Bell Numbers (Number of ways to Partition a Set)
Last Updated :
09 Nov, 2024
Given a set of n elements, find the number of ways of partitioning it.
Examples:
Input: n = 2
Output: 2
Explanation: Let the set be {1, 2}. The partitions are {{1},{2}} and {{1, 2}}.
Input: n = 3
Output: 5
Explanation: Let the set be {1, 2, 3}. The partitions are {{1},{2},{3}}, {{1},{2, 3}}, {{2},{1, 3}}, {{3},{1, 2}}, {{1, 2, 3}}.
What is a Bell Number?
Let S(n, k) be total number of partitions of n elements into k sets. The value of the n'th Bell Number is the sum of S(n, k) for k = 1 to n.
Bell(n)=∑​S (n,k) for k ranges from [1,n]
Value of S(n, k) can be defined recursively as, S(n+1, k) = k*S(n, k) + S(n, k-1)
How does above recursive formula work?
When we add a (n+1)'th element to k partitions, there are two possibilities.
1) It is added as a single element set to existing partitions, i.e, S(n, k-1)
2) It is added to all sets of every partition, i.e., k*S(n, k).
First few Bell numbers are 1, 1, 2, 5, 15, 52, 203, ....
Using recursion - O(2 ^ n) Time and O(n) Space
We can recursively calculate the number of ways to partition a set of n elements by considering each element and either placing it in an existing subset or creating a new subset. For each element, we calculate the number of ways to partition the remaining n-1 elements into k subsets and then sum these values for all possible k. This gives us the following recurrence relation for Stirling numbers of the second kind:
- S(n,k) = k * S( n - 1, k) + S(n - 1, k - 1)
Finally, to find the Bell number, we sum S(n, k) for all values of k from 1 to n. This gives us the total number of ways to partition the set.
C++
// C++ code of finding the bellNumber
// using recursion
#include <iostream>
#include <vector>
using namespace std;
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
int stirling(int n, int k) {
// Base cases
if (n == 0 && k == 0) return 1;
if (k == 0 || n == 0) return 0;
if (n == k) return 1;
if (k == 1) return 1;
// Recursive formula
return k * stirling(n - 1, k) + stirling(n - 1, k - 1);
}
// Function to calculate the total number of
// ways to partition a set of `n` elements
int bellNumber(int n) {
int result = 0;
// Sum up Stirling numbers S(n, k) for all
// k from 1 to n
for (int k = 1; k <= n; ++k) {
result += stirling(n, k);
}
return result;
}
int main() {
int n = 5;
int result = bellNumber(n);
cout << result << endl;
return 0;
}
// Java program to find the Bell Number
// using recursion
class GfG {
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
static int stirling(int n, int k) {
// Base cases
if (n == 0 && k == 0) return 1;
if (k == 0 || n == 0) return 0;
if (n == k) return 1;
if (k == 1) return 1;
// Recursive formula
return k * stirling(n - 1, k) + stirling(n - 1, k - 1);
}
// Function to calculate the total number of
// ways to partition a set of `n` elements
static int bellNumber(int n) {
int result = 0;
// Sum up Stirling numbers S(n, k) for
// all k from 1 to n
for (int k = 1; k <= n; ++k) {
result += stirling(n, k);
}
return result;
}
public static void main(String[] args) {
int n = 5;
int result = bellNumber(n);
System.out.println(result);
}
}
# Python program to find the Bell Number using recursion
# Function to compute Stirling numbers of
# the second kind S(n, k) with memoization
def stirling(n, k):
# Base cases
if n == 0 and k == 0: return 1
if k == 0 or n == 0: return 0
if n == k: return 1
if k == 1: return 1
# Recursive formula
return k * stirling(n - 1, k) + stirling(n - 1, k - 1)
# Function to calculate the total number of
# ways to partition a set of `n` elements
def bellNumber(n):
result = 0
# Sum up Stirling numbers S(n, k) for
# all k from 1 to n
for k in range(1, n + 1):
result += stirling(n, k)
return result
n = 5
result = bellNumber(n)
print(result)
// C# program to find the Bell Number
// using recursion
using System;
class GfG {
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
static int Stirling(int n, int k) {
