GeeksforGeeks App
Open App
Browser
Continue

# Prefix Factorials of a Prefix Sum Array

Given an array arr[] consisting of N positive integers, the task is to find the prefix factorials of a prefix sum array of the given array i.e., .

Examples:

Input: arr[] = {1, 2, 3, 4}
Output: 1 6 720 3628800
Explanation:
The prefix sum of the given array is {1, 3, 6, 10}. Therefore, prefix factorials of the obtained prefix sum array is {1!, (1+2)!, (1+2+3)!, (1+2+3+4)!} = {1!, 3!, 6!, 10!} = {1 6 720 3628800}.

Input: arr[] = {2, 4, 3, 1}
Output: 2 720 362880 3628800

Naive Approach: The simplest approach to solve the given problem is to find the prefix sum of the given array and then find the factorial of each array element in the prefix sum array. After calculating the prefix sum print the factorial array.

Below is the implementation of the above approach.

## C++

 // C++ program for the above approach #include using namespace std; // Function to find the factorial of// a number Nint fact(int N){    // Base Case    if (N == 1 || N == 0)        return 1;     // Find the factorial recursively    return N * fact(N - 1);} // Function to find the prefix// factorial arrayvoid prefixFactorialArray(int arr,                          int N){    // Find the prefix sum array    for (int i = 1; i < N; i++) {        arr[i] += arr[i - 1];    }     // Find the factorials of each    // array element    for (int i = 0; i < N; i++) {        arr[i] = fact(arr[i]);    }     // Print the resultant array    for (int i = 0; i < N; i++) {        cout << arr[i] << " ";    }} // Driver Codeint main(){    int arr[] = { 1, 2, 3, 4 };    int N = sizeof(arr) / sizeof(arr[0]);    prefixFactorialArray(arr, N);     return 0;}

## Java

 // Java program for the above approachclass GFG{ // Function to find the factorial of// a number Nstatic int fact(int N){         // Base Case    if (N == 1 || N == 0)        return 1;     // Find the factorial recursively    return N * fact(N - 1);} // Function to find the prefix// factorial arraystatic void prefixFactorialArray(int[] arr, int N){     // Find the prefix sum array    for(int i = 1; i < N; i++)    {        arr[i] += arr[i - 1];    }     // Find the factorials of each    // array element    for(int i = 0; i < N; i++)    {        arr[i] = fact(arr[i]);    }     // Print the resultant array    for(int i = 0; i < N; i++)    {        System.out.print(arr[i] + " ");    }} // Driver Codepublic static void main(String[] args){    int[] arr = { 1, 2, 3, 4 };    int N = arr.length;     prefixFactorialArray(arr, N);}} // This code is contributed by ukasp

## Python3

 # Python implementation of the approachdef fact(N):       # Base Case    if (N == 1 or N == 0):        return 1     # Find the factorial recursively    return N * fact(N - 1)  # Function to find the prefix# factorial arraydef prefixFactorialArray(arr, N):     # Find the prefix sum array    for i in range(1, N):        arr[i] += arr[i - 1]     # Find the factorials of each    # array element    for i in range(N):        arr[i] = fact(arr[i])         # Print the resultant array    for i in range(N):        print(arr[i], end=" ") # Driver Codeif __name__ == "__main__":     arr = [1, 2, 3, 4]    N = len(arr)    prefixFactorialArray(arr, N) # This code is contributed by kirtishsurangalikar

## C#

 // C# program for the above approachusing System;         class GFG{ // Function to find the factorial of// a number Nstatic int fact(int N){         // Base Case    if (N == 1 || N == 0)        return 1;     // Find the factorial recursively    return N * fact(N - 1);} // Function to find the prefix// factorial arraystatic void prefixFactorialArray(int[] arr,                                 int N){         // Find the prefix sum array    for(int i = 1; i < N; i++)    {        arr[i] += arr[i - 1];    }     // Find the factorials of each    // array element    for(int i = 0; i < N; i++)    {        arr[i] = fact(arr[i]);    }     // Print the resultant array    for(int i = 0; i < N; i++)    {        Console.Write(arr[i] + " ");    }}     // Driver Codepublic static void Main(){    int[] arr = { 1, 2, 3, 4 };    int N = arr.Length;         prefixFactorialArray(arr, N);}} // This code is contributed by code_hunt

## Javascript

 

Output:

1 6 720 3628800

Time Complexity: O(N*M), where M is the sum of the array elements
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by precalculating the factorial of sum of the array elements so that the factorial calculation at each index is can be calculated in O(1) time.

