Check if a subarray of length K with sum equal to factorial of a number exists or not

Last Updated : 12 Jul, 2021

Given an array arr[] of N integers and an integer K, the task is to find a subarray of length K with a sum of elements equal to factorial of any number. If no such subarray exists, print “-1”.

Examples:

Input: arr[] = {23, 45, 2, 4, 6, 9, 3, 32}, K = 5
Output: 2 4 6 9 3
Explanation:
Subarray {2, 4, 6, 9, 3} with sum 24 (= 4!) satisfies the required condition.

Input: arr[] = {23, 45, 2, 4, 6, 9, 3, 32}, K = 3
Output: -1
Explanation:
No such subarray of length K (= 3) exists.

Naive Approach: The simplest approach to solve the problem is to calculate the sum of all subarrays of length K and check if any of those sums is factorial of any number. If found to be true for any subarray, print that subarray. Otherwise, print "-1".

Time Complexity: O(N*K)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the Sliding Window technique to calculate the sum of all subarrays of length K and then check if the sum is a factorial or not. Below are the steps:

Calculate the sum of first K array elements and store the sum in a variable, say sum.
Then traverse the remaining array and keep updating sum to get the sum of the current subarray of size K by subtracting the first element from the previous subarray and adding the current array element.
To check whether the sum is a factorial of a number or not, divide the sum by 2, 3, and so on until it cannot be divided further. If the number reduces to 1, the sum is the factorial of a number.
If the sum in the above step is a factorial of a number, store the starting and ending index of that subarray to print the subarray.
After completing the above steps, if no such subarray is found, print "-1". Otherwise, print the subarray whose start and end indices are stored.

Below is the implementation of the above approach:

C++

// C++ program for the above approach  #include <bits/stdc++.h> using namespace std;  // Function to check if a number // is factorial of a number or not int isFactorial(int n) {     int i = 2;     while (n != 1) {          // If n is not a factorial         if (n % i != 0) {             return 0;         }         n /= i;         i++;     }     return i - 1; }  // Function to return the index of // the valid subarray pair<int, int> sumFactorial(     vector<int> arr, int K) {     int i, sum = 0, ans;      // Calculate the sum of     // first subarray of length K     for (i = 0; i < K; i++) {          sum += arr[i];     }      // Check if sum is a factorial     // of any number or not     ans = isFactorial(sum);      // If sum of first K length subarray     // is factorial of a number     if (ans != 0) {         return make_pair(ans, 0);     }      // Find the number formed from the     // subarray which is a factorial     for (int j = i; j < arr.size(); j++) {          // Update sum of current subarray         sum += arr[j] - arr[j - K];          // Check if sum is a factorial         // of any number or not         ans = isFactorial(sum);          // If ans is true, then return         // index of the current subarray         if (ans != 0) {             return make_pair(ans,                              j - K + 1);         }     }      // If the required subarray is     // not possible     return make_pair(-1, 0); }  // Function to print the subarray whose // sum is a factorial of any number void printRes(pair<int, int> answer,               vector<int> arr, int K) {      // If no such subarray exists     if (answer.first == -1) {          cout << -1 << endl;     }      // Otherwise     else {          int i = 0;         int j = answer.second;          // Iterate to print subarray         while (i < K) {              cout << arr[j] << " ";             i++;             j++;         }     } }  // Driver Code int main() {     vector<int> arr         = { 23, 45, 2, 4,             6, 9, 3, 32 };      // Given sum K     int K = 5;      // Function Call     pair<int, int> answer         = sumFactorial(arr, K);      // Print the result     printRes(answer, arr, K);      return 0; }

