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: 5
Explanation:
There 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++ implementation to count the // number of ways to divide N in // groups such that each group // has K number of elements #include <bits/stdc++.h> 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 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 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# 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 |
<script> // Javascript 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 function 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;
let answer = 0;
// put all possible values
// greater equal to prev
for (let 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 function countWaystoDivide(n, k)
{ return calculate(0, 1, n, k);
} // Driver Code let N = 8;
let K = 4;
document.write(countWaystoDivide(N, K));
// This code is contributed by Mayank Tyagi </script> |
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++ implementation to count the // number of ways to divide N in // groups such that each group // has K number of elements #include <bits/stdc++.h> using namespace std;
// DP Table int dp[100][100][100];
// 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 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 [ 100 ][ 100 ][ 100 ];
// 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 < 100 ; i++)
{
for ( int j = 0 ; j < 100 ; j++)
{
for ( int l = 0 ; l < 100 ; 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 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# 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 |
<script> // JavaScript implementation to count the // number of ways to divide N in // groups such that each group // has K number of elements // DP Table let dp = new Array(500);
for (let i=0;i<500;i++)
{ dp[i]= new Array(500);
for (let j=0;j<500;j++)
{
dp[i][j]= new Array(500);
for (let k=0;k<500;k++)
dp[i][j][k]=0;
}
} // Function to count the number // of ways to divide the number N // in groups such that each group // has K number of elements function 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];
let answer = 0;
// put all possible values
// greater equal to prev
for (let 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 function countWaystoDivide(n,k)
{ // Initialize DP Table as -1
for (let i = 0; i < 500; i++)
{
for (let j = 0; j < 500; j++)
{
for (let l = 0; l < 500; l++)
dp[i][j][l] = -1;
}
}
return calculate(0, 1, n, k);
} // Driver Code let N = 8; let K = 4; document.write(countWaystoDivide(N, K)); // This code is contributed by unknown2108 </script> |
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
- If first element is 1 then the rest form a total of n-1 divide into k-1 so CountWaystoDivide( n-1 , k-1 )
- 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-k , k ).
Mathematical Explanation of DP:
- 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][1] + dp[n-k][2] + dp[n-k][3] + …. + dp[n-k][k].
- 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)][1] + dp[(n-1)-(k-1)][2] + … + 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][1] will be 1 since total ways to divide n into 1 is 1. also dp[0][0] 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++ implementation to count the // number of ways to divide N in // groups such that each group // has K number of elements #include <bits/stdc++.h> 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<vector< int > > dp(n + 1, vector< int >(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<j so dp[i-j][j]
// becomes zero
}
}
return dp[n][k]; // returning number of ways to divide N
// in k groups
} // Driver Code int main()
{ int N = 8;
int K = 4;
cout << countWaystoDivide(N, K);
return 0;
} |
// 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<j so dp[i-j][j]
// becomes zero
}
}
return dp[n][k]; // returning number of ways to divide N
// in k groups
}
// Driver code
public static void main (String[] args)
{
int N = 8 ;
int K = 4 ;
System.out.println(countWaystoDivide(N, K));
}
} // This code is contributed by rohitsingh07052. |
# 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 |
// C# implementation to count the // number of ways to divide N in // groups such that each group // has K number of elements using System;
using System.Collections.Generic;
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<j so dp[i-j][j]
// becomes zero
}
}
return dp[n,k]; // returning number of ways to divide N
// in k groups
}
static void Main() {
int N = 8;
int K = 4;
Console.Write(countWaystoDivide(N, K));
}
} // This code is contributed by rameshtravel07. |
<script> // Javascript 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
function countWaystoDivide(n, k)
{
if (n < k)
return 0; // When n is less than k, No way to divide
// into groups
let dp = new Array(n + 1);
for (let i = 0; i < n + 1; i++)
{
dp[i] = new Array(k + 1);
for (let j = 0; j < k + 1; j++)
{
dp[i][j] = 0;
}
}
for (let i = 1; i <= n; i++)
dp[i][1]
= 1; // exact one way to divide n to 1 group
dp[0][0] = 1;
for (let i = 1; i <= n; i++) {
for (let 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<j so dp[i-j][j]
// becomes zero
}
}
return dp[n][k]; // returning number of ways to divide N
// in k groups
}
let N = 8;
let K = 4;
document.write(countWaystoDivide(N, K));
// This code is contributed by mukesh07. </script> |
5
Time complexity: O(N * K)
Auxiliary Space: O(N * K)