Given a Geometric Progression series in arr[] and Q queries in the form of [L, R], where L is the left boundary of the range and R is the right boundary. The task is to find the sum of the Geometric Progression elements in the given range.
Note: The range is 1-indexed and1 ? L, R ? N, where N is the size of arr.
Examples:
Input: arr[] = {2, 4, 8, 16, 32, 64, 128, 256}, Q = [[2, 4], [2, 6], [5, 8]]
Output:
28
124
480
Explanation:
Range 1: arr = {4, 8, 16}. Therefore sum = 28
Range 2: arr = {4, 8, 16, 32, 64}. Therefore sum = 124
Range 3: arr = {32, 64, 128, 256}. Therefore sum = 480
Input: arr[] = {7, 7, 7, 7, 7, 7}, Q = [[1, 6], [2, 4], [3, 3]]
Output:
42
21
7
Explanation:
Range 1: arr = {7, 7, 7, 7, 7, 7}. Therefore sum = 42
Range 2: arr = {7, 7, 7}. Therefore sum = 21
Range 3: arr = {7}. Therefore sum = 7
Approach: Since the given sequence is an Geometric progression, the sum can be easily found out in two steps efficiently:
- Get the first element of the range.
- If d = 1, then multiply d*k to it, else multiply the (dk - 1)/(d - 1) to it, where d is the common ratio of the GP and k is number of elements in the range.
For example:
Suppose a[i] be the first element of the range, d be the common ratio of GP and k be the number of elements in the given range.
Then the sum of the range would be
= a[i] + a[i+1] + a[i+2] + ..... + a[i+k-1]
= a[i] + (a[i] * d) + (a[i] * d * d) + .... + (a[i] * dk)
= a[i] * (1 + d + ... + dk)
= a[i] * (dk - 1)/(d - 1)
Below is the implementation of the above approach:
// C++ program to find the sum
// of elements of an GP in the
// given range
#include <bits/stdc++.h>
using namespace std;
// Function to find sum in the given range
int findSum(int arr[], int n,
int left, int right)
{
// Find the value of k
int k = right - left + 1;
// Find the common difference
int d = arr[1] / arr[0];
// Find the sum
int ans = arr[left - 1];
if (d == 1)
ans = ans * d * k;
else
ans = ans * ((int)pow(d, k) - 1 /
(d - 1));
return ans;
}
// Driver Code
int main()
{
int arr[] = { 2, 4, 8, 16, 32,
64, 128, 256 };
int queries = 3;
int q[][2] = { { 2, 4 }, { 2, 6 },
{ 5, 8 } };
int n = sizeof(arr) / sizeof(arr[0]);
for(int i = 0; i < queries; i++)
cout << (findSum(arr, n, q[i][0], q[i][1]))
<< endl;
return 0;
}
// This code is contributed by divyeshrabadiya07
// Java program to find the sum
// of elements of an GP in the
// given range
import java.io.*;
import java.util.*;
class GFG{
// Function to find sum in the given range
static int findSum(int[] arr, int n,
int left, int right)
{
// Find the value of k
int k = right - left + 1;
// Find the common difference
int d = arr[1] / arr[0];
// Find the sum
int ans = arr[left - 1];
if (d == 1)
ans = ans * d * k;
else
ans = ans * ((int)Math.pow(d, k) - 1 /
(d - 1));
return ans;
}
// Driver Code
public static void main(String args[])
{
int[] arr = { 2, 4, 8, 16, 32,
64, 128, 256 };
int queries = 3;
int[][] q = { { 2, 4 }, { 2, 6 }, { 5, 8 } };
int n = arr.length;
for(int i = 0; i < queries; i++)
System.out.println(findSum(arr, n, q[i][0],
q[i][1]));
}
}
// This code is contributed by offbeat
# Python3 program to
# find the sum of elements
# of an GP in the given range
# Function to find sum in the given range
def findSum(arr, n, left, right):
# Find the value of k
k = right - left + 1
# Find the common difference
d = arr[1] // arr[0]
# Find the sum
ans = arr[left - 1]
if d == 1:
ans = ans * d * k
else:
ans = ans * (d ** k - 1) // (d -1)
return ans
# Driver code
if __name__ == '__main__':
arr = [ 2, 4, 8, 16, 32, 64, 128, 256 ]
queries = 3
q = [[ 2, 4 ], [ 2, 6 ], [ 5, 8 ]]
n = len(arr)
for i in range(queries):
print(findSum(arr, n, q[i][0], q[i][1]))
// C# program to find the sum
// of elements of an GP in the
// given range
using System;
class GFG{
// Function to find sum in the given range
static int findSum(int[] arr, int n,
int left, int right)
{
// Find the value of k
int k = right - left + 1;
// Find the common difference
int d = arr[1] / arr[0];
// Find the sum
int ans = arr[left - 1];
if (d == 1)
ans = ans * d * k;
else
ans = ans * ((int)Math.Pow(d, k) - 1 /
(d - 1));
return ans;
}
// Driver Code
public static void Main(string []args)
{
int[] arr = { 2, 4, 8, 16, 32,
64, 128, 256 };
int queries = 3;
int[,] q = { { 2, 4 }, { 2, 6 }, { 5, 8 } };
int n = arr.Length;
for(int i = 0; i < queries; i++)
Console.Write(findSum(arr, n, q[i, 0],
q[i, 1]) + "\n");
}
}
// This code is contributed by rutvik_56
<script>
// JavaScript program to find the sum
// of elements of an GP in the
// given range
// Function to find sum in the given range
function findSum(arr, n, left, right) {
// Find the value of k
let k = right - left + 1;
// Find the common difference
let d = arr[1] / arr[0];
// Find the sum
let ans = arr[left - 1];
if (d == 1)
ans = ans * d * k;
else
ans = ans * (Math.pow(d, k) - 1 / (d - 1));
return ans;
}
// Driver Code
let arr = [2, 4, 8, 16, 32,
64, 128, 256];
let queries = 3;
let q = [[2, 4], [2, 6],
[5, 8]];
let n = arr.length;
for (let i = 0; i < queries; i++)
document.write(findSum(arr, n, q[i][0], q[i][1]));
// This code is contributed by blalverma92
</script>
Output:
28 124 480
- Time complexity: O(Q)
- Space complexity: O(1)