// Base cases
if (n == 0 && k == 0) return 1;
if (k == 0 || n == 0) return 0;
if (n == k) return 1;
if (k == 1) return 1;
// Recursive formula
return k * Stirling(n - 1, k) + Stirling(n - 1, k - 1);
}
// Function to calculate the total number of
// ways to partition a set of `n` elements
static int BellNumber(int n) {
int result = 0;
// Sum up Stirling numbers S(n, k) for all
// k from 1 to n
for (int k = 1; k <= n; ++k) {
result += Stirling(n, k);
}
return result;
}
public static void Main(string[] args) {
int n = 5;
int result = BellNumber(n);
Console.WriteLine(result);
}
}
// JavaScript program to find the Bell Number using recursion
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
function stirling(n, k) {
// Base cases
if (n === 0 && k === 0) return 1;
if (k === 0 || n === 0) return 0;
if (n === k) return 1;
if (k === 1) return 1;
// Recursive formula
return k * stirling(n - 1, k) + stirling(n - 1, k - 1);
}
// Function to calculate the total number of
// ways to partition a set of `n` elements
function bellNumber(n) {
let result = 0;
// Sum up Stirling numbers S(n, k) for all
// k from 1 to n
for (let k = 1; k <= n; ++k) {
result += stirling(n, k);
}
return result;
}
let n = 5;
let result = bellNumber(n);
console.log(result);
Using Top-Down DP (Memoization) - O(n^2) Time and O(n^2) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: The number of ways to partition a set of n elements into k subsets depends on two smaller subproblems:
- The number of ways to partition the first n-1 elements into k subsets, and
- The number of ways to partition the first n-1 elements into k-1 subsets and then add the new element as its own subset. By combining these optimal solutions, we can efficiently calculate the total number of ways to partition the set.
2. Overlapping Subproblems: In the recursive approach, certain subproblems are recalculated multiple times. For example, when computing S(n, k), the subproblems S(n-1, k) and S(n-1, k-1) are recomputed multiple times. This redundancy leads to overlapping subproblems, which can be avoided using memoization or tabulation.
Follow the steps below to solve the problem:
- Use recursion with memoization to calculate Stirling numbers of the second kind, S(n,k), which represent the number of ways to partition n items into k non-empty subsets.
- Define base cases for S(0, 0), S(n, 0), S(n, n), and S(n, 1) to handle specific situations in the recursive formula.
- Use the formula S(n, k) = k × S(n-1, k) + S(n-1, k-1) to compute the Stirling numbers with previously stored results.
- For a given n, sum S(n, k) for all from 1 to n to calculate the Bell number, which represents the total ways to partition n elements.
- Output the computed Bell number for the given n.
C++
// C++ code of finding the bellNumber
// using Memoization
#include <iostream>
#include <vector>
using namespace std;
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
int stirling(int n, int k, vector<vector<int>>& memo) {
// Base cases
if (n == 0 && k == 0) return 1;
if (k == 0 || n == 0) return 0;
if (n == k) return 1;
if (k == 1) return 1;
// Check if result is already computed
if (memo[n][k] != -1) return memo[n][k];
// Recursive formula
memo[n][k] = k * stirling(n - 1, k, memo) + stirling(n - 1, k - 1, memo);
return memo[n][k];
}
// Function to calculate the total number of
// ways to partition a set of `n` elements
int bellNumber(int n) {
// Initialize memoization table with -1
vector<vector<int>> memo(n + 1, vector<int>(n + 1, -1));
int result = 0;
// Sum up Stirling numbers S(n, k) for all
// k from 1 to n
for (int k = 1; k <= n; ++k) {
result += stirling(n, k, memo);
}
return result;
}
int main() {
int n = 5;
int result = bellNumber(n);
cout << result << endl;
return 0;
}
// Java code of finding the bellNumber
// using Memoization
import java.util.Arrays;
class GfG {
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
static int stirling(int n, int k, int[][] memo) {
// Base cases
