Open In App

All unique combinations whose sum equals to K (Combination Sum II)

Improve
Improve
Like Article
Like
Save
Share
Report

Given an array arr[] of size N and an integer K. The task is to find all the unique combinations from the given array such that sum of the elements in each combination is equal to K.

Examples: 

Input: arr[] = {1, 2, 3}, K = 3 
Output: 
{1, 2} 
{3} 
Explanation:
These are the combinations whose sum equals to 3. 

Input: arr[] = {2, 2, 2}, K = 4 
Output: 
{2, 2} 

Approach: Some elements can be repeated in the given array. Make sure to iterate over the number of occurrences of those elements to avoid repeated combinations. Once you do that, things are fairly straightforward. Call a recursive function with the remaining sum and make the indices to move forward. When the sum reaches K, print all the elements which were selected to get this sum.

Below is the implementation of the above approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find all unique combination of
// given elements such that their sum is K
void unique_combination(int l, int sum, int K,
                        vector<int>& local,
                        vector<int>& A)
{
    // If a unique combination is found
    if (sum == K) {
        cout << "{";
        for (int i = 0; i < local.size(); i++)
        {
            if (i != 0)
                cout << " ";
            cout << local[i];
            if (i != local.size() - 1)
                cout << ", ";
        }
        cout << "}" << endl;
        return;
    }
 
    // For all other combinations
    for (int i = l; i < A.size(); i++)
    {
 
        // Check if the sum exceeds K
        if (sum + A[i] > K)
            continue;
 
        // Check if it is repeated or not
        if (i > l and A[i] == A[i - 1])
            continue;
 
        // Take the element into the combination
        local.push_back(A[i]);
 
        // Recursive call
        unique_combination(i + 1, sum + A[i], K, local, A);
 
        // Remove element from the combination
        local.pop_back();
    }
}
 
// Function to find all combination
// of the given elements
void Combination(vector<int> A, int K)
{
    // Sort the given elements
    sort(A.begin(), A.end());
 
    // To store combination
    vector<int> local;
 
    unique_combination(0, 0, K, local, A);
}
 
// Driver code
int main()
{
    vector<int> A = { 10, 1, 2, 7, 6, 1, 5 };
 
    int K = 8;
 
    // Function call
    Combination(A, K);
 
    return 0;
}


Java




// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to find all unique combination of
    // given elements such that their sum is K
    static void unique_combination(int l, int sum, int K,
                                   Vector<Integer> local,
                                   Vector<Integer> A)
    {
        // If a unique combination is found
        if (sum == K) {
            System.out.print("{");
            for (int i = 0; i < local.size(); i++) {
                if (i != 0)
                    System.out.print(" ");
                System.out.print(local.get(i));
                if (i != local.size() - 1)
                    System.out.print(", ");
            }
            System.out.println("}");
            return;
        }
 
        // For all other combinations
        for (int i = l; i < A.size(); i++) {
 
            // Check if the sum exceeds K
            if (sum + A.get(i) > K)
                continue;
 
            // Check if it is repeated or not
            if (i > l && A.get(i) == A.get(i - 1) )
                continue;
 
            // Take the element into the combination
            local.add(A.get(i));
 
            // Recursive call
            unique_combination(i + 1, sum + A.get(i), K,
                               local, A);
 
            // Remove element from the combination
            local.remove(local.size() - 1);
        }
    }
 
    // Function to find all combination
    // of the given elements
    static void Combination(Vector<Integer> A, int K)
    {
        // Sort the given elements
        Collections.sort(A);
 
        // To store combination
        Vector<Integer> local = new Vector<Integer>();
 
        unique_combination(0, 0, K, local, A);
    }
 
    // Driver code
    public static void main(String[] args)
    {
        Integer[] arr = { 10, 1, 2, 7, 6, 1, 5 };
        Vector<Integer> A
            = new Vector<>(Arrays.asList(arr));
 
        int K = 8;
 
        // Function call
        Combination(A, K);
    }
}
 
// This code is contributed by PrinciRaj1992


Python3




# Python 3 implementation of the approach
 
# Function to find all unique combination of
# given elements such that their sum is K
 
 
def unique_combination(l, sum, K, local, A):
 
    # If a unique combination is found
    if (sum == K):
        print("{", end="")
        for i in range(len(local)):
            if (i != 0):
                print(" ", end="")
            print(local[i], end="")
            if (i != len(local) - 1):
                print(", ", end="")
        print("}")
        return
 
