Sum of elements in range L-R where first half and second half is filled with odd and even numbers
Last Updated :
31 Aug, 2021
Given a number N, create an array such the first half of the array is filled with odd numbers till N, and the second half of the array is filled with even numbers. Also given are L and R indices, the task is to print the sum of elements in the array in the range [L, R].
Examples:
Input: N = 12, L = 1, R = 11
Output: 66
The array formed thus is {1, 3, 5, 7, 9, 11, 2, 4, 6, 8, 10, 12}
The sum between index 1 and index 11 is 66 {1+3+5+7+9+11+2+4+6+8+10}
Input: N = 11, L = 3 and R = 7
Output: 34
The array formed is {1, 3, 5, 7, 9, 11, 2, 4, 6, 8, 10}
The sum between index 3 and index 7 is 34 {5+7+9+11+2}
A naive approach will be to form the array of size N and iterate from L to R and return the sum.
C++
// C++ program to find the sum between L-R
// range by creating the array
// Naive Approach
#include<bits/stdc++.h>
using namespace std;
// Function to find the sum between L and R
int rangesum(int n, int l, int r)
{
// array created
int arr[n];
// fill the first half of array
int c = 1, i = 0;
while (c <= n) {
arr[i++] = c;
c += 2;
}
// fill the second half of array
c = 2;
while (c <= n) {
arr[i++] = c;
c += 2;
}
int sum = 0;
// find the sum between range
for (i = l - 1; i < r; i++) {
sum += arr[i];
}
return sum;
}
// Driver Code
int main()
{
int n = 12;
int l = 1, r = 11;
cout<<(rangesum(n, l, r));
}
// This code is contributed by
// Sanjit_Prasad
// Java program to find the sum between L-R
// range by creating the array
// Naive Approach
import java.io.*;
public class GFG {
// Function to find the sum between L and R
static int rangesum(int n, int l, int r)
{
// array created
int[] arr = new int[n];
// fill the first half of array
int c = 1, i = 0;
while (c <= n) {
arr[i++] = c;
c += 2;
}
// fill the second half of array
c = 2;
while (c <= n) {
arr[i++] = c;
c += 2;
}
int sum = 0;
// find the sum between range
for (i = l - 1; i < r; i++) {
sum += arr[i];
}
return sum;
}
// Driver Code
public static void main(String[] args)
{
int n = 12;
int l = 1, r = 11;
System.out.println(rangesum(n, l, r));
}
}
# Python3 program to find the sum between L-R
# range by creating the array
# Naive Approach
# Function to find the sum between L and R
def rangesum(n, l, r):
# array created
arr = [0] * n;
# fill the first half of array
c = 1; i = 0;
while (c <= n):
arr[i] = c;
i += 1;
c += 2;
# fill the second half of array
c = 2;
while (c <= n):
arr[i] = c;
i += 1;
c += 2;
sum = 0;
# find the sum between range
for i in range(l - 1, r, 1):
sum += arr[i];
return sum;
# Driver Code
if __name__ == '__main__':
n = 12;
l, r = 1, 11;
print(rangesum(n, l, r));
# This code contributed by PrinciRaj1992
// C# program to find the sum
// between L-R range by creating
// the array Naive Approach
using System;
class GFG
{
// Function to find the
// sum between L and R
static int rangesum(int n,
int l, int r)
{
// array created
int[] arr = new int[n];
// fill the first
// half of array
int c = 1, i = 0;
while (c <= n)
{
arr[i++] = c;
c += 2;
}
// fill the second
// half of array
c = 2;
while (c <= n)
{
arr[i++] = c;
c += 2;
}
int sum = 0;
// find the sum
// between range
for (i = l - 1; i < r; i++)
{
sum += arr[i];
}
return sum;
}
// Driver Code
public static void Main()
{
int n = 12;
int l = 1, r = 11;
Console.WriteLine(rangesum(n, l, r));
}
}
// This code is contributed
// by inder_verma.
