Given a set of n integers where n <= 40. Each of them is at most 1012, determine the maximum sum subset having sum less than or equal S where S <= 1018.
Example:
Input : set[] = {45, 34, 4, 12, 5, 2} and S = 42
Output : 41
Maximum possible subset sum is 41 which can be
obtained by summing 34, 5 and 2.
Input : Set[] = {3, 34, 4, 12, 5, 2} and S = 10
Output : 10
Maximum possible subset sum is 10 which can be
obtained by summing 2, 3 and 5.A Brute Force approach to solve this problem would be find all possible subset sums of N integers and check if it is less than or equal S and keep track of such a subset with maximum sum. The time complexity using this approach would be O(2n) and n is at most 40. 240 will be quite large and hence we need to find more optimal approach.
Meet in the middle is a search technique which is used when the input is small but not as small that brute force can be used. Like divide and conquer it splits the problem into two, solves them individually and then merge them. But we can’t apply meet in the middle like divide and conquer because we don’t have the same structure as the original problem.
- Split the set of integers into 2 subsets say A and B. A having first n/2 integers and B having rest.
- Find all possible subset sums of integers in set A and store in an array X. Similarly calculate all possible subset sums of integers in set B and store in array Y. Hence, Size of each of the array X and Y will be at most 2n/2.
- Now merge these 2 subproblems. Find combinations from array X and Y such that their sum is less than or equal to S.
- One way to do that is simply iterate over all elements of array Y for each element of array X to check the existence of such a combination. This will take O( (2n/2)2) which is equivalent to O(2n).
- To make it less complex, first sort array Y and then iterate over each element of X and for each element x in X use binary search to find maximum element y in Y such that x + y <= S.
- Binary search here helps in reducing complexity from 2nto 2n/2log(2n/2)which is equivalent to 2n/2n.
- Thus our final running time is O(2n/2n).
C++ // C++ program to demonstrate working of Meet in the // Middle algorithm for maximum subset sum problem. #include <bits/stdc++.h> using namespace std; typedef long long int ll; ll X[2000005],Y[2000005]; // Find all possible sum of elements of a[] and store // in x[] void calcsubarray(ll a[], ll x[], int n, int c) { for (int i=0; i<(1<<n); i++) { ll s = 0; for (int j=0; j<n; j++) if (i & (1<<j)) s += a[j+c]; x[i] = s; } } // Returns the maximum possible sum less or equal to S ll solveSubsetSum(ll a[], int n, ll S) { // Compute all subset sums of first and second // halves calcsubarray(a, X, n/2, 0); calcsubarray(a, Y, n-n/2, n/2); int size_X = 1<<(n/2); int size_Y = 1<<(n-n/2); // Sort Y (we need to do doing binary search in it) sort(Y, Y+size_Y); // To keep track of the maximum sum of a subset // such that the maximum sum is less than S ll max = 0; // Traverse all elements of X and do Binary Search // for a pair in Y with maximum sum less than S. for (int i=0; i<size_X; i++) { if (X[i] <= S) { // lower_bound() returns the first address // which has value greater than or equal to // S-X[i]. int p = lower_bound(Y, Y+size_Y, S-X[i]) - Y; // If S-X[i] was not in array Y then decrease // p by 1 if (p == size_Y || Y[p] != (S-X[i])) p--; if ((Y[p]+X[i]) > max) max = Y[p]+X[i]; } } return max; } // Driver code int main() { ll a[] = {3, 34, 4, 12, 5, 2}; int n=sizeof(a)/sizeof(a[0]); ll S = 10; printf("Largest value smaller than or equal to given " "sum is %lld\n", solveSubsetSum(a,n,S)); return 0; } import java.util.*; class GFG { static long X[] = new long[2000005]; static long Y[] = new long[2000005]; // Find all possible sum of elements of a[] // and store in x[] static void calcsubarray(long a[], long x[], int n, int c) { for (int i = 0; i < (1 << n); i++) { long s = 0; for (int j = 0; j < n; j++) { if ((i & (1 << j)) != 0) { // Check inclusion in subset s += a[j + c]; } } x[i] = s; } } // Returns the maximum possible sum // less or equal to S static long solveSubsetSum(long a[], int n, long S) { // Compute all subset sums of first and second // halves calcsubarray(a, X, n / 2, 0); calcsubarray(a, Y, n - n / 2, n / 2); int size_X = 1 << (n / 2); int size_Y = 1 << (n - n / 2); // Sort Y (we need to do doing // binary search in it) Arrays.sort(Y, 0, size_Y); // To keep track of the maximum sum // of a subset such that the maximum // sum is less than S long max = 0; // Traverse all elements of X and do // Binary Search for a pair in Y with // maximum sum less than S. for (int i = 0; i < size_X; i++) { if (X[i] <= S) { // lower_bound() returns the first address // which has value greater than or equal to // S-X[i]. int p = lower_bound(Y, S - X[i], size_Y); // If S-X[i] was not in array Y then // decrease p by 1 if (p == size_Y || Y[p] != (S - X[i])) p--; if ((Y[p] + X[i]) > max) max = Y[p] + X[i]; } } return max; } static int lower_bound(long a[], long x, int size_Y) { // x is