Sum of elements of an AP in the given range
Last Updated :
20 Sep, 2023
Given an arithmetic 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 AP elements in the given range.
Note: The range is 1-indexed and 1 ? L, R ? N, where N is the size of arr.
Examples:
Input: arr[] = {2, 4, 6, 8, 10, 12, 14, 16}, Q = [[2, 4], [2, 6], [5, 8]]
Output:
18
40
52
Explanation:
Range 1: arr = {4, 6, 8}. Therefore sum = 18
Range 2: arr = {4, 6, 8, 10, 12}. Therefore sum = 40
Range 3: arr = {10, 12, 14, 16}. Therefore sum = 52
Input: arr[] = {7, 14, 21, 28, 35, 42}, Q = [[1, 6], [2, 4], [3, 3]]
Output:
147
63
21
Explanation:
Range 1: arr = {7, 14, 21, 28, 35, 42}. Therefore sum = 147
Range 2: arr = {14, 21, 28}. Therefore sum = 63
Range 3: arr = {21}. Therefore sum = 21
Approach: Since the given sequence is an arithmetic progression, the sum can be easily found in two steps efficiently:
- Multiply the first element of the range by the number of elements in the range.
- Add (d*k*(k+1))/2 to it, where d is the common difference of the AP and k is (number of elements in the range - 1) which corresponds to the number of gaps.
For example:
Suppose a[i] be the first element of the range, d is the common difference of AP and k+1 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]
= a[i] + a[i] + d + a[i] + 2 * d + .... + a[i] + k * d
= a[i] * (k + 1) + d * (1 + 2 + ... + k)
= a[i] * (k + 1) + (d * k * (k+1))/2
Below is the implementation of the above approach:
CPP
// C++ program to find the sum of elements
// of an AP 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;
// Find the common difference
int d = arr[1] - arr[0];
// Find the sum
int ans = arr[left - 1] * (k + 1);
ans = ans + (d * (k * (k + 1))) / 2;
return ans;
}
// Driver code
int main()
{
int arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 };
int queries = 3;
int q[queries][2] = { { 2, 4 },
{ 2, 6 },
{ 5, 6 } };
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;
}
// Java program to find the sum of elements
// of an AP in the given range
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;
// Find the common difference
int d = arr[1] - arr[0];
// Find the sum
int ans = arr[left - 1] * (k + 1);
ans = ans + (d * (k * (k + 1))) / 2;
return ans;
}
// Driver code
public static void main(String[] args)
{
int arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 };
int queries = 3;
int q[][] = { { 2, 4 },
{ 2, 6 },
{ 5, 6 } };
int n = arr.length;
for (int i = 0; i < queries; i++)
System.out.print(findSum(arr, n, q[i][0], q[i][1])
+"\n");
}
}
// This code is contributed by Princi Singh
# Python program to find the sum of elements
# of an AP 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;
# Find the common difference
d = arr[1] - arr[0];
# Find the sum
ans = arr[left - 1] * (k + 1);
ans = ans + (d * (k * (k + 1))) // 2;
return ans;
# Driver code
if __name__ == '__main__':
arr = [ 2, 4, 6, 8, 10, 12, 14, 16 ];
queries = 3;
q = [[ 2, 4 ],[ 2, 6 ],[ 5, 6 ]];
n = len(arr);
for i in range(queries):
print(findSum(arr, n, q[i][0], q[i][1]));
# This code is contributed by sapnasingh4991
// C# program to find the sum of elements
// of an AP 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;
// Find the common difference
int d = arr[1] - arr[0];
// Find the sum
int ans = arr[left - 1] * (k + 1);
ans = ans + (d * (k * (k + 1))) / 2;
return ans;
}
// Driver code
public static void Main(String[] args)
{
int []arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
int queries = 3;
int [,]q = { { 2, 4 },
{ 2, 6 },
{ 5, 6 } };
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 PrinciRaj1992
<script>
// JavaScript program to find the sum of elements
// of an AP 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;
// Find the common difference
let d = arr[1] - arr[0];
// Find the sum
let ans = arr[left - 1] * (k + 1);
ans = ans + (d * (k * (k + 1))) / 2;
return ans;
}
// Driver Code
let arr = [ 2, 4, 6, 8, 10, 12, 14, 16 ];
let queries = 3;
let q = [[ 2, 4 ],
[ 2, 6 ],
[ 5, 6 ]];
let n = arr.length;
for (let i = 0; i < queries; i++)
document.write(findSum(arr, n, q[i][0], q[i][1])
+"<br/>");
</script>
Time complexity: O(q), where q is the length of the given queries array.
Auxiliary Space: O(1)
Method 2: Using Prefix Sum
Approach:
- Create a prefix sum array for the given arithmetic series.
- For each query, calculate the sum of the range by subtracting the prefix sum at L-1 from the prefix sum at R.
- Output the sum for each query.