<?php
// PHP program to find the sum between
// L-R range by creating the array
// Naive Approach
// Function to find the sum between L and R
function rangesum($n, $l, $r)
{
// array created
$arr = array_fill(0, $n, 0);
// fill the first half of array
$c = 1;
$i = 0;
while ($c <= $n)
{
$arr[$i++] = $c;
$c += 2;
}
// fill the second half of array
$c = 2;
while ($c <= $n)
{
$arr[$i++] = $c;
$c += 2;
}
$sum = 0;
// find the sum between range
for ($i = $l - 1; $i < $r; $i++)
{
$sum += $arr[$i];
}
return $sum;
}
// Driver Code
$n = 12;
$l = 1;
$r = 11;
echo(rangesum($n, $l, $r));
// This code is contributed by mits
?>
<script>
// Javascript program to find the sum between L-R
// range by creating the array
// Naive Approach
// Function to find the sum between L and R
function rangesum(n, l, r)
{
// array created
let arr = new Array(n);
// fill the first half of array
let c = 1, i = 0;
while (c <= n) {
arr[i++] = c;
c += 2;
}
// fill the second half of array
c = 2;
while (c <= n) {
arr[i++] = c;
c += 2;
}
let sum = 0;
// find the sum between range
for (i = l - 1; i < r; i++) {
sum += arr[i];
}
return sum;
}
// Driver Code
let n = 12;
let l = 1, r = 11;
document.write(rangesum(n, l, r));
// This code is contributed by rishavmahato348.
</script>
Time complexity: O(N)
Auxiliary Space: O(N)
Efficient Approach: Without building the array the problem can be solved in O(1) complexity. Consider the middle point of the sequence thus formed. Then there can be 3 possibilities for L and R.
- l and r both are on the right of the middle point.
- l and r both are on the left of the middle point.
- l is on the left of the middle point and r is on the left of the middle point.
For the 1st and 2nd case, just find the number which corresponds to the l-th position from the start or middle and the number which corresponds to the r-th position from start or middle. The below formula can be used to find out the sum:
sum = (no. of terms in the range)*(first term + last term)/2
For the third case, consider it as [L-mid] and [mid-R] and then apply case 1 and case 2 formulae as mentioned above.
Below is the implementation of the above approach:
C++
// C++ program to find the sum between L-R
// range by creating the array
// Naive Approach
#include<bits/stdc++.h>
using namespace std;
// // Function to calculate the sum if n is even
int sumeven(int n, int l, int r)
{
int sum = 0;
int mid = n / 2;
// both l and r are to the left of mid
if (r <= mid)
{
// first and last element
int first = (2 * l - 1);
int last = (2 * r - 1);
// Total number of terms in
// the sequence is r-l+1
int no_of_terms = r - l + 1;
// use of formula derived
sum = ((no_of_terms) * ((first + last))) / 2;
}
// both l and r are to the right of mid
else if (l >= mid)
{
// first and last element
int first = (2 * (l - n / 2));
int last = (2 * (r - n / 2));
int no_of_terms = r - l + 1;
// Use of formula derived
sum = ((no_of_terms) * ((first + last))) / 2;
}
// left is to the left of mid and
// right is to the right of mid
else
{
// Take two sums i.e left and
// right differently and add
int sumleft = 0, sumright = 0;
// first and last element
int first_term1 = (2 * l - 1);
int last_term1 = (2 * (n / 2) - 1);
// total terms
int no_of_terms1 = n / 2 - l + 1;
// no of terms
sumleft = ((no_of_terms1) *
((first_term1 + last_term1))) / 2;
// The first even number is 2
int first_term2 = 2;
// The last element is given by 2*(r-n/2)
int last_term2 = (2 * (r - n / 2));
int no_of_terms2 = r - mid;
// formula applied
sumright = ((no_of_terms2) *
((first_term2 + last_term2))) / 2;
sum = (sumleft + sumright);
}
return sum;
}
// Function to calculate the sum if n is odd
int sumodd(int n, int l, int r)
{
// take ceil value if n is odd
int mid = n / 2 + 1;
int sum = 0;
// // both l and r are to the left of mid
if (r <= mid)
{
// first and last element
int first = (2 * l - 1);
int last = (2 * r - 1);
// number of terms
int no_of_terms = r - l + 1;
// formula
sum = ((no_of_terms) *
((first + last))) / 2;
}
// // both l and r are to the right of mid
else if (l > mid)
{
// first and last term,
