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++ 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 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.
# 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# 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.
<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)