if (n == 0 && k == 0) return 1;
if (k == 0 || n == 0) return 0;
if (n == k) return 1;
if (k == 1) return 1;
// Check if result is already computed
if (memo[n][k] != -1) return memo[n][k];
// Recursive formula
memo[n][k] = k * stirling(n - 1, k, memo) + stirling(n - 1, k - 1, memo);
return memo[n][k];
}
// Function to calculate the total number of
// ways to partition a set of n elements
static int bellNumber(int n) {
// Initialize memoization table with -1
int[][] memo = new int[n + 1][n + 1];
for (int[] row : memo) Arrays.fill(row, -1);
int result = 0;
// Sum up Stirling numbers S(n, k) for all k from 1 to n
for (int k = 1; k <= n; ++k) {
result += stirling(n, k, memo);
}
return result;
}
public static void main(String[] args) {
int n = 5;
int result = bellNumber(n);
System.out.println(result);
}
}
# Python code of finding the bellNumber
# Using Memoization
# Function to compute Stirling numbers of
# the second kind S(n, k) with memoization
def stirling(n, k, memo):
# Base cases
if n == 0 and k == 0:
return 1
if k == 0 or n == 0:
return 0
if n == k:
return 1
if k == 1:
return 1
# Check if result is already
# computed
if memo[n][k] != -1:
return memo[n][k]
# Recursive formula
memo[n][k] = k * stirling(n - 1, k, memo) + stirling(n - 1, k - 1, memo)
return memo[n][k]
# Function to calculate the total number of
# ways to partition a set of n elements
def bellNumber(n):
# Initialize memoization table with -1
memo = [[-1 for _ in range(n + 1)] for _ in range(n + 1)]
result = 0
# Sum up Stirling numbers S(n, k) for all k from 1 to n
for k in range(1, n + 1):
result += stirling(n, k, memo)
return result
if __name__ == "__main__":
n = 5
result = bellNumber(n)
print(result)
//C# code of finding the bellNumber
// using Memoization
using System;
class GfG {
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
static int Stirling(int n, int k, int[, ] memo) {
// Base cases
if (n == 0 && k == 0)
return 1;
if (k == 0 || n == 0)
return 0;
if (n == k)
return 1;
if (k == 1)
return 1;
// Check if result is already
// computed
if (memo[n, k] != -1)
return memo[n, k];
// Recursive formula
memo[n, k] = k * Stirling(n - 1, k, memo)
+ Stirling(n - 1, k - 1, memo);
return memo[n, k];
}
// Function to calculate the total number of
// ways to partition a set of n elements
static int BellNumber(int n) {
// Initialize memoization table with -1
int[, ] memo = new int[n + 1, n + 1];
for (int i = 0; i <= n; i++)
for (int j = 0; j <= n; j++)
memo[i, j] = -1;
int result = 0;
// Sum up Stirling numbers S(n, k) for all k from 1
// to n
for (int k = 1; k <= n; ++k) {
result += Stirling(n, k, memo);
}
return result;
}
static void Main(string[] args) {
int n = 5;
int result = BellNumber(n);
Console.WriteLine(result);
}
}
// JavaScript code of finding the bellNumber
// using Memoization
// Function to compute Stirling numbers of
// the second kind S(n, k) with memoization
function stirling(n, k, memo) {
// Base cases
if (n === 0 && k === 0) return 1;
if (k === 0 || n === 0) return 0;
if (n === k) return 1;
if (k === 1) return 1;
// Check if result is already computed
if (memo[n][k] !== -1) return memo[n][k];
// Recursive formula
memo[n][k] = k * stirling(n - 1, k, memo) + stirling(n - 1, k - 1, memo);
return memo[n][k];
}
// Function to calculate the total number of
// ways to partition a set of n elements
function bellNumber(n) {
// Initialize memoization table with -1
let memo = Array.from({ length: n + 1 }, () => Array(n + 1).fill(-1));
let result = 0;
// Sum up Stirling numbers S(n, k) for all k from 1 to n
for (let k = 1; k <= n; ++k) {
result += stirling(n, k, memo);
}
return result;
}
let n = 5;
let result = bellNumber(n);
console.log(result);
Using Bottom-Up DP (Tabulation - 1) - O(n^2) Time and O(n^2) Space
A Simple Method to compute n'th Bell Number is to one by one compute S(n, k) for k = 1 to n and return sum of all computed values. Refer Count number of ways to partition a set into k subsets for computation of S(n, k).