Below is an implementation of the above approach:

## C++

 // C++ program for the above approach #include using namespace std; // Function to find the factorial of// prefix sum at every possible indexvoid prefixFactorialArray(int A[], int N){    // Find the prefix sum array    for (int i = 1; i < N; i++) {        A[i] += A[i - 1];    }     // Stores the factorial of all the    // element till the sum of array    int fact[A[N - 1] + 1];    fact[0] = 1;     // Find the factorial array    for (int i = 1; i <= A[N - 1]; i++) {        fact[i] = i * fact[i - 1];    }     // Find the factorials of    // each array element    for (int i = 0; i < N; i++) {        A[i] = fact[A[i]];    }     // Print the resultant array    for (int i = 0; i < N; i++) {        cout << A[i] << " ";    }} // Driver Codeint main(){    int arr[] = { 1, 2, 3, 4 };    int N = sizeof(arr) / sizeof(arr[0]);    prefixFactorialArray(arr, N);     return 0;}

## Java

 // Java program for the above approachclass GFG{ // Function to find the factorial of// prefix sum at every possible indexstatic void prefixFactorialArray(int A[], int N){         // Find the prefix sum array    for(int i = 1; i < N; i++)    {        A[i] += A[i - 1];    }     // Stores the factorial of all the    // element till the sum of array    int fact[] = new int[A[N - 1] + 1];    fact[0] = 1;     // Find the factorial array    for(int i = 1; i <= A[N - 1]; i++)    {        fact[i] = i * fact[i - 1];    }     // Find the factorials of    // each array element    for(int i = 0; i < N; i++)    {        A[i] = fact[A[i]];    }     // Print the resultant array    for(int i = 0; i < N; i++)    {        System.out.print(A[i] + " ");    }} // Driver codepublic static void main(String[] args){    int arr[] = { 1, 2, 3, 4 };    int N = arr.length;         prefixFactorialArray(arr, N);}} // This code is contributed by abhinavjain194

## Python3

 # // python program for the above approach # // Function to find the factorial of# // prefix sum at every possible indexdef prefixFactorialArray(A, N):       # // Find the prefix sum array    for i in range(1, N):        A[i] += A[i - 1]     # // Stores the factorial of all the    # // element till the sum of array    fact = [0 for x in range(A[N - 1] + 1)]    fact[0] = 1     # // Find the factorial array    for i in range(1, A[N-1]+1):        fact[i] = i * fact[i - 1]     # // Find the factorials of    # // each array element    for i in range(0, N):        A[i] = fact[A[i]]             # // Print the resultant array    for i in range(0, N):        print(A[i], end=" ") # Driver codearr = [1, 2, 3, 4]N = len(arr)prefixFactorialArray(arr, N) # This code is contributed by amreshkumar3.

## C#

 // C# program for the above approachusing System;class GFG {         // Function to find the factorial of    // prefix sum at every possible index    static void prefixFactorialArray(int[] A, int N)    {                  // Find the prefix sum array        for(int i = 1; i < N; i++)        {            A[i] += A[i - 1];        }              // Stores the factorial of all the        // element till the sum of array        int[] fact = new int[A[N - 1] + 1];        fact[0] = 1;              // Find the factorial array        for(int i = 1; i <= A[N - 1]; i++)        {            fact[i] = i * fact[i - 1];        }              // Find the factorials of        // each array element        for(int i = 0; i < N; i++)        {            A[i] = fact[A[i]];        }              // Print the resultant array        for(int i = 0; i < N; i++)        {            Console.Write(A[i] + " ");        }    }   // Driver code  static void Main() {    int[] arr = { 1, 2, 3, 4 };    int N = arr.Length;          prefixFactorialArray(arr, N);  }} // This code is contributed by divyeshrabadiya07.

## Javascript

 

Output:

1 6 720 3628800

Time Complexity: O(N + M), where M is the sum of the array elements.
Auxiliary Space: O(M), where M is the sum of the array elements.

My Personal Notes arrow_drop_up