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# Generate all unique partitions of an integer | Set 2

• Difficulty Level : Hard
• Last Updated : 09 Dec, 2021

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[200]; 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[200];       // 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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