# Number of ways to split N as sum of K numbers from the given range

• Last Updated : 09 Aug, 2021

Given four positive integer N, K, L, and R, the task is to split N as sum of K numbers lying in the range [L, R].
Note: Since the number of ways can be very large. Output answer modulo 1000000007.
Examples:

Input : N = 12, K = 3, L = 1, R = 5
Output : 10
{2, 5, 5}
{3, 4, 5}
{3, 5, 4}
{4, 3, 5}
{4, 4, 4}
{4, 5, 3}
{5, 2, 5}
{5, 3, 4}
{5, 4, 3}
{5, 5, 2}

Input : N = 23, K = 4, L = 2, R = 10
Output : 480

Naive Approach
We can solve the problem using recursion. At each step of recursion try to make a group of size L to R all
There will be two arguments in the recursion which changes:

• pos – the group number
• left – how much of N is left to be distributed

Below is the implementation of the above approach:

## C++

 `// C++ implementation to count the``// number of ways to divide N in K``// groups such that each group``// has elements in range [L, R]` `#include ` `using` `namespace` `std;` `const` `int` `mod = 1000000007;` `// Function to count the number``// of ways to divide the number N``// in K groups such that each group``// has number of elements in range [L, R]``int` `calculate(``int` `pos, ``int` `left,``              ``int` `k, ``int` `L, ``int` `R)``{``    ``// 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 = L; i <= R; i++) {``        ``if` `(i > left)``            ``break``;``        ``answer = (answer + calculate(pos + 1,``                                     ``left - i, k, L, R))``                 ``% mod;``    ``}``    ``return` `answer;``}` `// Function to count the number of``// ways to divide the number N``int` `countWaystoDivide(``int` `n, ``int` `k,``                      ``int` `L, ``int` `R)``{``    ``return` `calculate(0, n, k, L, R);``}` `// Driver Code``int` `main()``{``    ``int` `N = 12;``    ``int` `K = 3;``    ``int` `L = 1;``    ``int` `R = 5;` `    ``cout << countWaystoDivide(N, K, L, R);``    ``return` `0;``}`

## Java

 `// Java implementation to count the``// number of ways to divide N in K``// groups such that each group``// has elements in range [L, R]``class` `GFG{` `static` `int` `mod = ``1000000007``;` `// Function to count the number``// of ways to divide the number N``// in K groups such that each group``// has number of elements in range [L, R]``static` `int` `calculate(``int` `pos, ``int` `left,``                     ``int` `k, ``int` `L, ``int` `R)``{``    ``// 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 = L; i <= R; i++)``    ``{``       ``if` `(i > left)``           ``break``;``       ``answer = ((answer +``                  ``calculate(pos + ``1``,left - i,``                            ``k, L, R)) % mod);``    ``}``    ``return` `answer;``}` `// Function to count the number of``// ways to divide the number N``static` `int` `countWaystoDivide(``int` `n, ``int` `k,``                             ``int` `L, ``int` `R)``{``    ``return` `calculate(``0``, n, k, L, R);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``12``;``    ``int` `K = ``3``;``    ``int` `L = ``1``;``    ``int` `R = ``5``;` `    ``System.out.print(countWaystoDivide(N, K, L, R));``}``}` `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 implementation to count the``# number of ways to divide N in K``# groups such that each group``# has elements in range [L, R]` `mod ``=` `1000000007` `# Function to count the number``# of ways to divide the number N``# in K groups such that each group``# has number of elements in range [L, R]``def` `calculate(pos, left, k, L, R):``    ` `    ``# 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``(L, R ``+` `1``):``        ``if` `(i > left):``            ``break``        ``answer ``=` `(answer ``+``                  ``calculate(pos ``+` `1``,``                            ``left ``-` `i, k, L, R)) ``%` `mod``    ``return` `answer` `# Function to count the number of``# ways to divide the number N``def` `countWaystoDivide(n, k, L, R):``    ``return` `calculate(``0``, n, k, L, R)` `# Driver Code``N ``=` `12``K ``=` `3``L ``=` `1``R ``=` `5` `print``(countWaystoDivide(N, K, L, R))``    ` `# This code is contributed by divyamohan123`

