# Count the number of ways to divide N in k groups incrementally

• Difficulty Level : Hard
• Last Updated : 26 Oct, 2021

Given two integers N and K, the task is to count the number of ways to divide N into K groups of positive integers such that their sum is N and the number of elements in groups follows a non-decreasing order (i.e group[i] <= group[i+1]).
Examples:

Input: N = 8, K = 4
Output:
Explanation:
Their are 5 groups such that their sum is 8 and the number of positive integers in each group is 4.
[1, 1, 1, 5], [1, 1, 2, 4], [1, 1, 3, 3], [1, 2, 2, 3], [2, 2, 2, 2]
Input: N = 24, K = 5
Output: 164
Explanation:
There are 164 such groups such that their sum is 24 and number of positive integers in each group is 5

Different Approaches:
1. Naive Approach (Time: O(NK), Space: O(N))
2. Memoization  (Time: O((N3*K), Space: O(N2*K))
3. Bottom-Up Dynamic Programming (Time: O(N*K), Space: O(N*K))

Naive Approach: We can solve this problem using recursion. At each step of recursion put all the values greater than equal to the previously computed value.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements` `#include ` `using` `namespace` `std;` `// Function to count the number``// of ways to divide the number N``// in groups such that each group``// has K number of elements``int` `calculate(``int` `pos, ``int` `prev,``                 ``int` `left, ``int` `k)``{``    ``// Base Case``    ``if` `(pos == k) {``        ``if` `(left == 0)``            ``return` `1;``        ``else``            ``return` `0;``    ``}` `    ``// if N is divides completely``    ``// into less than k groups``    ``if` `(left == 0)``        ``return` `0;` `    ``int` `answer = 0;``    ` `    ``// put all possible values``    ``// greater equal to prev``    ``for` `(``int` `i = prev; i <= left; i++) {``        ``answer += calculate(pos + 1, i,``                          ``left - i, k);``    ``}``    ``return` `answer;``}` `// Function to count the number of``// ways to divide the number N``int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``return` `calculate(0, 1, n, k);``}` `// Driver Code``int` `main()``{``    ``int` `N = 8;``    ``int` `K = 4;` `    ``cout << countWaystoDivide(N, K);``    ``return` `0;``}`

## Java

 `// Java implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements``import` `java.util.*;``class` `GFG{` `// Function to count the number``// of ways to divide the number N``// in groups such that each group``// has K number of elements``static` `int` `calculate(``int` `pos, ``int` `prev,``                     ``int` `left, ``int` `k)``{``    ` `    ``// Base Case``    ``if` `(pos == k)``    ``{``        ``if` `(left == ``0``)``            ``return` `1``;``        ``else``            ``return` `0``;``    ``}` `    ``// If N is divides completely``    ``// into less than k groups``    ``if` `(left == ``0``)``        ``return` `0``;` `    ``int` `answer = ``0``;``    ` `    ``// Put all possible values``    ``// greater equal to prev``    ``for``(``int` `i = prev; i <= left; i++)``    ``{``       ``answer += calculate(pos + ``1``, i,``                           ``left - i, k);``    ``}``    ``return` `answer;``}` `// Function to count the number of``// ways to divide the number N``static` `int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``return` `calculate(``0``, ``1``, n, k);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``8``;``    ``int` `K = ``4``;` `    ``System.out.print(countWaystoDivide(N, K));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation to count the``# number of ways to divide N in``# groups such that each group``# has K number of elements` `# Function to count the number``# of ways to divide the number N``# in groups such that each group``# has K number of elements``def` `calculate(pos, prev, left, k):``    ` `    ``# Base Case``    ``if` `(pos ``=``=` `k):``        ``if` `(left ``=``=` `0``):``            ``return` `1``;``        ``else``:``            ``return` `0``;` `    ``# If N is divides completely``    ``# into less than k groups``    ``if` `(left ``=``=` `0``):``        ``return` `0``;` `    ``answer ``=` `0``;` `    ``# Put all possible values``    ``# greater equal to prev``    ``for` `i ``in` `range``(prev, left ``+` `1``):``        ``answer ``+``=` `calculate(pos ``+` `1``, i,``                           ``left ``-` `i, k);` `    ``return` `answer;` `# Function to count the number of``# ways to divide the number N``def` `countWaystoDivide(n, k):``    ` `    ``return` `calculate(``0``, ``1``, n, k);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``N ``=` `8``;``    ``K ``=` `4``;` `    ``print``(countWaystoDivide(N, K));` `# This code is contributed by 29AjayKumar`