the target value or key int l = -1, r = size_Y; while (l + 1 < r) { int m = (l + r) >>> 1; if (a[m] >= x) r = m; else l = m; } return r; } // Driver code public static void main(String[] args) { long a[] = { 3, 34, 4, 12, 5, 2 }; int n = a.length; long S = 10; System.out.println("Largest value smaller " + "than or equal to given " + "sum is " + solveSubsetSum(a, n, S)); } } // This code is contributed by jyoti369 # Python program to demonstrate working of Meet in the # Middle algorithm for maximum subset sum problem. from typing import List import bisect X = [0] * 2000005 Y = [0] * 2000005 # Find all possible sum of elements of a[] and store # in x[] def calcsubarray(a: List[int], x: List[int], n: int, c: int) -> None: for i in range((1 << n)): s = 0 for j in range(n): if (i & (1 << j)): s += a[j + c] x[i] = s # Returns the maximum possible sum less or equal to S def solveSubsetSum(a: List[int], n: int, S: int) -> int: global Y # Compute all subset sums of first and second # halves calcsubarray(a, X, n // 2, 0) calcsubarray(a, Y, n - n // 2, n // 2) size_X = 1 << (n // 2) size_Y = 1 << (n - n // 2) # Sort Y (we need to do doing binary search in it) YY = Y[:size_Y] YY.sort() Y = YY # To keep track of the maximum sum of a subset # such that the maximum sum is less than S maxx = 0 # Traverse all elements of X and do Binary Search # for a pair in Y with maximum sum less than S. for i in range(size_X): if (X[i] <= S): # lower_bound() returns the first address # which has value greater than or equal to # S-X[i]. p = bisect.bisect_left(Y, S - X[i]) # If S-X[i] was not in array Y then decrease # p by 1 if (p == size_Y or (p < size_Y and Y[p] != (S - X[i]))): p -= 1 if ((Y[p] + X[i]) > maxx): maxx = Y[p] + X[i] return maxx # Driver code if __name__ == "__main__": a = [3, 34, 4, 12, 5, 2] n = len(a) S = 10 print("Largest value smaller than or equal to given sum is {}".format( solveSubsetSum(a, n, S))) # This code is contributed by sanjeev2552 using System; public class GFG { static long[] X = new long[2000005]; static long[] Y = new long[2000005]; // Find all possible sum of elements of a[] // and store in x[] static void calcsubarray(long[] a, long[] x, int n, int c) { for (int i = 0; i < (1 << n); i++) { long s = 0; for (int j = 0; j < n; j++) { if ((i & (1 << j)) != 0) // Check inclusion in subset { s += a[j + c]; } } x[i] = s; } } // Returns the maximum possible sum // less or equal to S static long solveSubsetSum(long[] a, int n, long S) { // Compute all subset sums of first and second // halves calcsubarray(a, X, n / 2, 0); calcsubarray(a, Y, n - n / 2, n / 2); int size_X = 1 << (n / 2); int size_Y = 1 << (n - n / 2); // Sort Y (we need to do doing // binary search in it) Array.Sort(Y, 0, size_Y); // To keep track of the maximum sum // of a subset such that the maximum // sum is less than S long max = 0; // Traverse all elements of X and do // Binary Search for a pair in Y with // maximum sum less than S. for (int i = 0; i < size_X; i++) { if (X[i] <= S) { // lower_bound() returns the first address // which has value greater than or equal to // S-X[i]. int p = lower_bound(Y, S - X[i], size_Y); // If S-X[i] was not in array Y then // decrease p by 1 if (p == size_Y || Y[p] != (S - X[i])) p--; if ((Y[p] + X[i]) > max) max = Y[p] + X[i]; } } return max; } static int lower_bound(long[] a, long x, int size_Y) { // x is the target value or key int l = -1, r = size_Y; while (l + 1 < r) { int m = (l + r) / 2; if (a[m] >= x) r = m; else l = m; } return r; } // Driver code public static void Main(string[] args) { long[] a = { 3, 34, 4, 12, 5, 2 }; int n = a.Length; long S = 10; Console.WriteLine("Largest value smaller " + "than or equal to given " + "sum is " + solveSubsetSum(a, n, S)); } } function calcsubarray(a, x, n, c) { for (let i = 0; i < (1 << n); i++) { let s = 0; for (let j = 0; j < n; j++) { if (i & (1 << j)) { // Check if j-th bit is set in i s += a[j + c]; } } x[i] = s; } } function solveSubsetSum(a, n, S) { let X = new Array(1 << (n / 2)).fill(0); let Y = new Array(1 << (n - (n / 2))).fill(0); // Compute all subset sums of first and second halves calcsubarray(a, X, n / 2, 0); calcsubarray(a, Y, n - (n / 2), n / 2); // Sort Y (we need to do binary search in it) Y.sort((a, b) => a - b); let max = 0; // Traverse all elements of X and do // Binary Search for a pair in Y with // maximum sum less than S. for (let i = 0; i < X.length; i++) { if (X[i] <= S) { let p = lower_bound(Y, S - X[i]); if (p > 0 && Y[p] !== (S - X[i])) p--; if ((Y[p] + X[i]) > max) max = Y[p] + X[i]; } } return max; } function lower_bound(a, x) { let l = 0, r = a.length; while (l < r) { let m = Math.floor((l + r) / 2); if (a[m] >= x) r = m; else l = m + 1; } return l; } // Driver code let a = [3, 34, 4, 12, 5, 2]; let n = a.length; let S = 10; console.log("Largest value smaller than or equal to given sum is " + solveSubsetSum(a, n, S)); Output:
Largest value smaller than or equal to given sum is 10
Reference:
Time Complexity: O(2 ^{n/2} * log(2 ^{n/2}) )
Auxiliary Space: O(2 ^ {n/2} )