## C#

 `// C# implementation to count the``// number of ways to divide N in K``// groups such that each group``// has elements in range [L, R]``using` `System;` `class` `GFG{` `static` `int` `mod = 1000000007;` `// Function to count the number``// of ways to divide the number N``// in K groups such that each group``// has number of elements in range [L, R]``static` `int` `calculate(``int` `pos, ``int` `left,``                     ``int` `k, ``int` `L, ``int` `R)``{``    ` `    ``// 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 = L; i <= R; i++)``    ``{``       ``if` `(i > left)``           ``break``;``       ``answer = ((answer +``                  ``calculate(pos + 1,left - i,``                            ``k, L, R)) % mod);``    ``}``    ``return` `answer;``}` `// Function to count the number of``// ways to divide the number N``static` `int` `countWaystoDivide(``int` `n, ``int` `k,``                             ``int` `L, ``int` `R)``{``    ``return` `calculate(0, n, k, L, R);``}` `// Driver Code``public` `static` `void` `Main(String[] args)``{``    ``int` `N = 12;``    ``int` `K = 3;``    ``int` `L = 1;``    ``int` `R = 5;` `    ``Console.Write(countWaystoDivide(N, K, L, R));``}``}` `// This code is contributed by Amit Katiyar`

## Javascript

 ``

Output:

`10`

Time Complexity: O(NK).

Efficient 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 DP table.

Below is the implementation of the above approach:

## C++

 `// C++ implementation to count the``// number of ways to divide N in K``// groups such that each group``// has elements in range [L, R]` `#include ` `using` `namespace` `std;` `const` `int` `mod = 1000000007;` `// DP Table``int` `dp;` `// Function to count the number``// of ways to divide the number N``// in K groups such that each group``// has number of elements in range [L, R]``int` `calculate(``int` `pos, ``int` `left,``              ``int` `k, ``int` `L, ``int` `R)``{``    ``// 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][left] != -1)``        ``return` `dp[pos][left];` `    ``int` `answer = 0;` `    ``// put all possible values``    ``// greater equal to prev``    ``for` `(``int` `i = L; i <= R; i++) {``        ``if` `(i > left)``            ``break``;``        ``answer = (answer + calculate(pos + 1,``                                     ``left - i, k, L, R))``                 ``% mod;``    ``}``    ``return` `dp[pos][left] = answer;``}` `// Function to count the number of``// ways to divide the number N``int` `countWaystoDivide(``int` `n, ``int` `k,``                      ``int` `L, ``int` `R)``{``    ``// Initialize DP Table as -1``    ``memset``(dp, -1, ``sizeof``(dp));``    ``return` `calculate(0, n, k, L, R);``}` `// Driver Code``int` `main()``{``    ``int` `N = 12;``    ``int` `K = 3;``    ``int` `L = 1;``    ``int` `R = 5;` `    ``cout << countWaystoDivide(N, K, L, R);``    ``return` `0;``}`