C++
// C++ code to calculate the Bell number for a given
// integer `n`
#include <iostream>
#include <vector>
using namespace std;
// Function to calculate the Bell number for a given integer `n`
int bellNumber(int n) {
// Create a 2D vector for Stirling numbers of
// the second kind
vector<vector<int>> dp(n + 1, vector<int>(n + 1, 0));
// Fill the table using dynamic programming
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
// These are some base cases
if (j > i)
dp[i][j] = 0;
else if (i == j)
dp[i][j] = 1;
else if (i == 0 || j == 0)
dp[i][j] = 0;
else {
// Recurrence relation: Stirling number calculation
dp[i][j] = j * dp[i - 1][j] + dp[i - 1][j - 1];
}
}
}
// Sum up Stirling numbers for all j
// from 0 to n to get the Bell number
int ans = 0;
for (int i = 0; i <= n; i++)
{
ans += dp[n][i];
}
// Return the Bell number
return ans;
}
int main() {
int n = 5;
int result = bellNumber(n);
cout << result << endl;
return 0;
}
// Java code to calculate the Bell number for a given
// integer `n`
class GfG {
// Function to calculate the Bell number for a given
// integer `n`
static int bellNumber(int n) {
// Create a 2D vector for Stirling numbers of the
// second kind
int[][] dp = new int[n + 1][n + 1];
// Fill the table using dynamic programming
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
// These are some base cases
if (j > i)
dp[i][j] = 0;
else if (i == j)
dp[i][j] = 1;
else if (i == 0 || j == 0)
dp[i][j] = 0;
else {
// Recurrence relation: Stirling number
// calculation
dp[i][j]
= j * dp[i - 1][j] + dp[i - 1][j - 1];
}
}
}
// Sum up Stirling numbers for all j
// from 0 to n to get the Bell number
int ans = 0;
for (int i = 0; i <= n; i++) {
ans += dp[n][i];
}
// Return the Bell number
return ans;
}
public static void main(String[] args) {
int n = 5;
int result = bellNumber(n);
System.out.println(result);
}
}
# Python code to calculate the Bell number for a given integer `n`
# Function to calculate the Bell number for a
# given integer `n`
def bellNumber(n):
# Create a 2D list for Stirling numbers of
# the second kind
dp = [[0] * (n + 1) for _ in range(n + 1)]
# Fill the table using dynamic programming
for i in range(n + 1):
for j in range(n + 1):
# These are some base cases
if j > i:
dp[i][j] = 0
elif i == j:
dp[i][j] = 1
elif i == 0 or j == 0:
dp[i][j] = 0
else:
# Recurrence relation: Stirling number calculation
dp[i][j] = j * dp[i - 1][j] + dp[i - 1][j - 1]
# Sum up Stirling numbers for all j
# from 0 to n to get the Bell number
ans = 0
for i in range(n + 1):
ans += dp[n][i]
# Return the Bell number
return ans
if __name__ == "__main__":
n = 5
result = bellNumber(n)
print(result)
// C# code to calculate the Bell number for a given integer
// `n`
using System;
class GfG {
// Function to calculate the Bell number for a given
// integer `n`
static int BellNumber(int n) {
// Create a 2D array for Stirling numbers of the
// second kind
int[, ] dp = new int[n + 1, n + 1];
// Fill the table using dynamic programming
for (int i = 0; i <= n; i++) {
for (int j = 0; j <= n; j++) {
// These are some base cases
if (j > i)
dp[i, j] = 0;
else if (i == j)
dp[i, j] = 1;
else if (i == 0 || j == 0)
dp[i, j] = 0;
else {
// Recurrence relation: Stirling number
// calculation
dp[i, j]
= j * dp[i - 1, j] + dp[i - 1, j - 1];
}
}
}
// Sum up Stirling numbers for all j
// from 0 to n to get the Bell number
int ans = 0;
for (int i = 0; i <= n; i++) {
ans += dp[n, i];
}
// Return the Bell number
return ans;
}
static void Main(string[] args) {
int n = 5;
int result = BellNumber(n);
Console.WriteLine(result);
}
}
// JavaScript code to calculate the Bell number for a given
// integer `n`
// Function to calculate the Bell number for a given integer
// `n`
function bellNumber(n) {
// Create a 2D array for Stirling numbers of the second