    # For all other combinations
    for i in range(l, len(A), 1):
 
        # Check if the sum exceeds K
        if (sum + A[i] > K):
            continue
 
        # Check if it is repeated or not
        if (i > l and
                A[i] == A[i - 1]):
            continue
 
        # Take the element into the combination
        local.append(A[i])
 
        # Recursive call
        unique_combination(i + 1, sum + A[i],
                           K, local, A)
 
        # Remove element from the combination
        local.remove(local[len(local) - 1])
 
# Function to find all combination
# of the given elements
 
 
def Combination(A, K):
 
    # Sort the given elements
    A.sort(reverse=False)
 
    local = []
 
    unique_combination(0, 0, K, local, A)
 
 
# Driver code
if __name__ == '__main__':
    A = [10, 1, 2, 7, 6, 1, 5]
 
    K = 8
 
    # Function call
    Combination(A, K)
 
# This code is contributed by
# Surendra_Gangwar


C#




// C# implementation of the approach
using System;
using System.Collections.Generic;
 
class GFG {
 
    // Function to find all unique combination of
    // given elements such that their sum is K
    static void unique_combination(int l, int sum, int K,
                                   List<int> local,
                                   List<int> A)
    {
        // If a unique combination is found
        if (sum == K)
        {
            Console.Write("{");
            for (int i = 0; i < local.Count; i++)
            {
                if (i != 0)
                    Console.Write(" ");
                Console.Write(local[i]);
                if (i != local.Count - 1)
                    Console.Write(", ");
            }
            Console.WriteLine("}");
            return;
        }
 
        // For all other combinations
        for (int i = l; i < A.Count; i++)
        {
            // Check if the sum exceeds K
            if (sum + A[i] > K)
                continue;
 
            // Check if it is repeated or not
            if (i >l && A[i] == A[i - 1])
                continue;
 
            // Take the element into the combination
            local.Add(A[i]);
 
            // Recursive call
            unique_combination(i + 1, sum + A[i], K, local,
                               A);
 
            // Remove element from the combination
            local.RemoveAt(local.Count - 1);
        }
    }
 
    // Function to find all combination
    // of the given elements
    static void Combination(List<int> A, int K)
    {
        // Sort the given elements
        A.Sort();
 
        // To store combination
        List<int> local = new List<int>();
 
        unique_combination(0, 0, K, local, A);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[] arr = { 10, 1, 2, 7, 6, 1, 5 };
        List<int> A = new List<int>(arr);
 
        int K = 8;
 
        // Function call
        Combination(A, K);
    }
}
 
// This code is contributed by Rajput-Ji


Javascript




<script>
 
// JavaScript implementation of the approach
 
// Function to find all unique combination of
// given elements such that their sum is K
function unique_combination(l, sum, K, local, A) {
    // If a unique combination is found
    if (sum == K) {
        document.write("{");
        for (let i = 0; i < local.length; i++) {
            if (i != 0)
                document.write(" ");
            document.write(local[i]);
            if (i != local.length - 1)
                document.write(", ");
        }
        document.write("}" + "<br>");
        return;
    }
 
    // For all other combinations
    for (let i = l; i < A.length; i++) {
 
        // Check if the sum exceeds K
        if (sum + A[i] > K)
            continue;
 
        // Check if it is repeated or not
        if (i > l && A[i] == A[i - 1])
            continue;
 
        // Take the element into the combination
        local.push(A[i]);
 
        // Recursive call
        unique_combination(i + 1, sum + A[i], K, local, A);
 
        // Remove element from the combination
        local.pop();
    }
}
 
// Function to find all combination
// of the given elements
function Combination(A, K) {
    // Sort the given elements
    A.sort((a, b) => a - b);
 
    // To store combination
    let local = [];
 
    unique_combination(0, 0, K, local, A);
}
 
// Driver code
 
let A = [10, 1, 2, 7, 6, 1, 5];
 
let K = 8;
 
// Function call
Combination(A, K);
 
// This code is contributed by _saurabh_jaiswal
 
</script>


Output

{1,  1,  6}
{1,  2,  5}
{1,  7}
{2,  6}

Time Complexity: O(n log n) for sorting, O(2^n) for generating a number of combinations. If ‘k’ is the avg length of every combination then adding it to the resultant list would take O(k x 2^n). Total complexity is O(n log n)+O(k x 2^n)
Auxiliary Space: If ‘k’ is the avg length of every combination and ‘t’ is the no. of combinations then space complexity would be O(k x t)



Last Updated : 11 Sep, 2023
Like Article
Save Article
Previous
Next
Share your thoughts in the comments
Similar Reads