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# Generate all unique partitions of an integer | Set 2

Given a positive integer n, the task is to generate all possible unique ways to represent n as sum of positive integers.
Examples:

Input:
Output:

3 1
2 2
2 1 1
1 1 1 1
Input:
Output:

2 1
1 1 1

Approach: We have already discussed the implementation of generating unique partitions in this post. This post contains another and far more intuitive implementation for the above problem using recursion.
The idea is simple and is kind of same approach as used here. We have to move recursively from n to 1 and keep on appending the numbers used to form sum in the array. When the sum becomes equal to n then we print the array and return. The base cases considered in the implementation are: remSum == 0: Required n is formed so print the array.
Then we have started forming required sum using the max value number used in the previous partition. If that number becomes greater than n we ignore it else we append that number to the array and move recursively to next iteration to form sum = (remSum – i) and where max value
number that could be used is i or less than i. This way we can generate the required unique partitions.
Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Array to store the numbers used``// to form the required sum``int` `dp;``int` `count = 0;` `// Function to print the array which contains``// the unique partitions which are used``// to form the required sum``void` `print(``int` `idx)``{``    ``for` `(``int` `i = 1; i < idx; i++) {``        ``cout << dp[i] << ``" "``;``    ``}``    ``cout << endl;``}` `// Function to find all the unique partitions``// remSum = remaining sum to form``// maxVal is the maximum number that``// can be used to make the partition``void` `solve(``int` `remSum, ``int` `maxVal, ``int` `idx, ``int``& count)``{` `    ``// If remSum == 0 that means the sum``    ``// is achieved so print the array``    ``if` `(remSum == 0) {``        ``print(idx);``        ``count++;``        ``return``;``    ``}` `    ``// i will begin from maxVal which is the``    ``// maximum value which can be used to form the sum``    ``for` `(``int` `i = maxVal; i >= 1; i--) {``        ``if` `(i > remSum) {``            ``continue``;``        ``}``        ``else` `if` `(i <= remSum) {` `            ``// Store the number used in forming``            ``// sum gradually in the array``            ``dp[idx] = i;` `            ``// Since i used the rest of partition``            ``// cant have any number greater than i``            ``// hence second parameter is i``            ``solve(remSum - i, i, idx + 1, count);``        ``}``    ``}``}` `// Driver code``int` `main()``{``    ``int` `n = 4, count = 0;` `    ``solve(n, n, 1, count);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `    ``// Array to store the numbers used``    ``// to form the required sum``    ``static` `int``[] dp = ``new` `int``[``200``];``    ``static` `int` `count = ``0``;` `    ``// Function to print the array which contains``    ``// the unique partitions which are used``    ``// to form the required sum``    ``static` `void` `print(``int` `idx)``    ``{``        ``for` `(``int` `i = ``1``; i < idx; i++)``        ``{``            ``System.out.print(dp[i] + ``" "``);``        ``}``        ``System.out.println(``""``);``    ``}` `    ``// Function to find all the unique partitions``    ``// remSum = remaining sum to form``    ``// maxVal is the maximum number that``    ``// can be used to make the partition``    ``static` `void` `solve(``int` `remSum, ``int` `maxVal,``                        ``int` `idx, ``int` `count)``    ``{` `        ``// If remSum == 0 that means the sum``        ``// is achieved so print the array``        ``if` `(remSum == ``0``)``        ``{``            ``print(idx);``            ``count++;``            ``return``;``        ``}` `        ``// i will begin from maxVal which is the``        ``// maximum value which can be used to form the sum``        ``for` `(``int` `i = maxVal; i >= ``1``; i--)``        ``{``            ``if` `(i > remSum)``            ``{``                ``continue``;``            ``}``            ``else` `if` `(i <= remSum)``            ``{` `                ``// Store the number used in forming``                ``// sum gradually in the array``                ``dp[idx] = i;` `                ``// Since i used the rest of partition``                ``// cant have any number greater than i``                ``// hence second parameter is i``                ``solve(remSum - i, i, idx + ``1``, count);``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `n = ``4``, count = ``0``;` `        ``solve(n, n, ``1``, count);``    ``}``}` `// This code has been contributed by 29AjayKumar`

## Python3

 `# Python 3 implementation of the approach` `# Array to store the numbers used``# to form the required sum``dp ``=` `[``0` `for` `i ``in` `range``(``200``)]``count ``=` `0` `# Function to print the array which contains``# the unique partitions which are used``# to form the required sum``def` `print1(idx):``    ``for` `i ``in` `range``(``1``,idx,``1``):``        ``print``(dp[i],end ``=` `" "``)``    ``print``(``"\n"``,end ``=` `"")` `# Function to find all the unique partitions``# remSum = remaining sum to form``# maxVal is the maximum number that``# can be used to make the partition``def` `solve(remSum,maxVal,idx,count):``    ``# If remSum == 0 that means the sum``    ``# is achieved so print the array``    ``if` `(remSum ``=``=` `0``):``        ``print1(idx)``        ``count ``+``=` `1``        ``return``    ``# i will begin from maxVal which is the``    ``# maximum value which can be used to form the sum``    ``i ``=` `maxVal``    ``while``(i >``=` `1``):``        ``if` `(i > remSum):``            ``i ``-``=` `1``            ``continue``        ``elif` `(i <``=` `remSum):``            ``# Store the number used in forming``            ``# sum gradually in the array``            ``dp[idx] ``=` `i` `            ``# Since i used the rest of partition``            ``# cant have any number greater than i``            ``# hence second parameter is i``            ``solve(remSum ``-` `i, i, idx ``+` `1``, count)``            ``i ``-``=` `1` `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``n ``=` `4``    ``count ``=` `0` `    ``solve(n, n, ``1``, count)``    ` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{` `    ``// Array to store the numbers used``    ``// to form the required sum``    ``static` `int``[] dp = ``new` `int``;` `    ``// Function to print the array which contains``    ``// the unique partitions which are used``    ``// to form the required sum``    ``static` `void` `print(``int` `idx)``    ``{``        ``for` `(``int` `i = 1; i < idx; i++)``        ``{``            ``Console.Write(dp[i] + ``" "``);``        ``}``        ``Console.WriteLine(``""``);``    ``}` `    ``// Function to find all the unique partitions``    ``// remSum = remaining sum to form``    ``// maxVal is the maximum number that``    ``// can be used to make the partition``    ``static` `void` `solve(``int` `remSum, ``int` `maxVal,``                        ``int` `idx, ``int` `count)``    ``{` `        ``// If remSum == 0 that means the sum``        ``// is achieved so print the array``        ``if` `(remSum == 0)``        ``{``            ``print(idx);``            ``count++;``            ``return``;``        ``}` `        ``// i will begin from maxVal which is the``        ``// maximum value which can be used to form the sum``        ``for` `(``int` `i = maxVal; i >= 1; i--)``        ``{``            ``if` `(i > remSum)``            ``{``                ``continue``;``            ``}``            ``else` `if` `(i <= remSum)``            ``{` `                ``// Store the number used in forming``                ``// sum gradually in the array``                ``dp[idx] = i;` `                ``// Since i used the rest of partition``                ``// cant have any number greater than i``                ``// hence second parameter is i``                ``solve(remSum - i, i, idx + 1, count);``            ``}``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``int` `n = 4, count = 0;` `        ``solve(n, n, 1, count);``    ``}``}` `// This code is contributed by AnkitRai01`

## Javascript

 ``

Output:

```4
3 1
2 2
2 1 1
1 1 1 1```

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