Given three integers N, P, and K, the task is to find the total number of ways to divide N into K integers having sum N where each integer is ≥ P.
Examples:
Input: K = 3, N = 8, P = 2
Output: 6
Explanation:
Six possible solutions are: {2, 2, 4}, {2, 3, 3}, {2, 4, 2}, {3, 2, 3}, {3, 3, 2}, {4, 2, 2}
Each of the above integers in all combinations are ≥ P(= 2) and sum of all the combinations is N(=8).Input: K = 2, N = 7, P = 2
Output: 4
Explanation:
Four possible solutions are: {4, 3}, {3, 4}, {5, 2}, {2, 5}
Each of the above integers in all combinations are ≥ P(= 2) and sum of all the combinations is N(= 7).
Naive Approach: The simplest approach is to try all possible combinations of K integers having sum N and check if they satisfy the above conditions or not. There are K positions and in each position, N integers can be placed. Therefore, the total number of combinations to check is NK.
Time Complexity: O(NK)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, the idea is to use the approach discussed in this article and the given above conditions can be written in terms of a linear equation is:
To find the number of solutions for the equation X1 + X2 + X3 + … + XK = N where Xi ≥ P.
Assume Xi‘ = Xi – P.
Therefore, the equation reduces to:
X1‘ + X2‘ + X3‘ + … + XK‘ = N – K*P
As per the above equations, the given problem reduces to finding total number of ways to split N – K * P into K integers. Print the total number of ways found above as the required answer.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include <bits/stdc++.h>using namespace std;// Function that finds the value of// the Binomial Coefficient C(n, k)int binomialCoeff(int n, int k){ int C[n + 1][k + 1]; int i, j; // Stores the value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= min(i, k); j++) { // Base Case if (j == 0 || j == i) C[i][j] = 1; // Find the value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } // Return the value of C(N, K) return C[n][k];}// Function that count the number of// ways to divide N into K integers// >= P such that their sum is Nint waysToSplitN(int k, int n, int P){ // Update the value of N int new_N = n - k * P; // Find the binomial coeffient // recursively return binomialCoeff(new_N + k - 1, new_N);}// Driver Codeint main(){ // Given K, N, and P int K = 3, N = 8, P = 2; cout << waysToSplitN(K, N, P); return 0;} |
Java
// Java program for// the above approachimport java.util.*;class GFG{// Function that finds the value of// the Binomial Coefficient C(n, k)static int binomialCoeff(int n, int k){ int [][]C = new int[n + 1][k + 1]; int i, j; // Stores the value of Binomial // Coefficient in bottom up manner for (i = 0; i <= n; i++) { for (j = 0; j <= Math.min(i, k); j++) { // Base Case if (j == 0 || j == i) C[i][j] = 1; // Find the value using // previously stored values else C[i][j] = C[i - 1][j - 1] + C[i - 1][j]; } } // Return the value of C(N, K) return C[n][k];}// Function that count the number of// ways to divide N into K integers// >= P such that their sum is Nstatic int waysToSplitN(int k, int n, int P){ // Update the value of N int new_N = n - k * P; // Find the binomial coeffient // recursively return binomialCoeff(new_N + k - 1, new_N);}// Driver Codepublic static void main(String[] args){ // Given K, N, and P int K = 3, N = 8, P = 2; System.out.print(waysToSplitN(K, N, P));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program for the above approach# Function that finds the value of# the Binomial Coefficient C(n, k)def binomialCoeff(n, k): C = [[0 for x in range(k + 1)] for y in range(n + 1)] # Stores the value of Binomial # Coefficient in bottom up manner for i in range(n + 1): for j in range(min(i, k) + 1): # Base Case if(j == 0 or j == i): C[i][j] = 1 # Find the value using # previously stored values else: C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) # Return the value of C(N, K) return C[n][k]# Function that count the number of# ways to divide N into K integers# >= P such that their sum is Ndef waysToSplitN(k, n, P): # Update the value of N new_N = n - k * P # Find the binomial coeffient # recursively return binomialCoeff(new_N + k - 1, new_N)# Driver Code# Given K, N, and PK = 3N = 8P = 2print(waysToSplitN(K, N, P))# This code is contributed by Shivam Singh |
C#
// C# program for the above approachusing System;class GFG{ // Function that finds the value of// the Binomial Coefficient C(n, k)static int binomialCoeff(int n, int k){ int [,] C = new int[n + 1, k + 1]; int i, j; // Stores the value of Binomial // Coefficient in bottom up manner for(i = 0; i <= n; i++) { for(j = 0; j <= Math.Min(i, k); j++) { // Base Case if (j == 0 || j == i) C[i, j] = 1; // Find the value using // previously stored values else C[i, j] = C[i - 1, j - 1] + C[i - 1, j]; } } // Return the value of C(N, K) return C[n, k];} // Function that count the number of// ways to divide N into K integers// >= P such that their sum is Nstatic int waysToSplitN(int k, int n, int P){ // Update the value of N int new_N = n - k * P; // Find the binomial coeffient // recursively return binomialCoeff(new_N + k - 1, new_N);} // Driver Codepublic static void Main(){ // Given K, N, and P int K = 3, N = 8, P = 2; Console.Write(waysToSplitN(K, N, P));}}// This code is contributed by sanjoy_62 |
6
Time Complexity: O(N2)
Auxiliary Space: O(N2)
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