问题
Given a set of numbers: {1, 3, 2, 5, 4, 9}, find the number of subsets that sum to a particular value (say, 9 for this example).
This is similar to subset sum problem with the slight difference that instead of checking if the set has a subset that sums to 9, we have to find the number of such subsets. I am following the solution for subset sum problem here. But and I am wondering how I can modify it to return the count of subsets.
回答1:
def total_subsets_matching_sum(numbers, sum):
array = [1] + [0] * (sum)
for current_number in numbers:
for num in xrange(sum - current_number, -1, -1):
if array[num]:
array[num + current_number] += array[num]
return array[sum]
assert(total_subsets_matching_sum(range(1, 10), 9) == 8)
assert(total_subsets_matching_sum({1, 3, 2, 5, 4, 9}, 9) == 4)
Explanation
This is one of the classic problems. The idea is to find the number of possible sums with the current number. And its true that, there is exactly one way to bring sum to 0. At the beginning, we have only one number. We start from our target (variable Maximum in the solution) and subtract that number. If it is possible to get a sum of that number (array element corresponding to that number is not zero) then add it to the array element corresponding to the current number. The program would be easier to understand this way
for current_number in numbers:
for num in xrange(sum, current_number - 1, -1):
if array[num - current_number]:
array[num] += array[num - current_number]
When the number is 1, there is only one way in which you can come up with the sum of 1 (1-1 becomes 0 and the element corresponding to 0 is 1). So the array would be like this (remember element zero will have 1)
[1, 1, 0, 0, 0, 0, 0, 0, 0, 0]
Now, the second number is 2. We start subtracting 2 from 9 and its not valid (since array element of 7 is zero we skip that) we keep doing this till 3. When its 3, 3 - 2 is 1 and the array element corresponding to 1 is 1 and we add it to the array element of 3. and when its 2, 2 - 2 becomes 0 and we the value corresponding to 0 to array element of 2. After this iteration the array looks like this
[1, 1, 1, 1, 0, 0, 0, 0, 0, 0]
We keep doing this till we process all the numbers and the array after every iteration looks like this
[1, 1, 0, 0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 2, 1, 1, 1, 0, 0, 0]
[1, 1, 1, 2, 2, 2, 2, 2, 1, 1]
[1, 1, 1, 2, 2, 3, 3, 3, 3, 3]
[1, 1, 1, 2, 2, 3, 4, 4, 4, 5]
[1, 1, 1, 2, 2, 3, 4, 5, 5, 6]
[1, 1, 1, 2, 2, 3, 4, 5, 6, 7]
[1, 1, 1, 2, 2, 3, 4, 5, 6, 8]
After the last iteration, we would have considered all the numbers and the number of ways to get the target would be the array element corresponding to the target value. In our case, Array[9] after the last iteration is 8.
回答2:
You may use Dynamic Programmining. Algo complexity is O(Sum * N) and use O(Sum) memory.
Here's my implementation in C#:
private static int GetmNumberOfSubsets(int[] numbers, int sum)
{
int[] dp = new int[sum + 1];
dp[0] = 1;
int currentSum =0;
for (int i = 0; i < numbers.Length; i++)
{
currentSum += numbers[i];
for (int j = Math.Min(sum, currentSum); j >= numbers[i]; j--)
dp[j] += dp[j - numbers[i]];
}
return dp[sum];
}
Notes: As number of subsets may have value 2^N it easy may overflows int type.
Algo works only for positive numbers.
回答3:
Here is a Java Solution:
This is a classical Back tracking problem for finding all possible subsets of the integer array or set that is the input and then filtering those which sum to e given target
import java.util.HashSet;
import java.util.StringTokenizer;
/**
* Created by anirudh on 12/5/15.
*/
public class findSubsetsThatSumToATarget {
/**
* The collection for storing the unique sets that sum to a target.
*/
private static HashSet<String> allSubsets = new HashSet<>();
/**
* The String token
*/
private static final String token = " ";
/**
* The method for finding the subsets that sum to a target.
*
* @param input The input array to be processed for subset with particular sum
* @param target The target sum we are looking for
* @param ramp The Temporary String to be beefed up during recursive iterations(By default value an empty String)
* @param index The index used to traverse the array during recursive calls
*/
public static void findTargetSumSubsets(int[] input, int target, String ramp, int index) {
if(index > (input.length - 1)) {
if(getSum(ramp) == target) {
allSubsets.add(ramp);
}
return;
}
//First recursive call going ahead selecting the int at the currenct index value
findTargetSumSubsets(input, target, ramp + input[index] + token, index + 1);
//Second recursive call going ahead WITHOUT selecting the int at the currenct index value
findTargetSumSubsets(input, target, ramp, index + 1);
}
/**
* A helper Method for calculating the sum from a string of integers
*
* @param intString the string subset
* @return the sum of the string subset
*/
private static int getSum(String intString) {
int sum = 0;
StringTokenizer sTokens = new StringTokenizer(intString, token);
while (sTokens.hasMoreElements()) {
sum += Integer.parseInt((String) sTokens.nextElement());
}
return sum;
}
/**
* Cracking it down here : )
*
* @param args command line arguments.
