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++ 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;
} |
// 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 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# 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 |
<script> // Javascript implementation to count the // number of ways to divide N in K // groups such that each group // has elements in range [L, R] var 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] function 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;
var answer = 0;
// put all possible values
// greater equal to prev
for ( var 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 function countWaystoDivide(n, k, L, R)
{ return calculate(0, n, k, L, R);
} // Driver Code var N = 12;
var K = 3;
var L = 1;
var R = 5;
document.write( countWaystoDivide(N, K, L, R)); // This code is contributed by noob2000. </script> |
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++ 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)
{ // 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 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 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# 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 |
<script> // JavaScript implementation to count the // number of ways to divide N in K // groups such that each group // has elements in range [L, R] var mod = 1000000007;
// DP Table var dp = Array.from(Array(1000), ()=>Array(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] function 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];
var answer = 0;
// put all possible values
// greater equal to prev
for ( var 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 function countWaystoDivide(n, k, L, R)
{ // Initialize DP Table as -1
dp = Array.from(Array(1000), ()=>Array(1000).fill(-1));
return calculate(0, n, k, L, R);
} // Driver Code var N = 12;
var K = 3;
var L = 1;
var R = 5;
document.write( countWaystoDivide(N, K, L, R)); </script> |
10
Time Complexity: O(N*K).
Auxiliary Space: O(N*K).
Efficient approach : Using DP Tabulation method ( Iterative approach )
The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls.
Steps to solve this problem :
- Create a vector to store the solution of the subproblems.
- Initialize the table with base cases
- Fill up the table iteratively
- Return the final solution
Implementation :
#include <bits/stdc++.h> using namespace std;
const int mod = 1000000007;
int dp[1000][1000];
// Function to count the number // of ways to divide the number N // in K groups such that each group int countWaystoDivide( int n, int k, int L, int R) {
memset (dp, 0, sizeof (dp));
// Base Case
dp[0][0] = 1;
// Loop to traverse the matrix and compute the subproblems
for ( int i = 1; i <= k; i++) {
for ( int j = 0; j <= n; j++) {
for ( int l = L; l <= R; l++) {
if (l > j) break ;
// update Dp with its previous computations
dp[i][j] = (dp[i][j] + dp[i - 1][j - l]) % mod;
}
}
}
//return answer stored at last index
return dp[k][n];
} // Driver Code int main() {
int N = 12;
int K = 3;
int L = 1;
int R = 5;
// function call
cout << countWaystoDivide(N, K, L, R);
return 0;
} |
//GFG //Java code for the given approach import java.util.*;
public class Main {
static final int MOD = 1000000007 ;
public static int countWaystoDivide( int n, int k, int L, int R) {
int [][] dp = new int [k+ 1 ][n+ 1 ];
// Base Case
dp[ 0 ][ 0 ] = 1 ;
// Loop to traverse the matrix and compute the subproblems
for ( int i = 1 ; i <= k; i++) {
for ( int j = 0 ; j <= n; j++) {
for ( int l = L; l <= R; l++) {
if (l > j) break ;
// update Dp with its previous computations
dp[i][j] = (dp[i][j] + dp[i - 1 ][j - l]) % MOD;
}
}
}
//return answer stored at last index
return dp[k][n];
}
public static void main(String[] args) {
int N = 12 ;
int K = 3 ;
int L = 1 ;
int R = 5 ;
// function call
System.out.println(countWaystoDivide(N, K, L, R));
}
} //This code is written Sundaram |
mod = 1000000007
dp = []
# Function to count the number # of ways to divide the number N # in K groups such that each group def countWaystoDivide(n, k, L, R):
for i in range (k + 1 ):
dp.append([ 0 ] * (n + 1 ))
# Base Case
dp[ 0 ][ 0 ] = 1
# Loop to traverse the matrix and compute the subproblems
for i in range ( 1 , k + 1 ):
for j in range (n + 1 ):
for l in range (L, R + 1 ):
if l > j:
break
# update Dp with its previous computations
dp[i][j] = (dp[i][j] + dp[i - 1 ][j - l]) % mod
#return answer stored at last index
return dp[k][n]
# Driver Code N = 12
K = 3
L = 1
R = 5
# function call print (countWaystoDivide(N, K, L, R))
|
using System;
public class GFG {
const int mod = 1000000007;
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
static int countWaystoDivide( int n, int k, int L, int R)
{
Array.Clear(dp, 0, dp.Length);
// Base Case
dp[0, 0] = 1;
// Loop to traverse the matrix and compute the
// subproblems
for ( int i = 1; i <= k; i++) {
for ( int j = 0; j <= n; j++) {
for ( int l = L; l <= R; l++) {
if (l > j)
break ;
// update Dp with its previous
// computations
dp[i, j] = (dp[i, j] + dp[i - 1, j - l])
% mod;
}
}
}
// return answer stored at last index
return dp[k, n];
}
// Driver Code
public static void Main()
{
int N = 12;
int K = 3;
int L = 1;
int R = 5;
// function call
Console.WriteLine(countWaystoDivide(N, K, L, R));
}
} // This code is contributed by user_dtewbxkn77n |
const mod = 1000000007; const dp = []; // Function to count the number // of ways to divide the number N // in K groups such that each group function countWaystoDivide(n, k, L, R) {
for (let i = 0; i <= k; i++) {
dp[i] = [];
for (let j = 0; j <= n; j++) {
dp[i][j] = 0;
}
}
// Base Case
dp[0][0] = 1;
// Loop to traverse the matrix and compute the subproblems
for (let i = 1; i <= k; i++) {
for (let j = 0; j <= n; j++) {
for (let l = L; l <= R; l++) {
if (l > j) break ;
// update Dp with its previous computations
dp[i][j] = (dp[i][j] + dp[i - 1][j - l]) % mod;
}
}
}
//return answer stored at last index
return dp[k][n];
} // Driver Code const N = 12; const K = 3; const L = 1; const R = 5; // function call console.log(countWaystoDivide(N, K, L, R)); |
10
Time Complexity: O(N*K).
Auxiliary Space: O(N*K).