int first = (2 * (l - mid));
int last = (2 * (r - mid));
// no of terms
int no_of_terms = r - l + 1;
// formula used
sum = ((no_of_terms) *
((first + last))) / 2;
}
// If l is on left and r on right
else
{
// calculate separate sums
int sumleft = 0, sumright = 0;
// first half
int first_term1 = (2 * l - 1);
int last_term1 = (2 * mid - 1);
// calculate terms
int no_of_terms1 = mid - l + 1;
sumleft = ((no_of_terms1) *
((first_term1 + last_term1))) / 2;
// second half
int first_term2 = 2;
int last_term2 = (2 * (r - mid));
int no_of_terms2 = r - mid;
sumright = ((no_of_terms2) *
((first_term2 + last_term2))) / 2;
// add both halves
sum = (sumleft + sumright);
}
return sum;
}
// Function to find the sum between L and R
int rangesum(int n, int l, int r)
{
int sum = 0;
// If n is even
if (n % 2 == 0)
return sumeven(n, l, r);
// If n is odd
else
return sumodd(n, l, r);
}
// Driver Code
int main()
{
int n = 12;
int l = 1, r = 11;
cout << (rangesum(n, l, r));
}
// This code is contributed by 29AjayKumar
// Java program to find the sum between L-R
// range by creating the array
// Efficient Approach
import java.io.*;
public class GFG {
// // Function to calculate the sum if n is even
static int sumeven(int n, int l, int r)
{
int sum = 0;
int mid = n / 2;
// both l and r are to the left of mid
if (r <= mid) {
// first and last element
int first = (2 * l - 1);
int last = (2 * r - 1);
// Total number of terms in
// the sequence is r-l+1
int no_of_terms = r - l + 1;
// use of formula derived
sum = ((no_of_terms) * ((first + last))) / 2;
}
// both l and r are to the right of mid
else if (l >= mid) {
// // first and last element
int first = (2 * (l - n / 2));
int last = (2 * (r - n / 2));
int no_of_terms = r - l + 1;
// Use of formula derived
sum = ((no_of_terms) * ((first + last))) / 2;
}
// left is to the left of mid and
// right is to the right of mid
else {
// Take two sums i.e left and
// right differently and add
int sumleft = 0, sumright = 0;
// first and last element
int first_term1 = (2 * l - 1);
int last_term1 = (2 * (n / 2) - 1);
// total terms
int no_of_terms1 = n / 2 - l + 1;
// no of terms
sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2;
// The first even number is 2
int first_term2 = 2;
// The last element is given by 2*(r-n/2)
int last_term2 = (2 * (r - n / 2));
int no_of_terms2 = r - mid;
// formula applied
sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2;
sum = (sumleft + sumright);
}
return sum;
}
// Function to calculate the sum if n is odd
static int sumodd(int n, int l, int r)
{
// take ceil value if n is odd
int mid = n / 2 + 1;
int sum = 0;
// // both l and r are to the left of mid
if (r <= mid) {
// first and last element
int first = (2 * l - 1);
int last = (2 * r - 1);
// number of terms
int no_of_terms = r - l + 1;
// formula
sum = ((no_of_terms) * ((first + last))) / 2;
}
// // both l and r are to the right of mid
else if (l > mid) {
// first and last term,
int first = (2 * (l - mid));
int last = (2 * (r - mid));
// no of terms
int no_of_terms = r - l + 1;
// formula used
sum = ((no_of_terms) * ((first + last))) / 2;
}
// If l is on left and r on right
else {
// calculate separate sums
int sumleft = 0, sumright = 0;
// first half
int first_term1 = (2 * l - 1);
int last_term1 = (2 * mid - 1);
// calculate terms
int no_of_terms1 = mid - l + 1;
sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2;
// second half
int first_term2 = 2;
int last_term2 = (2 * (r - mid));
int no_of_terms2 = r - mid;
sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2;
// add both halves
sum = (sumleft + sumright);
}
return sum;
}
// Function to find the sum between L and R
static int rangesum(int n, int l, int r)
{
int sum = 0;
// If n is even
if (n % 2 == 0)
return sumeven(n, l, r);
// If n is odd
else
return sumodd(n, l, r);
}
// Driver Code
public static void main(String[] args)
{
int n = 12;
int l = 1, r = 11;
System.out.println(rangesum(n, l, r));
}
}
# Python3 program to find
# the sum between L-R range
# by creating the array
# Naive Approach
# Function to calculate
# the sum if n is even
def sumeven(n, l, r):