*/
public static void main(String[] args) {
int [] n = {24, 1, 15, 3, 4, 15, 3};
int counter = 1;
FindSubsetsThatSumToATarget.findTargetSumSubsets(n, 25, "", 0);
for (String str: allSubsets) {
System.out.println(counter + ") " + str);
counter++;
}
}
}
It gives space separated values of the subsets that sum to a target.
Would print out commma separated values for the subsets that sum to 25 in
{24, 1, 15, 3, 4, 15, 3}
1) 24 1
2) 3 4 15 3
3) 15 3 4 3
回答4:
The same site geeksforgeeks also discusses the solution to output all subsets that sum to a particular value: http://www.geeksforgeeks.org/backttracking-set-4-subset-sum/
In your case, instead of the output sets, you just need to count them. Be sure to check the optimized version in the same page, as it is an NP-complete problem.
This question also has been asked and answered before in stackoverflow without mentioning that it's a subset-sum problem: Finding all possible combinations of numbers to reach a given sum
回答5:
This my program in ruby . It will return arrays, each holding the subsequences summing to the provided target value.
array = [1, 3, 4, 2, 7, 8, 9]
0..array.size.times.each do |i|
@ary.combination(i).to_a.each { |a| print a if a.inject(:+) == 9}
end
回答6:
I have solved this by java. This solution is quite simple.
import java.util.*;
public class Recursion {
static void sum(int[] arr, int i, int sum, int target, String s)
{
for(int j = i+1; j<arr.length; j++){
if(sum+arr[j] == target){
System.out.println(s+" "+String.valueOf(arr[j]));
}else{
sum(arr, j, sum+arr[j], target, s+" "+String.valueOf(arr[j]));
}
}
}
public static void main(String[] args)
{
int[] numbers = {6,3,8,10,1};
for(int i =0; i<numbers.length; i++){
sum(numbers, i, numbers[i], 18, String.valueOf(numbers[i]));
}
}
}
回答7:
Here is an efficient solution when there is a large set of inputs (i.e. 25 to 30)
I made efficiency gains in two ways:
- Utilized a simple rolling wheel concept for binary counting through all possible iterations, rather than using mathematically expensive base conversion language features. This "wheel" is akin to an old mechanical counter or odometer. It recursively rolls forward as many positions as necessary until we surpass the number of binary digits we have (e.g., the count of numbers in our set).
- The main gain is from not summing the entire candidate set each time. Instead, it maintains a running sum and each time it "rolls the wheel", it adjusts the running sum only for the pieces that changed from the last candidate set it tested. This saves a lot of computation since most "wheel rolls" only change a number or two.
This solution works with negative numbers, decimal amounts, and repeated input values. Due to the wonky way that floating point decimal math works in most languages, you will want to keep your input set to only a few decimal places or you may have some unpredictable behavior.
On my old 2012-era desktop computer the given code processes 25 input values in about 0.8 seconds in javascript/node.js, and 3.4 seconds in C#.
javascript
let numbers = [-0.47, -0.35, -0.19, 0.23, 0.36, 0.47, 0.51, 0.59, 0.63, 0.79, 0.85,
0.91, 0.99, 1.02, 1.17, 1.25, 1.39, 1.44, 1.59, 1.60, 1.79, 1.88, 1.99, 2.14, 2.31];
let target = 24.16;
displaySubsetsThatSumTo(target, numbers);
function displaySubsetsThatSumTo(target, numbers)
{
let wheel = [0];
let resultsCount = 0;
let sum = 0;
const start = new Date();
do {
sum = incrementWheel(0, sum, numbers, wheel);
//Use subtraction comparison due to javascript float imprecision
if (sum != null && Math.abs(target - sum) < 0.000001) {
//Found a subset. Display the result.