## C#

 `// C# implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements``using` `System;` `class` `GFG{` `// Function to count the number``// of ways to divide the number N``// in groups such that each group``// has K number of elements``static` `int` `calculate(``int` `pos, ``int` `prev,``                     ``int` `left, ``int` `k)``{``    ` `    ``// Base Case``    ``if` `(pos == k)``    ``{``        ``if` `(left == 0)``            ``return` `1;``        ``else``            ``return` `0;``    ``}` `    ``// If N is divides completely``    ``// into less than k groups``    ``if` `(left == 0)``        ``return` `0;` `    ``int` `answer = 0;``    ` `    ``// Put all possible values``    ``// greater equal to prev``    ``for``(``int` `i = prev; i <= left; i++)``    ``{``       ``answer += calculate(pos + 1, i,``                           ``left - i, k);``    ``}``    ``return` `answer;``}` `// Function to count the number of``// ways to divide the number N``static` `int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``return` `calculate(0, 1, n, k);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 8;``    ``int` `K = 4;` `    ``Console.Write(countWaystoDivide(N, K));``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``
Output
`5`

Time complexity: O(NK)
Auxiliary Space: O(N).
Memoization Approach: In the previous approach we can see that we are solving the subproblems repeatedly, i.e. it follows the property of Overlapping Subproblems. So we can memoize the same using the DP table.
Below is the implementation of the above approach:

## C++

 `// C++ implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements` `#include ` `using` `namespace` `std;` `// DP Table``int` `dp;` `// Function to count the number``// of ways to divide the number N``// in groups such that each group``// has K number of elements``int` `calculate(``int` `pos, ``int` `prev,``                ``int` `left, ``int` `k)``{``    ``// Base Case``    ``if` `(pos == k) {``        ``if` `(left == 0)``            ``return` `1;``        ``else``            ``return` `0;``    ``}``    ``// if N is divides completely``    ``// into less than k groups``    ``if` `(left == 0)``        ``return` `0;` `    ``// If the subproblem has been``    ``// solved, use the value``    ``if` `(dp[pos][prev][left] != -1)``        ``return` `dp[pos][prev][left];` `    ``int` `answer = 0;``    ``// put all possible values``    ``// greater equal to prev``    ``for` `(``int` `i = prev; i <= left; i++) {``        ``answer += calculate(pos + 1, i,``                           ``left - i, k);``    ``}` `    ``return` `dp[pos][prev][left] = answer;``}` `// Function to count the number of``// ways to divide the number N in groups``int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``// Initialize DP Table as -1``    ``memset``(dp, -1, ``sizeof``(dp));` `    ``return` `calculate(0, 1, n, k);``}` `// Driver Code``int` `main()``{``    ``int` `N = 8;``    ``int` `K = 4;` `    ``cout << countWaystoDivide(N, K);``    ``return` `0;``}`

## Java

 `// Java implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements``import` `java.util.*;``class` `GFG{`` ` `// DP Table``static` `int` `[][][]dp = ``new` `int``[``500``][``500``][``500``];`` ` `// Function to count the number``// of ways to divide the number N``// in groups such that each group``// has K number of elements``static` `int` `calculate(``int` `pos, ``int` `prev,``                     ``int` `left, ``int` `k)``{``    ``// Base Case``    ``if` `(pos == k)``    ``{``        ``if` `(left == ``0``)``            ``return` `1``;``        ``else``            ``return` `0``;``    ``}``  ` `    ``// if N is divides completely``    ``// into less than k groups``    ``if` `(left == ``0``)``        ``return` `0``;`` ` `    ``// If the subproblem has been``    ``// solved, use the value``    ``if` `(dp[pos][prev][left] != -``1``)``        ``return` `dp[pos][prev][left];`` ` `    ``int` `answer = ``0``;``  ` `    ``// put all possible values``    ``// greater equal to prev``    ``for` `(``int` `i = prev; i <= left; i++)``    ``{``        ``answer += calculate(pos + ``1``, i,``                           ``left - i, k);``    ``}`` ` `    ``return` `dp[pos][prev][left] = answer;``}`` ` `// Function to count the number of``// ways to divide the number N in groups``static` `int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``// Initialize DP Table as -1``        ``for` `(``int` `i = ``0``; i < ``500``; i++)``        ``{``            ``for` `(``int` `j = ``0``; j < ``500``; j++)``            ``{``                ``for` `(``int` `l = ``0``; l < ``500``; l++)``                    ``dp[i][j][l] = -``1``;``            ``}``        ``}`` ` `    ``return` `calculate(``0``, ``1``, n, k);``}`` ` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``8``;``    ``int` `K = ``4``;`` ` `    ``System.out.print(countWaystoDivide(N, K));``}``}` `// This code is contributed by Rajput-Ji`

