Given a set of N integers. Find how many subsets of given array have sum between A and B(inclusive).
Constraints:
1 ≤ N ≤ 34,
-2 * 107 ≤ arri ≤ 2 * 107
-5 * 108 ≤ A, B ≤ 5 * 108
Examples:
Input : S[] = { 1, -2, 3 }, A = -1, B = 2
Output : 5
Explanation:
1) 0 = 0 (the empty subset)
2) {1} = 1
3) {1, -2} = 1 + (-2) = -1
4) {-2, 3} = (-2) + 3 = 1
5) {1, -2, 3} = 1 + (-2) + 3 = 2
Method 1 (Brute Force): We can generate all subsets of the given numbers i.e. Power Set and find the number of subsets that would give a sum between A and B. But this will have 234 operations atmost, which is not very efficient. Hence, below is an efficient approach to solve this problem.
Method 2 (Meet In The Middle): This basically reduces time complexity from O(2N) to O(2N/2)
We divide the set into two sets [0…N/2] and [(N/2 + 1)…(N-1)] and generate all subsets sums individually for the two sets which will be 2 * 217 operations. Now, what we can do is to find the combinations of these sets that would give the desired sum. This again can be done in an efficient way, sort one of the summed up set and binary search the values that will yield the sum for the particular value of the other set. Sort the second set and for each element in the first set, search for the lower bound of A – S2[i] (let say ‘low’) and upper bound of B – S2[i]
(let say ‘high’). Subtract (high – low) to get the desired answer.
For e.g S = { 1, 2, -1, 0 }, A = 1, B = -1.
After dividing S into two sets, S1 = { 1, 2 } and S2 = { -1, 0 }.
Power set of S1 = { {0}, {1}, {2}, {1, 2} } and Power set of S2 = { {0}, {-1}, {0}, {-1, 0} }
Subset Sum of S1 = { 0, 1, 2, 3 } and Subset Sum of S2 = { 0, -1, 0, -1 }
Now sort the S2 { -1, -1, 0, 0 } and for every value in S1, we binary search values that would yield the desired sum. For 0 we search for (-1) – 0 = -1 for lower bound and 1 – 0 = 1 for upper bound in S2, for 1 we search for (-1) – 1 = -2 and 1 – 1 = 0 in S2 and so on.
C++
// C++ program to find the Number of Subsets that // have sum between A and B#include <bits/stdc++.h>using namespace std; /* Function to Generate all subsets of a set start --> Starting Index of the Set for the first/second half Set setSize --> Number of element in half Set S --> Original Complete Set res --> Store the subsets sums */void generateSubsets(int start, int setSize, int S[], vector<int>& res){ // setSize of power set of a set with setSize // N is (2^n - 1) unsigned int pow_setSize = pow(2, setSize); // Store the sum of particular subset of set int sum; // Run from counter 000..0 to 111..1 for (int counter = 0; counter < pow_setSize; counter++) { // set the sum initially to zero sum = 0; for (int j = 0; j < setSize; j++) { // Check if jth bit in the counter is set // If set then pront jth element from set if (counter & (1 << j)) sum += S[j + start]; } // Store the sum in a vector res.push_back(sum); }} int numberOfSubsets(int S[], int N, int A, int B){ // Vectors to store the subsets sums // of two half sets individually vector<int> S1, S2; // Generate subset sums for the first half set generateSubsets(0, N / 2, S, S1); // Generate subset sums for the second half set if (N % 2 != 0) generateSubsets(N / 2, N / 2 + 1, S, S2); else generateSubsets(N / 2, N / 2, S, S2); // Sort the second half set sort(S2.begin(), S2.end()); // Vector Iterator for S1 and S2; vector<int>::iterator low, high; // number of required subsets with desired Sum int ans = 0; for (int i = 0; i < S1.size(); i++) { // search for lower bound low = lower_bound(S2.begin(), S2.end(), A - S1[i]); // search for upper bound high = upper_bound(S2.begin(), S2.end(), B - S1[i]); // Add up to get the desired answer ans += (high - low); } return ans;} // Driver Program to test above functionsint main(){ int S[] = { 1, -2, 3 }; int N = sizeof(S) / sizeof(S[0]); int A = -1, B = 2; // Find the number of subsets with desired Sum cout << numberOfSubsets(S, N, A, B) << endl; return 0;} |
Python3
# Python3 program to find the number of# subsets that have sum between A and B # Module for Bisection algorithmsimport bisect '''Function to Generate all subsets of a set start --> Starting Index of the Set for the first/second half Set setSize --> Number of element in half Set S --> Original Complete Set res --> Store the subsets sums '''def generateSubsets(start, setSize, S, res): # setSize of power set of a set with setSize # N is (2^n - 1) pow_setSize = pow(2, setSize) # Store the sum of particular subset of set add = 0 # Run from counter 000..0 to 111..1 for counter in range(pow_setSize): # set the sum initially to zero add = 0 for j in range(setSize): # Check if jth bit in the counter is set # If set then pront jth element from set if counter & (1 << j): add += S[j + start] # Store the sum in a vector res.append(add) def numberOfSubsets(S, N, A, B): # Vectors to store the subsets sums # of two half sets individually S1 = [] S2 = [] # Generate subset sums for the first half set generateSubsets(0, N // 2, S, S1) # Generate subset sums for the second half set if (N % 2 != 0): generateSubsets(N // 2, N // 2 + 1, S, S2) else: generateSubsets(N // 2, N // 2, S, S2) # Sort the second half set S2.sort() # Number of required subsets # with desired Sum ans = 0 for i in range(len(S1)): # Search for lower bound low = bisect.bisect_left(S2, A - S1[i]) # Search for upper bound high = bisect.bisect_right(S2, B - S1[i]) # Add up to get the desired answer ans += (high - low) return ans # Driver codeif __name__=="__main__": S = [ 1, -2, 3 ] N = len(S) A = -1 B = 2 # Find the number of subsets # with desired Sum print(numberOfSubsets(S, N, A, B)) # This code is contributed by vinaylingam |
5
Time Complexity: O(2 * 2N/2), where N is the size of set.