Java

// Java program for the above approach  import java.util.*; import java.io.*;  class GFG{      // Pair class public static class Pair {     int x;     int y;          Pair(int x, int y)      {         this.x = x;         this.y = y;     } }  // Function to check if a number  // is factorial of a number or not  static int isFactorial(int n)  {      int i = 2;      while (n != 1)     {           // If n is not a factorial          if (n % i != 0)          {              return 0;          }          n /= i;          i++;      }      return i - 1;  }  // Function to return the index of  // the valid subarray  static ArrayList<Pair> sumFactorial(int arr[],                                     int K)  {      ArrayList<Pair> pair = new ArrayList<>();          int i, sum = 0, ans;       // Calculate the sum of      // first subarray of length K      for(i = 0; i < K; i++)     {          sum += arr[i];      }           // Check if sum is a factorial      // of any number or not      ans = isFactorial(sum);           // If sum of first K length subarray      // is factorial of a number      if (ans != 0)      {         Pair p = new Pair(ans, 0);         pair.add(p);         return pair;      }       // Find the number formed from the      // subarray which is a factorial      for(int j = i; j < arr.length; j++)     {           // Update sum of current subarray          sum += arr[j] - arr[j - K];           // Check if sum is a factorial          // of any number or not          ans = isFactorial(sum);           // If ans is true, then return          // index of the current subarray          if (ans != 0)         {              Pair p = new Pair(ans, j - K + 1);             pair.add(p);             return pair;         }      }       // If the required subarray is      // not possible      Pair p = new Pair(-1, 0);     pair.add(p);     return pair; }   // Function to print the subarray whose  // sum is a factorial of any number  static void printRes(ArrayList<Pair> answer,                       int arr[], int K)  {           // If no such subarray exists      if (answer.get(0).x == -1)      {                   // cout << -1 << endl;         System.out.println("-1");     }       // Otherwise      else      {          int i = 0;          int j = answer.get(0).y;           // Iterate to print subarray          while (i < K)          {              System.out.print(arr[j] + " ");             i++;              j++;          }      }  }   // Driver Code  public static void main(String args[])  {           // Given array arr[] and brr[]      int arr[] = { 23, 45, 2, 4,                    6, 9, 3, 32 };                         int K = 5;     ArrayList<Pair> answer = new ArrayList<>();          // Function call      answer = sumFactorial(arr,K);                          // Print the result      printRes(answer, arr, K); }  }  // This code is contributed by bikram2001jha

Python3

# Python3 program for the above approach  # Function to check if a number # is factorial of a number or not def isFactorial(n):      i = 2          while (n != 1):                  # If n is not a factorial         if (n % i != 0):             return 0                  n = n // i         i += 1          return i - 1  # Function to return the index of # the valid subarray def sumFactorial(arr, K):      i, Sum = 0, 0      # Calculate the sum of     # first subarray of length K     while(i < K):         Sum += arr[i]         i += 1      # Check if sum is a factorial     # of any number or not     ans = isFactorial(Sum)      # If sum of first K length subarray     # is factorial of a number     if (ans != 0):         return (ans, 0)      # Find the number formed from the     # subarray which is a factorial     for j in range(i, len(arr)):              # Update sum of current subarray         Sum = Sum + arr[j] - arr[j - K]          # Check if sum is a factorial         # of any number or not         ans = isFactorial(Sum)          # If ans is true, then return         # index of the current subarray         if (ans != 0):             return (ans, j - K + 1)      # If the required subarray is     # not possible     return (-1, 0)  # Function to print the subarray whose # sum is a factorial of any number def printRes(answer, arr, K):      # If no such subarray exists     if (answer[0] == -1):         print(-1)      # Otherwise     else:         i = 0         j = answer[1]          # Iterate to print subarray         while (i < K):             print(arr[j], end = " ")             i += 1             j += 1  # Driver code  arr = [ 23, 45, 2, 4, 6, 9, 3, 32 ]  # Given sum K K = 5  # Function call answer = sumFactorial(arr, K)  # Print the result printRes(answer, arr, K)  # This code is contributed by divyeshrabadiya07

