Count the number of ways to divide N in k groups incrementally
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:
Their 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
Naive Approach: We can solve this problem using recursion. At each step of recursion put all the values greater than equal to the previous computed value.
Below is the implementation of the above approach:
C++
// 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; } |
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Java
// 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 |
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C#
// 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 |
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Output:
5
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 // groups such that each group // has K number of elements #include <bits/stdc++.h> using namespace std; // DP Table int dp[500][500][500]; // 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) { // Intialize 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; } |
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Time complexity : O(N2 * K)
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