Find position i to split Array such that prefix sum till i-1, i and suffix sum till i+1 are in GP with common ratio K

Given an array, arr[] and a positive integer K. The task is to find the position say i of the element in arr[] such that prefix sum till i-1, i and suffix sum till i+1 are in Geometric Progression with common ratio K. 

Examples:

Input: arr[] = { 5, 1, 4, 20, 6, 15, 9, 10 }, K = 2
Output: 4
Explanation:  The following operations are performed to get required GP.
Sum of element from position 1 to 3 is 5 + 1 + 4 = 10 and from 5 to 8 is 6 + 15 + 9 + 10 = 40.
And element at position 4 is 20.
Therefore10, 20, 40 is a Geometric Progression series with common ratio K.

Input: arr[] ={ -3, 5, 0, 2, 1, 25, 25, 100 }, K = 5 
Output: 6 

Approach:  The given problem can be solved by using Linear Search and basic prefix sum. Follow the steps below to solve the given problem.

• If the size of array is less than 3 then no sequence is possible so simply return -1.
• Initialize a variable say, arrSum to store sum of all elements of arr[].
• Calculate sum of array arr[] and store it in arrSum.
• if arrSum % R != 0, then return 0. Where R = K * K + 1 + K + 1.
• Initialize a variable say mid = K * (Sum / R) to store middle element of GP series with common ratio as K.
• Take a variable say temp to store temporary results.
• Iterate arr[] from index 1 to (size of arr[]) – 2 with variable i.
  • temp = temp + arr[i-1]
  • if arr[i] = mid
    • if temp = mid/k, return (i+1) as the answer.
    • else return 0.
• If loop terminates and no element in arr[] is equal to mid then simply return 0.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <iostream>
using namespace std;

// Function to check if there is
// an element forming G.P. series
// having common ratio k
int checkArray(int arr[], int N, int k)
{

    // If size of array is less than
    // three then return -1
    if (N < 3)
        return -1;

    // Initialize the variables
    int i, Sum = 0, temp = 0;

    // Calculate total sum of array
    for (i = 0; i < N; i++)
        Sum += arr[i];

    int R = (k * k + k + 1);

    if (Sum % R != 0)
        return 0;

    // Calculate Middle element of G.P. series
    int Mid = k * (Sum / R);

    // Iterate over the range
    for (i = 1; i < N - 1; i++) {

        // Store the first element of G.P.
        // series in the variable temp
        temp += arr[i - 1];

        if (arr[i] == Mid) {

            // Return position of middle element
            // of the G.P. series if the first
            // element is in G.P. of common ratio k
            if (temp == Mid / k)
                return i + 1;

            // Else return 0
            else
                return 0;
        }
    }

    // if middle element is not found in arr[]
    return 0;
}

// Driver Code
int main()
{
    // Given array
    int arr[] = { 5, 1, 4, 20, 6, 15, 9, 10 };

    int N = sizeof(arr) / sizeof(arr[0]);

    int K = 2;

    cout << checkArray(arr, N, K) << endl;

    return 0;
}

Java

// Java program for the above approach

import java.io.*;

class GFG {

// Function to check if there is
// an element forming G.P. series
// having common ratio k
static int checkArray(int arr[], int N, int k)
{

    // If size of array is less than
    // three then return -1
    if (N < 3)
        return -1;

    // Initialize the variables
    int i, Sum = 0, temp = 0;

    // Calculate total sum of array
    for (i = 0; i < N; i++)
        Sum += arr[i];

    int R = (k * k + k + 1);

    if (Sum % R != 0)
        return 0;

    // Calculate Middle element of G.P. series
    int Mid = k * (Sum / R);

    // Iterate over the range
    for (i = 1; i < N - 1; i++) {

        // Store the first element of G.P.
        // series in the variable temp
        temp += arr[i - 1];

        if (arr[i] == Mid) {

            // Return position of middle element
            // of the G.P. series if the first
            // element is in G.P. of common ratio k
            if (temp == Mid / k)
                return i + 1;

            // Else return 0
            else
                return 0;
        }
    }

    // if middle element is not found in arr[]
    return 0;
}

// Driver Code
public static void main (String[] args) {

    // Given array
    int arr[] = { 5, 1, 4, 20, 6, 15, 9, 10 };

    int N = arr.length;

    int K = 2;

       System.out.println(checkArray(arr, N, K));
}
}

// This code is contributed by Dharanendra L V.

