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++
// 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 Nint countWaystoDivide(int n, int k, int L, int R){ return calculate(0, n, k, L, R);}// Driver Codeint 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
// 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 Nstatic int countWaystoDivide(int n, int k, int L, int R){ return calculate(0, n, k, L, R);}// Driver Codepublic 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
# 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 = 12K = 3L = 1R = 5print(countWaystoDivide(N, K, L, R)) # This code is contributed by divyamohan123 |
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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 Nstatic int countWaystoDivide(int n, int k, int L, int R){ return calculate(0, n, k, L, R);}// Driver Codepublic 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++
// 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 Tableint 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 Nint 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 Codeint 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
// 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 Tablestatic 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 Nstatic 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 Codepublic 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
# 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 Tabledp = [[-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 Ndef countWaystoDivide(n, k, L, R): return calculate(0, n, k, L, R)# Driver codeif __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#
// 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 Tablestatic 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 Nstatic 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 Codepublic 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).
