Given a positive integer N, the task is to find out all the combinations of positive integers that add upto the given integer N. The program should print only combinations, not permutations and all the integers in a combination must be distinct. For example, for input 3, either 1, 2 or 2, 1 should be printed and 1, 1, 1 must not be printed as the integers are not distinct.
Examples:
Input: N = 3
Output:
1 2
3
Input: N = 7
Output:
1 2 4
1 6
2 5
3 4
7
Approach: The approach is an extension of the approach discussed here. The idea used to get all the distinct element is that first we find all the elements that add up to give sum N. Then we iterate over each of the elements and store the elements into the set. Storing the elements into set would remove all the duplicate elements, and after that we add up the sum of the elements of the set and check whether it is equal to N or not.
Below is the implementation of the above approach:
// C++ implementation of the approach #include <bits/stdc++.h> using namespace std;
/* arr[] to store all the distinct elements index - next location in array
num - given number
reducedNum - reduced number */
void findCombinationsUtil( int arr[], int index,
int n, int red_num)
{ // Set to store all the
// distinct elements
set< int > s;
int sum = 0;
// Base condition
if (red_num < 0) {
return ;
}
if (red_num == 0) {
// Iterate over all the elements
// and store it into the set
for ( int i = 0; i < index; i++) {
s.insert(arr[i]);
}
// Calculate the sum of all
// the elements of the set
for ( auto itr = s.begin();
itr != s.end(); itr++) {
sum = sum + (*itr);
}
// Compare whether the sum is equal to n or not,
// if it is equal to n print the numbers
if (sum == n) {
for ( auto i = s.begin();
i != s.end(); i++) {
cout << *i << " " ;
}
cout << endl;
return ;
}
}
// Find previous number stored in the array
int prev = (index == 0) ? 1 : arr[index - 1];
for ( int k = prev; k <= n; k++) {
// Store all the numbers recursively
// into the arr[]
arr[index] = k;
findCombinationsUtil(arr, index + 1,
n, red_num - k);
}
} // Function to find all the // distinct combinations of n void findCombinations( int n)
{ int a[n];
findCombinationsUtil(a, 0, n, n);
} // Driver code int main()
{ int n = 7;
findCombinations(n);
return 0;
} |
// Java implementation of the approach import java.util.*;
class GFG
{ /* arr[] to store all the distinct elements index - next location in array
num - given number
reducedNum - reduced number */
static void findCombinationsUtil( int arr[], int index,
int n, int red_num)
{ // Set to store all the
// distinct elements
HashSet<Integer> s = new HashSet<>();
int sum = 0 ;
// Base condition
if (red_num < 0 )
{
return ;
}
if (red_num == 0 )
{
// Iterate over all the elements
// and store it into the set
for ( int i = 0 ; i < index; i++)
{
s.add(arr[i]);
}
// Calculate the sum of all
// the elements of the set
for (Integer itr : s)
{
sum = sum + itr;
}
// Compare whether the sum is equal to n or not,
// if it is equal to n print the numbers
if (sum == n)
{
for (Integer i : s)
{
System.out.print(i+ " " );
}
System.out.println();
return ;
}
}
// Find previous number stored in the array
int prev = (index == 0 ) ? 1 : arr[index - 1 ];
for ( int k = prev; k <= n; k++)
{
// Store all the numbers recursively
// into the arr[]
if (index < n)
{
arr[index] = k;
findCombinationsUtil(arr, index + 1 ,
n, red_num - k);
}
}
} // Function to find all the // distinct combinations of n static void findCombinations( int n)
{ int []a = new int [n];
findCombinationsUtil(a, 0 , n, n);
} // Driver code public static void main(String arr[])
{ int n = 7 ;
findCombinations(n);
} } /* This code contributed by PrinciRaj1992 */ |
# Python3 implementation of the approach # arr[] to store all the distinct elements # index - next location in array # num - given number # reducedNum - reduced number def findCombinationsUtil(arr, index, n, red_num):
# Set to store all the
# distinct elements
s = set ()