Python3

# python program for the above approach

# Function to check if there is
# an element forming G.P. series
# having common ratio k
def checkArray(arr, N, k):

        # If size of array is less than
        # three then return -1
    if (N < 3):
        return -1

        # Initialize the variables
    Sum = 0
    temp = 0

    # Calculate total sum of array
    for i in range(0, N):
        Sum += arr[i]

    R = (k * k + k + 1)

    if (Sum % R != 0):
        return 0

        # Calculate Middle element of G.P. series
    Mid = k * (Sum // R)

    # Iterate over the range
    for i in range(1, N-1):

                # Store the first element of G.P.
                # series in the variable temp
        temp += arr[i - 1]

        if (arr[i] == Mid):

                        # Return position of middle element
                        # of the G.P. series if the first
                        # element is in G.P. of common ratio k
            if (temp == Mid // k):
                return i + 1

                # Else return 0
            else:
                return 0

        # if middle element is not found in arr[]
    return 0

# Driver Code
if __name__ == "__main__":

    # Given array
    arr = [5, 1, 4, 20, 6, 15, 9, 10]
    N = len(arr)
    K = 2

    print(checkArray(arr, N, K))

    # This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;
class GFG {

    // Function to check if there is
    // an element forming G.P. series
    // having common ratio k
    static int checkArray(int[] arr, int N, int k)
    {

        // If size of array is less than
        // three then return -1
        if (N < 3)
            return -1;

        // Initialize the variables
        int i, Sum = 0, temp = 0;

        // Calculate total sum of array
        for (i = 0; i < N; i++)
            Sum += arr[i];

        int R = (k * k + k + 1);

        if (Sum % R != 0)
            return 0;

        // Calculate Middle element of G.P. series
        int Mid = k * (Sum / R);

        // Iterate over the range
        for (i = 1; i < N - 1; i++) {

            // Store the first element of G.P.
            // series in the variable temp
            temp += arr[i - 1];

            if (arr[i] == Mid) {

                // Return position of middle element
                // of the G.P. series if the first
                // element is in G.P. of common ratio k
                if (temp == Mid / k)
                    return i + 1;

                // Else return 0
                else
                    return 0;
            }
        }

        // if middle element is not found in arr[]
        return 0;
    }

    // Driver Code
    public static void Main(string[] args)
    {

        // Given array
        int[] arr = { 5, 1, 4, 20, 6, 15, 9, 10 };

        int N = arr.Length;

        int K = 2;

        Console.WriteLine(checkArray(arr, N, K));
    }
}

// This code is contributed by ukasp.

Javascript

<script>
        // JavaScript Program to implement
        // the above approach

        // Function to check if there is
        // an element forming G.P. series
        // having common ratio k
        function checkArray(arr, N, k) {

            // If size of array is less than
            // three then return -1
            if (N < 3)
                return -1;

            // Initialize the variables
            let i, Sum = 0, temp = 0;

            // Calculate total sum of array
            for (i = 0; i < N; i++)
                Sum += arr[i];

            let R = (k * k + k + 1);

            if (Sum % R != 0)
                return 0;

            // Calculate Middle element of G.P. series
            let Mid = k * (Sum / R);

            // Iterate over the range
            for (i = 1; i < N - 1; i++) {

                // Store the first element of G.P.
                // series in the variable temp
                temp += arr[i - 1];

                if (arr[i] == Mid) {

                    // Return position of middle element
                    // of the G.P. series if the first
                    // element is in G.P. of common ratio k
                    if (temp == Mid / k)
                        return i + 1;

                    // Else return 0
                    else
                        return 0;
                }
            }

            // if middle element is not found in arr[]
            return 0;
        }

        // Driver Code

        // Given array
        let arr = [5, 1, 4, 20, 6, 15, 9, 10];
        let N = arr.length;
        let K = 2;
        document.write(checkArray(arr, N, K) + "<br>");

// This code is contributed by Potta Lokesh
    </script>

Output

4

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