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Combinational Sum

  • Difficulty Level : Medium
  • Last Updated : 22 Nov, 2021

Given an array of positive integers arr[] and a sum x, find all unique combinations in arr[] where the sum is equal to x. The same repeated number may be chosen from arr[] unlimited number of times. Elements in a combination (a1, a2, …, ak) must be printed in non-descending order. (ie, a1 <= a2 <= … <= ak). 
The combinations themselves must be sorted in ascending order, i.e., the combination with smallest first element should be printed first. If there is no combination possible the print “Empty” (without quotes).

Examples: 

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Input : arr[] = 2, 4, 6, 8 
            x = 8
Output : [2, 2, 2, 2]
         [2, 2, 4]
         [2, 6]
         [4, 4]
         [8]

Since the problem is to get all the possible results, not the best or the number of result, thus we don’t need to consider DP(dynamic programming), recursion is needed to handle it.

We should use the following algorithm. 



1. Sort the array(non-decreasing).
2. First remove all the duplicates from array.
3. Then use recursion and backtracking to solve 
   the problem.
   (A) If at any time sub-problem sum == 0 then 
       add that array to the result (vector of 
       vectors).
   (B) Else if sum is negative then ignore that 
       sub-problem.
   (C) Else insert the present index in that 
       array to the current vector and call 
       the function with sum = sum-ar[index] and
       index = index, then pop that element from 
       current index (backtrack) and call the 
       function with sum = sum and index = index+1

Below is the C++ implementation of the above steps.

C++




// C++ program to find all combinations that
// sum to a given value
#include <bits/stdc++.h>
using namespace std;
 
// Print all members of ar[] that have given
void findNumbers(vector<int>& ar, int sum,
                 vector<vector<int> >& res, vector<int>& r,
                 int i)
{
    // if we get exact answer
    if (sum == 0) {
        res.push_back(r);
        return;
    }
 
    // Recur for all remaining elements that
    // have value smaller than sum.
    while (i < ar.size() && sum - ar[i] >= 0) {
 
        // Till every element in the array starting
        // from i which can contribute to the sum
        r.push_back(ar[i]); // add them to list
 
        // recur for next numbers
        findNumbers(ar, sum - ar[i], res, r, i);
        i++;
 
        // Remove number from list (backtracking)
        r.pop_back();
    }
}
 
// Returns all combinations
// of ar[] that have given
// sum.
vector<vector<int> > combinationSum(vector<int>& ar,
                                    int sum)
{
    // sort input array
    sort(ar.begin(), ar.end());
 
    // remove duplicates
    ar.erase(unique(ar.begin(), ar.end()), ar.end());
 
    vector<int> r;
    vector<vector<int> > res;
    findNumbers(ar, sum, res, r, 0);
 
    return res;
}
 
// Driver code
int main()
{
    vector<int> ar;
    ar.push_back(2);
    ar.push_back(4);
    ar.push_back(6);
    ar.push_back(8);
    int n = ar.size();
 
    int sum = 8; // set value of sum
    vector<vector<int> > res = combinationSum(ar, sum);
 
    // If result is empty, then
    if (res.size() == 0) {
        cout << "Emptyn";
        return 0;
    }
 
    // Print all combinations stored in res.
    for (int i = 0; i < res.size(); i++) {
        if (res[i].size() > 0) {
            cout << " ( ";
            for (int j = 0; j < res[i].size(); j++)
                cout << res[i][j] << " ";
            cout << ")";
        }
    }
}

Java




// Java program to find all combinations that
// sum to a given value
import java.io.*;
import java.util.*;
 
class GFG {
 
    static ArrayList<ArrayList<Integer> >
    combinationSum(ArrayList<Integer> arr, int sum)
    {
        ArrayList<ArrayList<Integer> > ans
            = new ArrayList<>();
        ArrayList<Integer> temp = new ArrayList<>();
 
        // first do hashing since hashset does not always
        // sort
        //  removing the duplicates using HashSet and
        // Sorting the arrayList
 
        Set<Integer> set = new HashSet<>(arr);
        arr.clear();
        arr.addAll(set);
        Collections.sort(arr);
 
        findNumbers(ans, arr, sum, 0, temp);
        return ans;
    }
 
    static void
    findNumbers(ArrayList<ArrayList<Integer> > ans,
                ArrayList<Integer> arr, int sum, int index,
                ArrayList<Integer> temp)
    {
 
        if (sum == 0) {
 
            // Adding deep copy of list to ans
 
            ans.add(new ArrayList<>(temp));
            return;
        }
 
        for (int i = index; i < arr.size(); i++) {
 
            // checking that sum does not become negative
 
            if ((sum - arr.get(i)) >= 0) {
 
                // adding element which can contribute to
                // sum
 
                temp.add(arr.get(i));
 
                findNumbers(ans, arr, sum - arr.get(i), i,
                            temp);
 
