Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.



Subset Sum Problem | DP-25
Given a set of non-negative integers, and a value sum, determine if there is a subset of the given set with sum equal to given sum.
Example:

Input:  set[] = {3, 34, 4, 12, 5, 2}, sum = 9
Output:  True  //There is a subset (4, 5) with sum 9.

// A recursive solution for subset sum problem
#include <stdio.h>
  
// Returns true if there is a subset of set[] with sun equal to given sum
bool isSubsetSum(int set[], int n, int sum)
{
   // Base Cases
   if (sum == 0)
     return true;
   if (n == 0 && sum != 0)
     return false;
  
   // If last element is greater than sum, then ignore it
   if (set[n-1] > sum)
     return isSubsetSum(set, n-1, sum);
  
   /* else, check if sum can be obtained by any of the following
      (a) including the last element
      (b) excluding the last element   */
   return isSubsetSum(set, n-1, sum) || 
                        isSubsetSum(set, n-1, sum-set[n-1]);
}
  
// Driver program to test above function
int main()
{
  int set[] = {3, 34, 4, 12, 5, 2};
  int sum = 9;
  int n = sizeof(set)/sizeof(set[0]);
  if (isSubsetSum(set, n, sum) == true)
     printf("Found a subset with given sum");
  else
     printf("No subset with given sum");
  return 0;
}

Output:
Found a subset with given sum
// A recursive solution for subset sum
// problem
class GFG {
      
    // Returns true if there is a subset
    // of set[] with sum equal to given sum
    static boolean isSubsetSum(int set[],
                            int n, int sum)
    {
        // Base Cases
        if (sum == 0)
            return true;
        if (n == 0 && sum != 0)
            return false;
          
        // If last element is greater than 
        // sum, then ignore it
        if (set[n-1] > sum)
            return isSubsetSum(set, n-1, sum);
          
        /* else, check if sum can be obtained 
        by any of the following
            (a) including the last element
            (b) excluding the last element */
        return isSubsetSum(set, n-1, sum) || 
            isSubsetSum(set, n-1, sum-set[n-1]);
    }
      
    /* Driver program to test above function */
    public static void main (String args[])
    {
        int set[] = {3, 34, 4, 12, 5, 2};
        int sum = 9;
        int n = set.length;
        if (isSubsetSum(set, n, sum) == true)
            System.out.println("Found a subset"
                          + " with given sum");
        else
            System.out.println("No subset with"
                               + " given sum");
    }
}
  
/* This code is contributed by Rajat Mishra */
Output:
Found a subset with given sum

# A recursive solution for subset sum
# problem
  
# Returns true if there is a subset 
# of set[] with sun equal to given sum
def isSubsetSum(set,n, sum) :
    
    # Base Cases
    if (sum == 0) :
        return True
    if (n == 0 and sum != 0) :
        return False
   
    # If last element is greater than
    # sum, then ignore it
    if (set[n - 1] > sum) :
        return isSubsetSum(set, n - 1, sum);
   
    # else, check if sum can be obtained
    # by any of the following
    # (a) including the last element
    # (b) excluding the last element   
    return isSubsetSum(set, n-1, sum) or isSubsetSum(set, n-1, sum-set[n-1])
      
      
# Driver program to test above function
set = [3, 34, 4, 12, 5, 2]
sum = 9
n = len(set)
if (isSubsetSum(set, n, sum) == True) :
    print("Found a subset with given sum")
else :
    print("No subset with given sum")
      














































































Comments

Post a Comment

Popular posts from this blog

Kth most frequent Character in a given String

  Given a string   str   and an integer  K , the task is to find the K-th most frequent character in the string. If there are multiple characters that can account as K-th most frequent character then, print any one of them. Examples: Input:  str = “GeeksforGeeks”, K = 3 Output:  f Explanation: K = 3, here ‘e’ appears 4 times & ‘g’, ‘k’, ‘s’ appears 2 times & ‘o’, ‘f’, ‘r’ appears 1 time. Any output from ‘o’ (or) ‘f’ (or) ‘r’ will be correct. Input:  str = “trichotillomania”, K = 2 Output:  l The idea is to Use the Characters as key in  Hashmap  and store their  occurrences in the string . Sort the Hashmap  and find the K-th character. Below is the implementation of the above approach. // C++ program to find kth most frequent // character in a string #include <bits/stdc++.h> using namespace std;     // Used for sorting by frequency. bool sortByVal( const pair< char , int >& a,                 const pair< char , int >& b) {      return a.second >
Read more

[16 Feb 2020] Given an array where every element occurs three times, except one element which occurs only once.

Find the element that appears once Given an array where every element occurs three times, except one element which occurs only once. Find the element that occurs once. Expected time complexity is O(n) and O(1) extra space. Examples : Input: arr[] = {12, 1, 12, 3, 12, 1, 1, 2, 3, 3} Output: 2 In the given array all element appear three times except 2 which appears once. Input: arr[] = {10, 20, 10, 30, 10, 30, 30} Output: 20 In the given array all element appear three times except 20 which appears once. Following is another O(n) time complexity and O(1) extra space method suggested by aj. We can sum the bits in same positions for all the numbers and take modulo with 3. The bits for which sum is not multiple of 3, are the bits of number with single occurrence. Let us consider the example array {5, 5, 5, 8}. The 101, 101, 101, 1000 Sum of first bits%3 = (1 + 1 + 1 + 0)%3 = 0; Sum of second bits%3 = (0 + 0 + 0 + 0)%0 = 0; Sum of third bits%3 = (1 + 1 + 1 + 0)%3 = 0; Sum of
Read more

[17 Feb 2020] Given an array of integers, find the nearest smaller number for every element such that the smaller element is on left side.

Find the nearest smaller numbers on left side in an array Given an array of integers, find the nearest smaller number for every element such that the smaller element is on left side. Examples: Input:  arr[] = {1, 6, 4, 10, 2, 5} Output:         {_, 1, 1,  4, 1, 2} First element ('1') has no element on left side. For 6, there is only one smaller element on left side '1'. For 10, there are three smaller elements on left side (1, 6 and 4), nearest among the three elements is 4. Input: arr[] = {1, 3, 0, 2, 5} Output:        {_, 1, _, 0, 2} Expected time complexity is O(n). // C++ implementation of efficient algorithm to find // smaller element on left side #include <iostream> #include <stack> using namespace std;   // Prints smaller elements on left side of every element void printPrevSmaller(int arr[], int n) {     // Create an empty stack     stack<int> S;       // Traverse all array elements     for (int i=0; i<n; i++)  
Read more