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
vector<int> arithmetic_series_sum(int arr[], int N, vector<vector<int>> queries) {
// precompute prefix sum
int prefixSum[N];
prefixSum[0] = arr[0];
for (int i = 1; i < N; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
// process queries
vector<int> ans;
for (int i = 0; i < queries.size(); i++) {
int L = queries[i][0];
int R = queries[i][1];
int sum = prefixSum[R - 1] - prefixSum[L - 2];
ans.push_back(sum);
}
return ans;
}
int main() {
int arr[] = {2, 4, 6, 8, 10, 12, 14, 16};
int N = sizeof(arr) / sizeof(arr[0]);
// define queries
vector<vector<int>> queries = {{2, 4},{2, 6},{5, 6}};
// call function to compute arithmetic series sum for queries
vector<int> ans = arithmetic_series_sum(arr, N, queries);
// print answers
for (int i = 0; i < ans.size(); i++) {
cout << ans[i] << endl;
}
return 0;
}
import java.util.ArrayList;
public class GFG {
public static ArrayList<Integer> arithmeticSeriesSum(
int[] arr, int N,
ArrayList<ArrayList<Integer> > queries)
{
// Precompute prefix sum
int[] prefixSum = new int[N];
prefixSum[0] = arr[0];
for (int i = 1; i < N; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
// Process queries
ArrayList<Integer> ans = new ArrayList<>();
for (int i = 0; i < queries.size(); i++) {
int L = queries.get(i).get(0);
int R = queries.get(i).get(1);
int sum = prefixSum[R - 1]
- (L > 1 ? prefixSum[L - 2] : 0);
ans.add(sum);
}
return ans;
}
public static void main(String[] args)
{
int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
int N = arr.length;
// Define queries
ArrayList<ArrayList<Integer> > queries
= new ArrayList<>();
queries.add(
new ArrayList<>(java.util.Arrays.asList(2, 4)));
queries.add(
new ArrayList<>(java.util.Arrays.asList(2, 6)));
queries.add(
new ArrayList<>(java.util.Arrays.asList(5, 6)));
// Call function to compute arithmetic series sum
// for queries
ArrayList<Integer> ans
= arithmeticSeriesSum(arr, N, queries);
// Print answers
for (int i = 0; i < ans.size(); i++) {
System.out.println(ans.get(i));
}
}
}
def arithmetic_series_sum(arr, N, queries):
# precompute prefix sum
prefixSum = [0] * N
prefixSum[0] = arr[0]
for i in range(1, N):
prefixSum[i] = prefixSum[i - 1] + arr[i]
# process queries
ans = []
for i in range(len(queries)):
L = queries[i][0]
R = queries[i][1]
# Compute the sum using prefix sums
sum = prefixSum[R - 1] - prefixSum[L - 2]
ans.append(sum)
return ans
if __name__ == "__main__":
arr = [2, 4, 6, 8, 10, 12, 14, 16]
N = len(arr)
# define queries
queries = [[2, 4], [2, 6], [5, 6]]
# call function to compute arithmetic series sum for queries
ans = arithmetic_series_sum(arr, N, queries)
# print answers
for i in range(len(ans)):
print(ans[i])
using System;
using System.Collections.Generic;
class GFG
{
static List<int> ArithmeticSeriesSum(int[] arr, int N, List<List<int>> queries)
{
// precompute prefix sum
int[] prefixSum = new int[N];
prefixSum[0] = arr[0];
for (int i = 1; i < N; i++)
{
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
// process queries
List<int> ans = new List<int>();
for (int i = 0; i < queries.Count; i++)
{
int L = queries[i][0];
int R = queries[i][1];
int sum = prefixSum[R - 1] - (L - 2 >= 0 ? prefixSum[L - 2] : 0);
ans.Add(sum);
}
return ans;
}
static void Main()
{
int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16 };
int N = arr.Length;
// define queries
List<List<int>> queries = new List<List<int>> {
new List<int> { 2, 4 },
new List<int> { 2, 6 },
new List<int> { 5, 6 }
};
// call function to compute arithmetic series sum for queries
List<int> ans = ArithmeticSeriesSum(arr, N, queries);
// print answers
foreach (int value in ans)
{
Console.WriteLine(value);
}
}
}
function arithmeticSeriesSum(arr, N, queries) {
// precompute prefix sum
let prefixSum = [];
prefixSum[0] = arr[0];
for (let i = 1; i < N; i++) {
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
// process queries
let ans = [];
for (let i = 0; i < queries.length; i++) {
let L = queries[i][0];
let R = queries[i][1];
let sum = prefixSum[R - 1] - (L > 1 ? prefixSum[L - 2] : 0);
ans.push(sum);
}
return ans;
}
let arr = [2, 4, 6, 8, 10, 12, 14, 16];
let N = arr.length;
// define queries
let queries = [[2, 4], [2, 6], [5, 6]];
// call function to compute arithmetic series sum for queries
let ans = arithmeticSeriesSum(arr, N, queries);
// print answers
for (let i = 0; i < ans.length; i++) {
console.log(ans[i]);
}
Time complexity: O(N + Q),N is the size of the arithmetic series and Q is the number of queries.
Auxiliary Space: O(N) for the prefix sum array.
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