console.log(numbers.filter(function(num, index) {
return wheel[index] === 1;
}).join(' + ') + ' = ' + target);
resultsCount++;
}
} while (sum != null);
const end = new Date();
console.log('--------------------------');
console.log(`Processed ${numbers.length} numbers in ${(end - start) / 1000} seconds (${resultsCount} results)`);
}
function incrementWheel(position, sum, numbers, wheel) {
if (position === numbers.length || sum === null) {
return null;
}
wheel[position]++;
if (wheel[position] === 2) {
wheel[position] = 0;
sum -= numbers[position];
if (wheel.length < position + 2) {
wheel.push(0);
}
sum = incrementWheel(position + 1, sum, numbers, wheel);
}
else {
sum += numbers[position];
}
return sum;
}
C#
public class Program
{
static void Main(string[] args)
{
double[] numbers = { -0.47, -0.35, -0.19, 0.23, 0.36, 0.47, 0.51, 0.59, 0.63, 0.79, 0.85,
0.91, 0.99, 1.02, 1.17, 1.25, 1.39, 1.44, 1.59, 1.60, 1.79, 1.88, 1.99, 2.14, 2.31 };
double target = 24.16;
DisplaySubsetsThatSumTo(target, numbers);
}
private static void DisplaySubsetsThatSumTo(double Target, double[] numbers)
{
var stopwatch = new System.Diagnostics.Stopwatch();
bool[] wheel = new bool[numbers.Length];
int resultsCount = 0;
double? sum = 0;
stopwatch.Start();
do
{
sum = IncrementWheel(0, sum, numbers, wheel);
//Use subtraction comparison due to double type imprecision
if (sum.HasValue && Math.Abs(sum.Value - Target) < 0.000001F)
{
//Found a subset. Display the result.
Console.WriteLine(string.Join(" + ", numbers.Where((n, idx) => wheel[idx])) + " = " + Target);
resultsCount++;
}
} while (sum != null);
stopwatch.Stop();
Console.WriteLine("--------------------------");
Console.WriteLine($"Processed {numbers.Length} numbers in {stopwatch.ElapsedMilliseconds / 1000.0} seconds ({resultsCount} results). Press any key to exit.");
Console.ReadKey();
}
private static double? IncrementWheel(int Position, double? Sum, double[] numbers, bool[] wheel)
{
if (Position == numbers.Length || !Sum.HasValue)
{
return null;
}
wheel[Position] = !wheel[Position];
if (!wheel[Position])
{
Sum -= numbers[Position];
Sum = IncrementWheel(Position + 1, Sum, numbers, wheel);
}
else
{
Sum += numbers[Position];
}
return Sum;
}
}
Output
-0.35 + 0.23 + 0.36 + 0.47 + 0.51 + 0.59 + 0.63 + 0.79 + 0.85 + 0.91 + 0.99 + 1.02 + 1.17 + 1.25 + 1.44 + 1.59 + 1.6 + 1.79 + 1.88 + 1.99 + 2.14 + 2.31 = 24.16
0.23 + 0.51 + 0.59 + 0.63 + 0.79 + 0.85 + 0.99 + 1.02 + 1.17 + 1.25 + 1.39 + 1.44 + 1.59 + 1.6 + 1.79 + 1.88 + 1.99 + 2.14 + 2.31 = 24.16
-0.47 + 0.23 + 0.47 + 0.51 + 0.59 + 0.63 + 0.79 + 0.85 + 0.99 + 1.02 + 1.17 + 1.25 + 1.39 + 1.44 + 1.59 + 1.6 + 1.79 + 1.88 + 1.99 + 2.14 + 2.31 = 24.16
-0.19 + 0.36 + 0.51 + 0.59 + 0.63 + 0.79 + 0.91 + 0.99 + 1.02 + 1.17 + 1.25 + 1.39 + 1.44 + 1.59 + 1.6 + 1.79 + 1.88 + 1.99 + 2.14 + 2.31 = 24.16
-0.47 + -0.19 + 0.36 + 0.47 + 0.51 + 0.59 + 0.63 + 0.79 + 0.91 + 0.99 + 1.02 + 1.17 + 1.25 + 1.39 + 1.44 + 1.59 + 1.6 + 1.79 + 1.88 + 1.99 + 2.14 + 2.31 = 24.16
0.23 + 0.47 + 0.51 + 0.63 + 0.85 + 0.91 + 0.99 + 1.02 + 1.17 + 1.25 + 1.39 + 1.44 + 1.59 + 1.6 + 1.79 + 1.88 + 1.99 + 2.14 + 2.31 = 24.16
--------------------------
Processed 25 numbers in 0.823 seconds (6 results)
回答8:
The usual DP solution is true for the problem.
One optimization you may do, is to keep a count of how many solutions exist for the particular sum rather than the actual sets that make up that sum...