sum = 0
# Base condition
if (red_num < 0 ):
return
if (red_num = = 0 ):
# Iterate over all the elements
# and store it into the set
for i in range (index):
s.add(arr[i])
# Calculate the sum of all
# the elements of the set
for itr in s:
sum = sum + (itr)
# Compare whether the sum is equal to n or not,
# if it is equal to n print the numbers
if ( sum = = n):
for i in s:
print (i, end = " " )
print ( "\n" , end = "")
return
# Find previous number stored in the array
if (index = = 0 ):
prev = 1
else :
prev = arr[index - 1 ]
for k in range (prev, n + 1 , 1 ):
# Store all the numbers recursively
# into the arr[]
arr[index] = k
findCombinationsUtil(arr, index + 1 ,
n, red_num - k)
# Function to find all the # distinct combinations of n def findCombinations(n):
a = [ 0 for i in range (n + 1 )]
findCombinationsUtil(a, 0 , n, n)
# Driver code if __name__ = = '__main__' :
n = 7
findCombinations(n)
# This code is contributed by Surendra_Gangwar |
// C# implementation of the approach using System;
using System.Collections.Generic;
class GFG
{ /* arr[] to store all the distinct elements index - next location in array
num - given number
reducedNum - reduced number */
static void findCombinationsUtil( int []arr, int index,
int n, int red_num)
{ // Set to store all the
// distinct elements
HashSet< int > s = new HashSet< int >();
int sum = 0;
// Base condition
if (red_num < 0)
{
return ;
}
if (red_num == 0)
{
// Iterate over all the elements
// and store it into the set
for ( int i = 0; i < index; i++)
{
s.Add(arr[i]);
}
// Calculate the sum of all
// the elements of the set
foreach ( int itr in s)
{
sum = sum + itr;
}
// Compare whether the sum is equal to n or not,
// if it is equal to n print the numbers
if (sum == n)
{
foreach ( int i in s)
{
Console.Write(i+ " " );
}
Console.WriteLine();
return ;
}
}
// Find previous number stored in the array
int prev = (index == 0) ? 1 : arr[index - 1];
for ( int k = prev; k <= n; k++)
{
// Store all the numbers recursively
// into the arr[]
if (index < n)
{
arr[index] = k;
findCombinationsUtil(arr, index + 1,
n, red_num - k);
}
}
} // Function to find all the // distinct combinations of n static void findCombinations( int n)
{ int []a = new int [n];
findCombinationsUtil(a, 0, n, n);
} // Driver code public static void Main(String []arr)
{ int n = 7;
findCombinations(n);
} } // This code contributed by Rajput-Ji |
<script> // JavaScript implementation of the approach
/* arr[] to store all the distinct elements
index - next location in array
num - given number
reducedNum - reduced number */
function findCombinationsUtil(arr, index, n, red_num) {
// Set to store all the
// distinct elements
var s = new Set();
var sum = 0;
// Base condition
if (red_num < 0) {
return ;
}
if (red_num === 0) {
// Iterate over all the elements
// and store it into the set
for ( var i = 0; i < index; i++) {
s.add(arr[i]);
}
// Calculate the sum of all
// the elements of the set
for (const itr of s) {
sum = sum + itr;
}
// Compare whether the sum is equal to n or not,
// if it is equal to n print the numbers
if (sum === n) {
for (const i of s) {
document.write(i + " " );
}
document.write( "<br>" );
return ;
}
}
// Find previous number stored in the array
var prev = index === 0 ? 1 : arr[index - 1];
for ( var k = prev; k <= n; k++) {
// Store all the numbers recursively
// into the arr[]
if (index < n) {
arr[index] = k;
findCombinationsUtil(arr, index + 1, n, red_num - k);
}
}
}
// Function to find all the
// distinct combinations of n
function findCombinations(n) {
var a = new Array(n).fill(0);
findCombinationsUtil(a, 0, n, n);
}
// Driver code
var n = 7;
findCombinations(n);
</script> |
1 2 4 1 6 2 5 3 4 7
Time Complexity: O(nlogn)
Auxiliary Space: O(n)