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