sum = 0
mid = n // 2
# Both l and r are
# to the left of mid
if (r <= mid):
# First and last element
first = (2 * l - 1)
last = (2 * r - 1)
# Total number of terms in
# the sequence is r-l+1
no_of_terms = r - l + 1
# Use of formula derived
sum = ((no_of_terms) *
((first + last))) // 2
# Both l and r are to the right of mid
elif (l >= mid):
# First and last element
first = (2 * (l - n // 2))
last = (2 * (r - n // 2))
no_of_terms = r - l + 1
# Use of formula derived
sum = ((no_of_terms) *
((first + last))) // 2
# Left is to the left of mid and
# right is to the right of mid
else :
# Take two sums i.e left and
# right differently and add
sumleft, sumright = 0, 0
# First and last element
first_term1 = (2 * l - 1)
last_term1 = (2 * (n // 2) - 1)
# total terms
no_of_terms1 = n // 2 - l + 1
# no of terms
sumleft = (((no_of_terms1) *
((first_term1 +
last_term1))) // 2)
# The first even number is 2
first_term2 = 2
# The last element is
#
last_term2 = (2 * (r - n // 2))
no_of_terms2 = r - mid
# formula applied
sumright = (((no_of_terms2) *
((first_term2 +
last_term2))) // 2)
sum = (sumleft + sumright);
return sum
# Function to calculate
# the sum if n is odd
def sumodd(n, l, r):
# Take ceil value if n is odd
mid = n // 2 + 1;
sum = 0
# Both l and r are to
# the left of mid
if (r <= mid):
# First and last element
first = (2 * l - 1)
last = (2 * r - 1)
# number of terms
no_of_terms = r - l + 1
# formula
sum = (((no_of_terms) *
((first + last))) // 2)
# both l and r are to the right of mid
elif (l > mid):
# first and last term,
first = (2 * (l - mid))
last = (2 * (r - mid))
# no of terms
no_of_terms = r - l + 1
# formula used
sum = (((no_of_terms) *
((first + last))) // 2)
# If l is on left and r on right
else :
# calculate separate sums
sumleft, sumright = 0, 0
# first half
first_term1 = (2 * l - 1)
last_term1 = (2 * mid - 1)
# calculate terms
no_of_terms1 = mid - l + 1
sumleft = (((no_of_terms1) *
((first_term1 +
last_term1))) // 2)
# second half
first_term2 = 2
last_term2 = (2 * (r - mid))
no_of_terms2 = r - mid
sumright = (((no_of_terms2) *
((first_term2 +
last_term2))) // 2)
# add both halves
sum = (sumleft + sumright)
return sum
# Function to find the
# sum between L and R
def rangesum(n, l, r):
sum = 0
# If n is even
if (n % 2 == 0):
return sumeven(n, l, r);
# If n is odd
else:
return sumodd(n, l, r)
# Driver Code
if __name__ == "__main__":
n = 12
l = 1
r = 11;
print (rangesum(n, l, r))
# This code is contributed by Chitranayal
// C# program to find the sum
// between L-R range by creating
// the array Efficient Approach
using System;
class GFG
{
// Function to calculate
// the sum if n is even
static int sumeven(int n,
int l, int r)
{
int sum = 0;
int mid = n / 2;
// both l and r are
// to the left of mid
if (r <= mid)
{
// first and last element
int first = (2 * l - 1);
int last = (2 * r - 1);
// Total number of terms in
// the sequence is r-l+1
int no_of_terms = r - l + 1;
// use of formula derived
sum = ((no_of_terms) *
((first + last))) / 2;
}
// both l and r are
// to the right of mid
else if (l >= mid)
{
// first and last element
int first = (2 * (l - n / 2));
int last = (2 * (r - n / 2));
int no_of_terms = r - l + 1;
// Use of formula derived
sum = ((no_of_terms) *
((first + last))) / 2;
}
// left is to the left of
// mid and right is to the
// right of mid
else
{
// Take two sums i.e left and
// right differently and add
int sumleft = 0, sumright = 0;
// first and last element
int first_term1 = (2 * l - 1);
int last_term1 = (2 * (n / 2) - 1);
// total terms
int no_of_terms1 = n / 2 - l + 1;
// no of terms
sumleft = ((no_of_terms1) *
((first_term1 +
last_term1))) / 2;
// The first even
// number is 2
int first_term2 = 2;
// The last element is
// given by 2*(r-n/2)
int last_term2 = (2 * (r - n / 2));
int no_of_terms2 = r - mid;
// formula applied
sumright = ((no_of_terms2) *
((first_term2 +
last_term2))) / 2;
sum = (sumleft + sumright);
}
return sum;
}
// Function to calculate
// the sum if n is odd
static int sumodd(int n,
int l, int r)
{
// take ceil value