// kind
let dp = Array.from({length : n + 1},
() => Array(n + 1).fill(0));
// Fill the table using dynamic programming
for (let i = 0; i <= n; i++) {
for (let j = 0; j <= n; j++) {
// These are some base cases
if (j > i)
dp[i][j] = 0;
else if (i === j)
dp[i][j] = 1;
else if (i === 0 || j === 0)
dp[i][j] = 0;
else {
// Recurrence relation: Stirling number
// calculation
dp[i][j] = j * dp[i - 1][j] + dp[i - 1][j - 1];
}
}
}
// Sum up Stirling numbers for all j
// from 0 to n to get the Bell number
let ans = 0;
for (let i = 0; i <= n; i++) {
ans += dp[n][i];
}
// Return the Bell number
return ans;
}
let n = 5;
let result = bellNumber(n);
console.log(result);
Using Bottom-Up DP (Tabulation - 2) - O(n^2) Time and O(n^2) Space
A Better Method is to use Bell Triangle. Below is a sample Bell Triangle for first few Bell Numbers.
1
1 2
2 3 5
5 7 10 15
15 20 27 37 52
Follow the steps below to solve the problem:
- Create a 2D array to store Bell numbers, starting with bell[0][0] = 1 as the base case.
- Use dynamic programming to fill the table based on the recurrence relation. For each i, set bell[i][0] = bell[i-1][i-1] and compute bell[i][j] using the formula bell[i][j] = bell[i-1][j-1] + bell[i][j-1] for all valid j.
- The formula computes the Bell numbers by using the Stirling numbers of the second kind, which count ways to partition a set of i elements into j non-empty subsets.
- The Bell number for n is stored in bell[n][0].
C++14
// A C++ program to find n'th Bell number
#include <iostream>
using namespace std;
int bellNumber(int n) {
int dp[n + 1][n + 1];
dp[0][0] = 1;
for (int i = 1; i <= n; i++) {
// Explicitly fill for j = 0
dp[i][0] = dp[i - 1][i - 1];
// Fill for remaining values of j
for (int j = 1; j <= i; j++)
dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1];
}
return dp[n][0];
}
int main() {
int n = 5;
int result = bellNumber(n);
cout << result << endl;
return 0;
}
// Java program to find n'th Bell number
class GfG {
// Function to calculate the Bell number for a given
// integer `n`
static int bellNumber(int n) {
int[][] dp = new int[n + 1][n + 1];
dp[0][0] = 1;
for (int i = 1; i <= n; i++) {
// Explicitly fill for j = 0
dp[i][0] = dp[i - 1][i - 1];
// Fill for remaining values of j
for (int j = 1; j <= i; j++)
dp[i][j]
= dp[i - 1][j - 1] + dp[i][j - 1];
}
return dp[n][0];
}
public static void main(String[] args) {
int n = 5;
int result = bellNumber(n);
System.out.println(result);
}
}
# Python program to find n'th Bell number
# Function to calculate the Bell number for a given integer `n`
def bellNumber(n):
dp = [[0] * (n + 1) for _ in range(n + 1)]
dp[0][0] = 1
for i in range(1, n + 1):
# Explicitly fill for j = 0
dp[i][0] = dp[i - 1][i - 1]
# Fill for remaining values of j
for j in range(1, i + 1):
dp[i][j] = dp[i - 1][j - 1] + dp[i][j - 1]
return dp[n][0]
if __name__ == "__main__":
n = 5
result = bellNumber(n)
print(result)
// C# program to find n'th Bell number
using System;
class GfG {
// Function to calculate the Bell number for a given
// integer `n`
static int BellNumber(int n) {
int[, ] dp = new int[n + 1, n + 1];
dp[0, 0] = 1;
for (int i = 1; i <= n; i++) {
// Explicitly fill for j = 0
dp[i, 0] = dp[i - 1, i - 1];
// Fill for remaining values of j
for (int j = 1; j <= i; j++)
dp[i, j]
= dp[i - 1, j - 1] + dp[i, j - 1];
}
return dp[n, 0];
}
static void Main(string[] args) {
int n = 5;
int result = BellNumber(n);
Console.WriteLine(result);
}
}
// JavaScript program to find n'th Bell number
// Function to calculate the Bell number for a given integer
// `n`
function bellNumber(n) {
let dp = Array.from({length : n + 1},
() => Array(n + 1).fill(0));
dp[0][0] = 1;
for (let i = 1; i <= n; i++) {
// Explicitly fill for j = 0
dp[i][0] = dp[i - 1][i - 1];
