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

Last Updated : 24 Mar, 2022

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>