## Python3

 `# Python3 implementation to count the``# number of ways to divide N in``# groups such that each group``# has K number of elements`` ` `# DP Table``dp ``=` `[[[``0` `for` `i ``in` `range``(``50``)]``             ``for` `j ``in` `range``(``50``)]``          ``for` `j ``in` `range``(``50``)]` `# Function to count the number``# of ways to divide the number N``# in groups such that each group``# has K number of elements``def` `calculate(pos, prev, left, k):``   ` `    ``# Base Case``    ``if` `(pos ``=``=` `k):``        ``if` `(left ``=``=` `0``):``            ``return` `1``;``        ``else``:``            ``return` `0``;`` ` `    ``# if N is divides completely``    ``# into less than k groups``    ``if` `(left ``=``=` `0``):``        ``return` `0``;`` ` `    ``# If the subproblem has been``    ``# solved, use the value``    ``if` `(dp[pos][prev][left] !``=` `-``1``):``        ``return` `dp[pos][prev][left];`` ` `    ``answer ``=` `0``;`` ` `    ``# put all possible values``    ``# greater equal to prev``    ``for` `i ``in` `range``(prev,left``+``1``):``        ``answer ``+``=` `calculate(pos ``+` `1``, i,``                            ``left ``-` `i, k);``    ``dp[pos][prev][left] ``=` `answer;``    ``return` `dp[pos][prev][left];`` ` `# Function to count the number of``# ways to divide the number N in groups``def` `countWaystoDivide(n, k):``  ` `    ``# Initialize DP Table as -1``    ``for` `i ``in` `range``(``50``):``        ``for` `j ``in` `range``(``50``):``            ``for` `l ``in` `range``(``50``):``                ``dp[i][j][l] ``=` `-``1``;`` ` `    ``return` `calculate(``0``, ``1``, n, k);`` ` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `8``;``    ``K ``=` `4``;`` ` `    ``print``(countWaystoDivide(N, K));`` ` `# This code is contributed by Rajput-Ji`

## C#

 `// C# implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements``using` `System;``class` `GFG{`` ` `// DP Table``static` `int` `[,,]dp = ``new` `int``[50, 50, 50];`` ` `// Function to count the number``// of ways to divide the number N``// in groups such that each group``// has K number of elements``static` `int` `calculate(``int` `pos, ``int` `prev,``                     ``int` `left, ``int` `k)``{``    ``// Base Case``    ``if` `(pos == k)``    ``{``        ``if` `(left == 0)``            ``return` `1;``        ``else``            ``return` `0;``    ``}``  ` `    ``// if N is divides completely``    ``// into less than k groups``    ``if` `(left == 0)``        ``return` `0;`` ` `    ``// If the subproblem has been``    ``// solved, use the value``    ``if` `(dp[pos, prev, left] != -1)``        ``return` `dp[pos, prev, left];`` ` `    ``int` `answer = 0;``  ` `    ``// put all possible values``    ``// greater equal to prev``    ``for` `(``int` `i = prev; i <= left; i++)``    ``{``        ``answer += calculate(pos + 1, i,``                           ``left - i, k);``    ``}`` ` `    ``return` `dp[pos, prev, left] = answer;``}`` ` `// Function to count the number of``// ways to divide the number N in groups``static` `int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``// Initialize DP Table as -1``        ``for` `(``int` `i = 0; i < 50; i++)``        ``{``            ``for` `(``int` `j = 0; j < 50; j++)``            ``{``                ``for` `(``int` `l = 0; l < 50; l++)``                    ``dp[i, j, l] = -1;``            ``}``        ``}`` ` `    ``return` `calculate(0, 1, n, k);``}`` ` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 8;``    ``int` `K = 4;`` ` `    ``Console.Write(countWaystoDivide(N, K));``}``}` `// This code is contributed by gauravrajput1`

## Javascript

 ``

Output:

`5`

Time complexity: O(N^3 * K)
Auxiliary Space: O(N^2 * K).

Bottom-Up DP:  We are asked to find CountWaystoDivide(n,k) So the recurrence approach and explanation of DP is:

CountWaystoDivide( n , k ) = CountWaystoDivide( n-k , k ) + CountWaystoDivide( n-1 , k-1 )

Explanation:
Divide CountWaystoDivide( n , k ) into two parts where

1. If first element is 1 then the rest form a total  of n-1 divide into k-1 so CountWaystoDivide( n-1 , k-1 )
2. If first element is greater than 1 then, we can subtract 1 from every element and get a valid partition of n-k into k parts, hence CountWaystoDivide( n-1 , k-1 ).