// Fill for remaining values of j
for (let j = 1; j <= i; j++) {
dp[i][j]
= dp[i - 1][j - 1] + dp[i][j - 1];
}
}
return dp[n][0];
}
let n = 5;
let result = bellNumber(n);
console.log(result);
Using Space Optimized DP - O(n^2) Time and O(n) Space
We can use a 1-D array to represent the previous row of the Bell triangle. We initialize dp[0] to 1, since there is only one way to partition an empty set.
To compute the Bell numbers for n > 0, we first set dp[0] = dp[i-1], since the first element in each row is the same as the last element in the previous row. Then, we use the recurrence relation dp[j] = prev + dp[j-1] to compute the Bell number for each partition, where prev is the value of dp[j] in the previous iteration of the inner loop. We update prev to the temporary variable temp before updating dp[j]. Finally, we return dp[0], which is the Bell number for the partition of a set with n elements into non-empty subsets.
C++
// C++ program to find n'th Bell number using
// tabulation
#include <iostream>
#include <vector>
using namespace std;
// Function to calculate the Bell number for 'n'
int bellNumber(int n) {
// Initialize the previous row of the Bell triangle with
// dp[0] = 1
vector<int> dp(n + 1, 0);
dp[0] = 1;
for (int i = 1; i <= n; i++) {
// The first element in each row is the same as the
// last element in the previous row
int prev = dp[0];
dp[0] = dp[i - 1];
for (int j = 1; j <= i; j++) {
// The Bell number for n is the sum of the Bell
// numbers for all previous partitions
int temp = dp[j];
dp[j] = prev + dp[j - 1];
prev = temp;
}
}
return dp[0];
}
int main() {
int n = 5;
cout << bellNumber(n) << std::endl;
return 0;
}
// Java program to find n'th Bell number using
// tabulation
import java.util.Arrays;
class GfG {
// Function to calculate the Bell number for 'n'
static int bellNumbers(int n) {
// Initialize the previous row of the Bell triangle
// with dp[0] = 1
int[] dp = new int[n + 1];
Arrays.fill(dp, 0);
dp[0] = 1;
for (int i = 1; i <= n; i++) {
// The first element in each row is the same as
// the last element in the previous row
int prev = dp[0];
dp[0] = dp[i - 1];
for (int j = 1; j <= i; j++) {
// The Bell number for n is the sum of the
// Bell numbers for all previous partitions
int temp = dp[j];
dp[j] = prev + dp[j - 1];
prev = temp;
}
}
return dp[0];
}
public static void main(String[] args) {
int n = 5;
System.out.println(bellNumbers(n));
}
}
# Python program to find n'th Bell number using
# tabulation
def bell_numbers(n):
# Initialize the previous row of the
# Bell triangle with dp[0] = 1
dp = [1] + [0] * n
for i in range(1, n + 1):
# The first element in each row is the same
# as the last element in the previous row
prev = dp[0]
dp[0] = dp[i - 1]
for j in range(1, i + 1):
# The Bell number for n is the sum of the
# Bell numbers for all previous partitions
temp = dp[j]
dp[j] = prev + dp[j - 1]
prev = temp
return dp[0]
n = 5
print(bell_numbers(n))
// C# program to find n'th Bell number using
// tabulation
using System;
class GfG {
// Function to calculate the Bell number for 'n'
static int BellNumbers(int n) {
// Initialize the previous row of the Bell triangle
// with dp[0] = 1
int[] dp = new int[n + 1];
dp[0] = 1;
for (int i = 1; i <= n; i++) {
// The first element in each row is the same as
// the last element in the previous row
int prev = dp[0];
dp[0] = dp[i - 1];
for (int j = 1; j <= i; j++) {
// The Bell number for n is the sum of the
// Bell numbers for all previous partitions
int temp = dp[j];
dp[j] = prev + dp[j - 1];
prev = temp;
}
}
return dp[0];
}
static void Main() {
int n = 5;
Console.WriteLine(BellNumbers(n));
}
}
// JavaScript program to find n'th Bell number using
// tabulation
function bellNumbers(n) {
// Create an array to store intermediate values,
// initialized with zeros
