ||
                                       dp[i-1][j-arr[i]]
                                     : dp[i - 1][j];
    if (dp[n-1][sum] == false)
    {
        printf("There are no subsets with sum %dn", sum);
        return;
    }

    // Now recursively traverse dp[][] to find all
    // paths from dp[n-1][sum]
    vector<int> p;
    printSubsetsRec(arr, n-1, sum, p);
}

// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5};
    int n = sizeof(arr)/sizeof(arr[0]);
    int sum = 10;
    printAllSubsets(arr, n, sum);
    return 0;
}

Output:

4 3 2 1
5 3 2
5 4 1

This article is contributed by Jas Kaur. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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