## Java

 `// Java implementation to count the``// number of ways to divide N in K``// groups such that each group``// has elements in range [L, R]``class` `GFG{` `static` `int` `mod = ``1000000007``;` `// DP Table``static` `int` `[][]dp = ``new` `int``[``1000``][``1000``];` `// Function to count the number``// of ways to divide the number N``// in K groups such that each group``// has number of elements in range [L, R]``static` `int` `calculate(``int` `pos, ``int` `left,``                     ``int` `k, ``int` `L, ``int` `R)``{``    ` `    ``// 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][left] != -``1``)``        ``return` `dp[pos][left];` `    ``int` `answer = ``0``;` `    ``// Put all possible values``    ``// greater equal to prev``    ``for``(``int` `i = L; i <= R; i++)``    ``{``       ``if` `(i > left)``           ``break``;``       ``answer = (answer + calculate(pos + ``1``, left - i,``                                      ``k, L, R)) % mod;``    ``}``    ``return` `dp[pos][left] = answer;``}` `// Function to count the number of``// ways to divide the number N``static` `int` `countWaystoDivide(``int` `n, ``int` `k,``                             ``int` `L, ``int` `R)``{``    ` `    ``// Initialize DP Table as -1``    ``for``(``int` `i = ``0``; i < ``1000``; i++)``    ``{``       ``for``(``int` `j = ``0``; j < ``1000``; j++)``       ``{``          ``dp[i][j] = -``1``;``       ``}``    ``}``    ``return` `calculate(``0``, n, k, L, R);``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `N = ``12``;``    ``int` `K = ``3``;``    ``int` `L = ``1``;``    ``int` `R = ``5``;` `    ``System.out.print(countWaystoDivide(N, K, L, R));``}``}` `// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 implementation to count the``# number of ways to divide N in K``# groups such that each group``# has elements in range [L, R]``mod ``=` `1000000007`` ` `# DP Table``dp ``=` `[[``-``1` `for` `j ``in` `range``(``1000``)]``          ``for` `i ``in` `range``(``1000``)]`` ` `# Function to count the number``# of ways to divide the number N``# in K groups such that each group``# has number of elements in range [L, R]``def` `calculate(pos, left, k, L, R):` `    ``# 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][left] !``=` `-``1``):``        ``return` `dp[pos][left]`` ` `    ``answer ``=` `0`` ` `    ``# put all possible values``    ``# greater equal to prev``    ``for` `i ``in` `range``(L, R ``+` `1``):``        ``if` `(i > left):``            ``break``        ` `        ``answer ``=` `(answer ``+``                  ``calculate(pos ``+` `1``,``                           ``left ``-` `i,``                           ``k, L, R)) ``%` `mod``    ` `    ``dp[pos][left] ``=` `answer``    ` `    ``return` `answer` `# Function to count the number of``# ways to divide the number N``def` `countWaystoDivide(n, k, L, R):``    ` `    ``return` `calculate(``0``, n, k, L, R)` `# Driver code``if` `__name__``=``=``"__main__"``:``    ` `    ``N ``=` `12``    ``K ``=` `3``    ``L ``=` `1``    ``R ``=` `5`` ` `    ``print``(countWaystoDivide(N, K, L, R))``    ` `# This code is contributed by rutvik_56`

## C#

 `// C# implementation to count the``// number of ways to divide N in K``// groups such that each group``// has elements in range [L, R]``using` `System;``class` `GFG{` `static` `int` `mod = 1000000007;` `// DP Table``static` `int` `[,]dp = ``new` `int``[1000, 1000];` `// Function to count the number``// of ways to divide the number N``// in K groups such that each group``// has number of elements in range [L, R]``static` `int` `calculate(``int` `pos, ``int` `left,``                     ``int` `k, ``int` `L, ``int` `R)``{``    ` `    ``// 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, left] != -1)``        ``return` `dp[pos, left];` `    ``int` `answer = 0;` `    ``// Put all possible values``    ``// greater equal to prev``    ``for``(``int` `i = L; i <= R; i++)``    ``{``        ``if` `(i > left)``            ``break``;``        ``answer = (answer + calculate(pos + 1, left - i,``                                       ``k, L, R)) % mod;``    ``}``    ``return` `dp[pos, left] = answer;``}` `// Function to count the number of``// ways to divide the number N``static` `int` `countWaystoDivide(``int` `n, ``int` `k,``                             ``int` `L, ``int` `R)``{``    ` `    ``// Initialize DP Table as -1``    ``for``(``int` `i = 0; i < 1000; i++)``    ``{``        ``for``(``int` `j = 0; j < 1000; j++)``        ``{``            ``dp[i, j] = -1;``        ``}``    ``}``    ``return` `calculate(0, n, k, L, R);``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int` `N = 12;``    ``int` `K = 3;``    ``int` `L = 1;``    ``int` `R = 5;` `    ``Console.Write(countWaystoDivide(N, K, L, R));``}``}` `// This code is contributed by Code_Mech`

## Javascript

 ``

Output:

`10`

Time Complexity: O(N*K).
Auxiliary Space: O(N*K).

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