let dp = new Array(n + 1).fill(0);
// The first element represents the Bell number for 0,
// which is 1
dp[0] = 1;
// Iterate through each row of the Bell triangle up to
// 'n'
for (let i = 1; i <= n; i++) {
// Store the value of the first element in the
// current row
let prev = dp[0];
// Update the first element of the row using the
// last element of the previous row
dp[0] = dp[i - 1];
// Iterate through each element in the current row
for (let j = 1; j <= i; j++) {
// Store the current value of dp[j] in a
// temporary variable
let temp = dp[j];
// Update dp[j] by adding the previous value
// (prev) and the value at dp[j-1]
dp[j] = prev + dp[j - 1];
// Update the 'prev' variable for the next
// iteration of the inner loop
prev = temp;
}
}
// Return the Bell number for 'n'
return dp[0];
}
let n = 5;
console.log(bellNumbers(n));
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Print all prime factors of a given numberGiven a number n, the task is to find all prime factors of n.Examples:Input: n = 24Output: 2 2 2 3Explanation: The prime factorization of 24 is 23×3.Input: n = 13195Output: 5 7 13 29Explanation: The prime factorization of 13195 is 5×7×13×29.Approach:Every composite number has at least one prime fact
6 min read
Smith NumberGiven a number n, the task is to find out whether this number is smith or not. A Smith Number is a composite number whose sum of digits is equal to the sum of digits in its prime factorization. Examples: Input : n = 4Output : YesPrime factorization = 2, 2 and 2 + 2 = 4Therefore, 4 is a smith numberI
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Sphenic NumberA Sphenic Number is a positive integer n which is a product of exactly three distinct primes. The first few sphenic numbers are 30, 42, 66, 70, 78, 102, 105, 110, 114, ... Given a number n, determine whether it is a Sphenic Number or not. Examples: Input: 30Output : YesExplanation: 30 is the smalles
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Hoax NumberGiven a number 'n', check whether it is a hoax number or not. A Hoax Number is defined as a composite number, whose sum of digits is equal to the sum of digits of its distinct prime factors. It may be noted here that, 1 is not considered a prime number, hence it is not included in the sum of digits
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k-th prime factor of a given numberGiven two numbers n and k, print k-th prime factor among all prime factors of n. For example, if the input number is 15 and k is 2, then output should be "5". And if the k is 3, then output should be "-1" (there are less than k prime factors). Examples: Input : n = 225, k = 2 Output : 3 Prime factor
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Pollard's Rho Algorithm for Prime FactorizationGiven a positive integer n, and that it is composite, find a divisor of it.Example:Input: n = 12;Output: 2 [OR 3 OR 4]Input: n = 187;Output: 11 [OR 17]Brute approach: Test all integers less than n until a divisor is found. Improvisation: Test all integers less than ?nA large enough number will still
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Finding power of prime number p in n!Given a number 'n' and a prime number 'p'. We need to find out the power of 'p' in the prime factorization of n!Examples: Input : n = 4, p = 2 Output : 3 Power of 2 in the prime factorization of 2 in 4! = 24 is 3 Input : n = 24, p = 2 Output : 22 Naive approach The naive approach is to find the powe
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Find all factors of a Positive NumberGiven a positive integer n, find all the distinct divisors of n.Examples:Input: n = 10 Output: [1, 2, 5, 10]Explanation: 1, 2, 5 and 10 are the divisors of 10. Input: n = 100Output: [1, 2, 4, 5, 10, 20, 25, 50, 100]Explanation: 1, 2, 4, 5, 10, 20, 25, 50 and 100 are divisors of 100.Table of Content[