回答9:
This my dynamical programming implementation in JS. It will return an array of arrays, each holding the subsequences summing to the provided target value.
function getSummingItems(a,t){
return a.reduce((h,n) => Object.keys(h)
.reduceRight((m,k) => +k+n <= t ? (m[+k+n] = m[+k+n] ? m[+k+n].concat(m[k].map(sa => sa.concat(n)))
: m[k].map(sa => sa.concat(n)),m)
: m, h), {0:[[]]})[t];
}
var arr = Array(20).fill().map((_,i) => i+1), // [1,2,..,20]
tgt = 42,
res = [];
console.time("test");
res = getSummingItems(arr,tgt);
console.timeEnd("test");
console.log("found",res.length,"subsequences summing to",tgt);
console.log(JSON.stringify(res));回答10:
RUBY
This code will reject the empty arrays and returns the proper array with values.
def find_sequence(val, num)
b = val.length
(0..b - 1).map {|n| val.uniq.combination(n).each.find_all {|value| value.reduce(:+) == num}}.reject(&:empty?)
end
val = [-10, 1, -1, 2, 0]
num = 2
Output will be [[2],[2,0],[-1,1,2],[-1,1,2,0]]
回答11:
public class SumOfSubSet {
public static void main(String[] args) {
// TODO Auto-generated method stub
int a[] = {1,2};
int sum=0;
if(a.length<=0) {
System.out.println(sum);
}else {
for(int i=0;i<a.length;i++) {
sum=sum+a[i];
for(int j=i+1;j<a.length;j++) {
sum=sum+a[i]+a[j];
}
}
System.out.println(sum);
}
}
}
回答12:
My backtracking solution :- Sort the array , then apply the backtracking.
void _find(int arr[],int end,vector<int> &v,int start,int target){
if(target==0){
for(int i = 0;i<v.size();i++){
cout<<v[i]<<" ";
}
cout<<endl;
}
else{
for(int i = start;i<=end && target >= arr[i];i++){
v.push_back(arr[i]);
_find(arr,end,v,i+1,target-arr[i]);
v.pop_back();
}
}
}
回答13:
While it is straightforward to find if their is a subset or not that sums to the target, implementation gets tricky when you need to keep track of the partial subsets under consideration.
If you use a linked list, a hash set or any another generic collection, you would be tempted to add an item to this collection before the call that includes the item, and then remove it before the call that excludes the item. This does not work as expected, as the stack frames in which the add will occur is not the same as the one in which remove will occur.
Solution is to use a string to keep track of the sequence. Appends to the string can be done inline in the function call; thereby maintaining the same stack frame and your answer would then conform beautifully to the original hasSubSetSum recursive structure.
import java.util.ArrayList;
public class Solution {
public static boolean hasSubSet(int [] A, int target) {
ArrayList<String> subsets = new ArrayList<>();
helper(A, target, 0, 0, subsets, "");
// Printing the contents of subsets is straightforward
return !subsets.isEmpty();
}
private static void helper(int[] A, int target, int sumSoFar, int i, ArrayList<String> subsets, String curr) {
if(i == A.length) {
if(sumSoFar == target) {
subsets.add(curr);
}
return;
}
helper(A, target, sumSoFar, i+1, subsets, curr);
helper(A, target, sumSoFar + A[i], i+1, subsets, curr + A[i]);
}
public static void main(String [] args) {
System.out.println(hasSubSet(new int[] {1,2,4,5,6}, 8));
}
}
回答14:
Subset sum problem can be solved in O(sum*n) using dynamic programming. Optimal substructure for subset sum is as follows:
SubsetSum(A, n, sum) = SubsetSum(A, n-1, sum) || SubsetSum(A, n-1, sum-set[n-1])
SubsetSum(A, n, sum) = 0, if sum > 0 and n == 0 SubsetSum(A, n, sum) = 1, if sum == 0
Here A is array of elements, n is the number of elements of array A and sum is the sum of elements in the subset.
Using this dp, you can solve for the number of subsets for the sum.
For getting subset elements, we can use following algorithm:
After filling dp[n][sum] by calling SubsetSum(A, n, sum), we recursively traverse it from dp[n][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.
void findAllSubsets(int dp[], int A[], int i, int sum, vector<int>& p) {
if (sum == 0) {
print(p);
return;
}
// If sum can be formed without including current element
if (dp[i-1][sum])
{
// Create a new vector to store new subset
vector<int> b = p;
findAllSubsets(dp, A, i-1, sum, b);
}
// If given sum can be formed after including
// current element.
if (sum >= A[i] && dp[i-1][sum-A[i]])
{
p.push_back(A[i]);
findAllSubsets(dp, A, i-1, sum-A[i], p);
}
}
回答15:
Following solution also provide array of subset which provide specific sum (here sum = 9)
array = [1, 3, 4, 2, 7, 8, 9]
(0..array.size).map { |i| array.combination(i).to_a.select { |a| a.sum == 9 } }.flatten(1)
return array of subsets which return sum of 9
=> [[9], [1, 8], [2, 7], [3, 4, 2]]
来源：https://stackoverflow.com/questions/18305843/find-all-subsets-that-sum-to-a-particular-value