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

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++

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// 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 <bits/stdc++.h>
 
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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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++

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// 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 <bits/stdc++.h>
 
using namespace std;
 
const int mod = 1000000007;
 
// DP Table
int dp[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]
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)
{
    // Intialize 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;
}

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Java

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// 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)
{
     
    // Intialize 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

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Python3

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# 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

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C#

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// 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)
{
     
    // Intialize 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

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Output: 

10


 

Time Complexity: O(N2).
 

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