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Find numbers with n-divisors in a given rangeGiven three integers a, b, n .Your task is to print number of numbers between a and b including them also which have n-divisors. A number is called n-divisor if it has total n divisors including 1 and itself. Examples: Input : a = 1, b = 7, n = 2 Output : 4 There are four numbers with 2 divisors in
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Modular Arithmetic
Modular Exponentiation (Power in Modular Arithmetic)Given three integers x, n, and M, compute (x^n) % M (remainder when x raised to the power n is divided by M).Examples : Input: x = 3, n = 2, M = 4Output: 1Explanation: 32 % 4 = 9 % 4 = 1.Input: x = 2, n = 6, M = 10Output: 4Explanation: 26 % 10 = 64 % 10 = 4.Table of Content[Naive Approach] Repeated
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Modular multiplicative inverseGiven two integers A and M, find the modular multiplicative inverse of A under modulo M.The modular multiplicative inverse is an integer X such that:A X ≡ 1 (mod M) Note: The value of X should be in the range {1, 2, ... M-1}, i.e., in the range of integer modulo M. ( Note that X cannot be 0 as A*0 m
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Modular DivisionGiven three positive integers a, b, and M, the objective is to find (a/b) % M i.e., find the value of (a × b-1 ) % M, where b-1 is the modular inverse of b modulo M.Examples: Input: a = 10, b = 2, M = 13Output: 5Explanation: The modular inverse of 2 modulo 13 is 7, so (10 / 2) % 13 = (10 × 7) % 13 =
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Euler's criterion (Check if square root under modulo p exists)Given a number 'n' and a prime p, find if square root of n under modulo p exists or not. A number x is square root of n under modulo p if (x*x)%p = n%p. Examples : Input: n = 2, p = 5 Output: false There doesn't exist a number x such that (x*x)%5 is 2 Input: n = 2, p = 7 Output: true There exists a
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Find sum of modulo K of first N natural numberGiven two integer N ans K, the task is to find sum of modulo K of first N natural numbers i.e 1%K + 2%K + ..... + N%K. Examples : Input : N = 10 and K = 2. Output : 5 Sum = 1%2 + 2%2 + 3%2 + 4%2 + 5%2 + 6%2 + 7%2 + 8%2 + 9%2 + 10%2 = 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 = 5.Recommended PracticeReve
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How to compute mod of a big number?Given a big number 'num' represented as string and an integer x, find value of "num % a" or "num mod a". Output is expected as an integer. Examples : Input: num = "12316767678678", a = 10 Output: num (mod a) ? 8 The idea is to process all digits one by one and use the property that xy (mod a) ? ((x
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Exponential Squaring (Fast Modulo Multiplication)Given two numbers base and exp, we need to compute baseexp under Modulo 10^9+7 Examples: Input : base = 2, exp = 2Output : 4Input : base = 5, exp = 100000Output : 754573817In competitions, for calculating large powers of a number we are given a modulus value(a large prime number) because as the valu
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Trick for modular division ( (x1 * x2 .... xn) / b ) mod (m)Given integers x1, x2, x3......xn, b, and m, we are supposed to find the result of ((x1*x2....xn)/b)mod(m). Example 1: Suppose that we are required to find (55C5)%(1000000007) i.e ((55*54*53*52*51)/120)%1000000007 Naive Method : Simply calculate the product (55*54*53*52*51)= say x,Divide x by 120 a
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