Mathematical Explanation of DP:

1. As each group must have at least one person, so, give each group one person, then we are left with n-k persons, which can we divided into 1,2,3..or k groups. Thus our dp will be: dp[n][k] = dp[n-k] + dp[n-k] + dp[n-k] + …. + dp[n-k][k].
2. At first look, the previous might give O(N3) vibes, but with a little manipulation we can optimize it:
dp[n][k] = dp[(n-1)-(k-1)] + dp[(n-1)-(k-1)] + … + dp[(n-1)-(k-1)][k-1] + dp[(n-1)-(k-1)][k]
From the recurrence, we can write:
dp[n][k] = dp[n-1][k-1] + dp[n-k][k]

Steps to solve the problem using DP:

• Initialize a 2-D array dp[][] of size n+1, k+1 where dp[i][j] will store the optimal solution to divide n into k groups.
• For each value from i=0 to n, dp[n] will be 1 since total ways to divide n into 1 is 1. also dp will be 1.

DP states are updated as follows:

• If i>=j then dp[i][j] = dp[i-1][j-1] + dp[i-j][j]
• otherwise i-j<0 and dp[i-j][j] become zero so dp[i][j] = dp[i-1][j-1]

## C++

 `// C++ implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements` `#include ` `using` `namespace` `std;` `// Function to count the number of``// ways to divide the number N in groups``int` `countWaystoDivide(``int` `n, ``int` `k)``{``    ``if` `(n < k)``        ``return` `0; ``// When n is less than k, No way to divide``                  ``// into groups``    ``vector > dp(n + 1, vector<``int``>(k + 1));` `    ``for` `(``int` `i = 1; i <= n; i++)``        ``dp[i]``            ``= 1; ``// exact one way to divide n to 1 group``    ``dp = 1;` `    ``for` `(``int` `i = 1; i <= n; i++) {``        ``for` `(``int` `j = 2; j <= k; j++) {``            ``if` `(i >= j)``                ``dp[i][j] = dp[i - j][j] + dp[i - 1][j - 1];``            ``else``                ``dp[i][j]``                    ``= dp[i - 1][j - 1]; ``// i

## Java

 `// Java implementation to count the``// number of ways to divide N in``// groups such that each group``// has K number of elements``import` `java.io.*;` `class` `GFG {` `  ``static` `int` `countWaystoDivide(``int` `n, ``int` `k)``  ``{``    ``if` `(n < k)``      ``return` `0``; ``// When n is less than k, No way to divide``    ``// into groups` `    ``int` `[][]dp = ``new` `int``[n+``1``][k+``1``];``    ``for` `(``int` `i = ``1``; i <= n; i++)``      ``dp[i][``1``]``      ``= ``1``; ``// exact one way to divide n to 1 group``    ``dp[``0``][``0``] = ``1``;` `    ``for` `(``int` `i = ``1``; i <= n; i++) {``      ``for` `(``int` `j = ``2``; j <= k; j++) {``        ``if` `(i >= j)``          ``dp[i][j] = dp[i - j][j] + dp[i - ``1``][j - ``1``];``        ``else``          ``dp[i][j]``          ``= dp[i - ``1``][j - ``1``]; ``// i

## Python3

 `# Python3 implementation to count the``# number of ways to divide N in``# groups such that each group``# has K number of elements` `# DP Table``# Function to count the number of``# ways to divide the number N in groups`  `def` `countWaystoDivide(n, k):``    ``if``(n < k):``        ``return` `0` `    ``dp ``=` `[[``0` `for` `i ``in` `range``(k``+``1``)] ``for` `i ``in` `range``(n``+``1``)]``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``dp[i][``1``] ``=` `1``    ``dp[``0``][``0``] ``=` `1``    ``for` `i ``in` `range``(``1``, n``+``1``):``        ``for` `j ``in` `range``(``2``, k``+``1``):``            ``if``(i >``=` `j):``                ``dp[i][j] ``=` `dp[i``-``1``][j``-``1``] ``+` `dp[i``-``j][j]` `            ``else``:``                ``dp[i][j] ``=` `dp[i``-``1``][j``-``1``]``    ``return` `dp[n][k]`  `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``N ``=` `8``    ``K ``=` `4` `    ``print``(countWaystoDivide(N, K))` `# This code is contributed by Rajput-Ji`

## Javascript

 ``
Output
`5`

Time complexity: O(N * K)
Auxiliary Space:  O(N * K)

My Personal Notes arrow_drop_up