                // removing element from list (backtracking)
                temp.remove(arr.get(i));
            }
        }
    }
 
    // Driver Code
 
    public static void main(String[] args)
    {
        ArrayList<Integer> arr = new ArrayList<>();
 
        arr.add(2);
        arr.add(4);
        arr.add(6);
        arr.add(8);
 
        int sum = 8;
 
        ArrayList<ArrayList<Integer> > ans
            = combinationSum(arr, sum);
 
        // If result is empty, then
        if (ans.size() == 0) {
            System.out.println("Empty");
            return;
        }
 
        // print all combinations stored in ans
 
        for (int i = 0; i < ans.size(); i++) {
 
            System.out.print("(");
            for (int j = 0; j < ans.get(i).size(); j++) {
                System.out.print(ans.get(i).get(j) + " ");
            }
            System.out.print(") ");
        }
    }
}

Python3




# Python3 program to find all combinations that
# sum to a given value
 
def combinationSum(arr, sum):
    ans = []
    temp = []
 
    # first do hashing nothing but set{}
    # since set does not always sort
    # removing the duplicates using Set and
    # Sorting the List
    arr = sorted(list(set(arr)))
    findNumbers(ans, arr, temp, sum, 0)
    return ans
 
def findNumbers(ans, arr, temp, sum, index):
     
    if(sum == 0):
         
        # Adding deep copy of list to ans
        ans.append(list(temp))
        return
       
    # Iterate from index to len(arr) - 1
    for i in range(index, len(arr)):
 
        # checking that sum does not become negative
        if(sum - arr[i]) >= 0:
 
            # adding element which can contribute to
            # sum
            temp.append(arr[i])
            findNumbers(ans, arr, temp, sum-arr[i], i)
 
            # removing element from list (backtracking)
            temp.remove(arr[i])
 
 
# Driver Code
arr = [2, 4, 6, 8]
sum = 8
ans = combinationSum(arr, sum)
 
# If result is empty, then
if len(ans) <= 0:
    print("empty")
     
# print all combinations stored in ans
for i in range(len(ans)):
 
    print("(", end=' ')
    for j in range(len(ans[i])):
        print(str(ans[i][j])+" ", end=' ')
    print(")", end=' ')
 
 
# This Code Is Contributed by Rakesh(vijayarigela09)

C#




// C# program for the above approach
using System;
using System.Collections.Generic;
using System.Collections;
 
class GFG {
 
    static List<List<int> >
    combinationSum(List<int> arr, int sum)
    {
        List<List<int> > ans
            = new List<List<int> >();
        List<int> temp = new List<int>();
 
        // first do hashing since hashset does not always
        // sort
        //  removing the duplicates using HashSet and
        // Sorting the List
 
        HashSet<int> set = new HashSet<int>(arr);
        arr.Clear();
        arr.AddRange(set);
        arr.Sort();
 
        findNumbers(ans, arr, sum, 0, temp);
        return ans;
    }
 
    static void
    findNumbers(List<List<int> > ans,
                List<int> arr, int sum, int index,
                List<int> temp)
    {
 
        if (sum == 0) {
 
            // Adding deep copy of list to ans
 
            ans.Add(new List<int>(temp));
            return;
        }
 
        for (int i = index; i < arr.Count; i++) {
 
            // checking that sum does not become negative
 
            if ((sum - arr[i]) >= 0) {
 
                // Adding element which can contribute to
                // sum
 
                temp.Add(arr[i]);
 
                findNumbers(ans, arr, sum - arr[i], i,
                            temp);
 
                // removing element from list (backtracking)
                temp.Remove(arr[i]);
            }
        }
    }
 
    // Driver Code
 
    public static void Main()
    {
        List<int> arr = new List<int>();
         
        arr.Add(2);
        arr.Add(4);
        arr.Add(6);
        arr.Add(8);
 
        int sum = 8;
 
        List<List<int> > ans
            = combinationSum(arr, sum);
 
        // If result is empty, then
        if (ans.Count == 0) {
            Console.WriteLine("Empty");
            return;
        }
 
        // print all combinations stored in ans
 
        for (int i = 0; i < ans.Count; i++) {
 
            Console.Write("(");
            for (int j = 0; j < ans[i].Count; j++) {
                Console.Write(ans[i][j] + " ");
            }
            Console.Write(") ");
        }
    }
}
 
// This code is contributed by ShubhamSingh10
Output
 ( 2 2 2 2 ) ( 2 2 4 ) ( 2 6 ) ( 4 4 ) ( 8 )

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