// C# program for  // the above approach  using System; using System.Collections.Generic; class GFG{      // Pair class public class Pair {   public int x;   public int y;    public Pair(int x, int y)    {     this.x = x;     this.y = y;   } }  // Function to check if a number  // is factorial of a number or not  static int isFactorial(int n)  {    int i = 2;    while (n != 1)   {      // If n is not      // a factorial      if (n % i != 0)      {        return 0;      }      n /= i;      i++;    }    return i - 1;  }  // Function to return the index of  // the valid subarray  static List<Pair> sumFactorial(int []arr,                                int K)  {    List<Pair> pair = new List<Pair>();   int i, sum = 0, ans;     // Calculate the sum of    // first subarray of length K    for(i = 0; i < K; i++)   {      sum += arr[i];    }     // Check if sum is a factorial    // of any number or not    ans = isFactorial(sum);     // If sum of first K length subarray    // is factorial of a number    if (ans != 0)    {     Pair p = new Pair(ans, 0);     pair.Add(p);     return pair;    }     // Find the number formed from the    // subarray which is a factorial    for(int j = i; j < arr.Length; j++)   {      // Update sum of current subarray      sum += arr[j] - arr[j - K];       // Check if sum is a factorial      // of any number or not      ans = isFactorial(sum);       // If ans is true, then return      // index of the current subarray      if (ans != 0)     {        Pair p = new Pair(ans, j - K + 1);       pair.Add(p);       return pair;     }    }     // If the required subarray is    // not possible    Pair p1 = new Pair(-1, 0);   pair.Add(p1);   return pair; }   // Function to print the subarray whose  // sum is a factorial of any number  static void printRes(List<Pair> answer,                       int []arr, int K)  {    // If no such subarray exists    if (answer[0].x == -1)    {      // cout << -1 << endl;     Console.WriteLine("-1");   }     // Otherwise    else    {      int i = 0;      int j = answer[0].y;       // Iterate to print subarray      while (i < K)      {        Console.Write(arr[j] + " ");       i++;        j++;      }    }  }   // Driver Code  public static void Main(String []args)  {        // Given array []arr and brr[]    int []arr = {23, 45, 2, 4,                 6, 9, 3, 32};    int K = 5;   List<Pair> answer = new List<Pair>();      // Function call    answer = sumFactorial(arr, K);    // Print the result    printRes(answer, arr, K); }  }  // This code is contributed by shikhasingrajput

JavaScript

<script>      // JavaScript program for the above approach     // Function to check if a number     // is factorial of a number or not     function isFactorial(n)     {         let i = 2;         while (n != 1) {              // If n is not a factorial             if (n % i != 0) {                 return 0;             }             n = parseInt(n / i, 10);             i++;         }         return i - 1;     }      // Function to return the index of     // the valid subarray     function sumFactorial(arr,K)     {         let i, sum = 0, ans;          // Calculate the sum of         // first subarray of length K         for (i = 0; i < K; i++) {              sum += arr[i];         }          // Check if sum is a factorial         // of any number or not         ans = isFactorial(sum);          // If sum of first K length subarray         // is factorial of a number         if (ans != 0) {             return [ans, 0];         }          // Find the number formed from the         // subarray which is a factorial         for (let j = i; j < arr.length; j++) {              // Update sum of current subarray             sum += arr[j] - arr[j - K];              // Check if sum is a factorial             // of any number or not             ans = isFactorial(sum);              // If ans is true, then return             // index of the current subarray             if (ans != 0) {                 return [ans, j - K + 1];             }         }          // If the required subarray is         // not possible         return [-1, 0];     }      // Function to print the subarray whose     // sum is a factorial of any number     function printRes(answer,arr,K)     {          // If no such subarray exists         if (answer[0] == -1) {              document.write(-1);         }          // Otherwise         else {              let i = 0;             let j = answer[1];              // Iterate to print subarray             while (i < K) {                  document.write(arr[j] + " ");                 i++;                 j++;             }         }     }      // Driver Code      let arr= [ 23, 45, 2, 4,             6, 9, 3, 32 ];      // Given sum K     let K = 5;      // Function Call     let answer= sumFactorial(arr, K);      // Print the result     printRes(answer, arr, K);  </script>

Output:

2 4 6 9 3

Time Complexity: O(N)
Auxiliary Space: O(1)

Check if a Subarray exists with sums as a multiple of k

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Check if a subarray of length K with sum equal to factorial of a number exists or not

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