// if n is odd
int mid = n / 2 + 1;
int sum = 0;
// both l and r are
// to the left of mid
if (r <= mid)
{
// first and last element
int first = (2 * l - 1);
int last = (2 * r - 1);
// number of terms
int no_of_terms = r - l + 1;
// formula
sum = ((no_of_terms) *
((first + last))) / 2;
}
// both l and r are
// to the right of mid
else if (l > mid)
{
// first and last term,
int first = (2 * (l - mid));
int last = (2 * (r - mid));
// no of terms
int no_of_terms = r - l + 1;
// formula used
sum = ((no_of_terms) *
((first + last))) / 2;
}
// If l is on left
// and r on right
else
{
// calculate separate sums
int sumleft = 0, sumright = 0;
// first half
int first_term1 = (2 * l - 1);
int last_term1 = (2 * mid - 1);
// calculate terms
int no_of_terms1 = mid - l + 1;
sumleft = ((no_of_terms1) *
((first_term1 +
last_term1))) / 2;
// second half
int first_term2 = 2;
int last_term2 = (2 * (r - mid));
int no_of_terms2 = r - mid;
sumright = ((no_of_terms2) *
((first_term2 +
last_term2))) / 2;
// add both halves
sum = (sumleft + sumright);
}
return sum;
}
// Function to find the
// sum between L and R
static int rangesum(int n,
int l, int r)
{
// If n is even
if (n % 2 == 0)
return sumeven(n, l, r);
// If n is odd
else
return sumodd(n, l, r);
}
// Driver Code
public static void Main()
{
int n = 12;
int l = 1, r = 11;
Console.WriteLine(rangesum(n, l, r));
}
}
// This code is contributed
// by chandan_jnu.
<script>
// Javascript program to find the sum between L-R
// range by creating the array
// Efficient Approach
// Function to calculate the sum if n is even
function sumeven(n,l,r)
{
let sum = 0;
let mid = Math.floor(n / 2);
// both l and r are to the left of mid
if (r <= mid) {
// first and last element
let first = (2 * l - 1);
let last = (2 * r - 1);
// Total number of terms in
// the sequence is r-l+1
let no_of_terms = r - l + 1;
// use of formula derived
sum = ((no_of_terms) * ((first + last))) / 2;
}
// both l and r are to the right of mid
else if (l >= mid) {
// // first and last element
let first = (2 * (l - n / 2));
let last = (2 * (r - n / 2));
let no_of_terms = r - l + 1;
// Use of formula derived
sum = ((no_of_terms) * ((first + last))) / 2;
}
// left is to the left of mid and
// right is to the right of mid
else {
// Take two sums i.e left and
// right differently and add
let sumleft = 0, sumright = 0;
// first and last element
let first_term1 = (2 * l - 1);
let last_term1 = (2 * (n / 2) - 1);
// total terms
let no_of_terms1 = n / 2 - l + 1;
// no of terms
sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2;
// The first even number is 2
let first_term2 = 2;
// The last element is given by 2*(r-n/2)
let last_term2 = (2 * (r - n / 2));
let no_of_terms2 = r - mid;
// formula applied
sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2;
sum = (sumleft + sumright);
}
return sum;
}
// Function to calculate the sum if n is odd
function sumodd(n,l,r)
{
// take ceil value if n is odd
let mid = Math.floor(n / 2) + 1;
let sum = 0;
// // both l and r are to the left of mid
if (r <= mid) {
// first and last element
let first = (2 * l - 1);
let last = (2 * r - 1);
// number of terms
let no_of_terms = r - l + 1;
// formula
sum = ((no_of_terms) * ((first + last))) / 2;
}
// // both l and r are to the right of mid
else if (l > mid) {
// first and last term ,
let first = (2 * (l - mid));
let last = (2 * (r - mid));
// no of terms
let no_of_terms = r - l + 1;
// formula used
sum = ((no_of_terms) * ((first + last))) / 2;
}
// If l is on left and r on right
else {
// calculate separate sums
let sumleft = 0, sumright = 0;
// first half
let first_term1 = (2 * l - 1);
let last_term1 = (2 * mid - 1);
// calculate terms
let no_of_terms1 = mid - l + 1;
sumleft = ((no_of_terms1) * ((first_term1 + last_term1))) / 2;
// second half
let first_term2 = 2;
let last_term2 = (2 * (r - mid));
let no_of_terms2 = r - mid;
sumright = ((no_of_terms2) * ((first_term2 + last_term2))) / 2;
// add both halves
sum = (sumleft + sumright);
}
return sum;
}
// Function to find the sum between L and R
function rangesum(n,l,r)
{
let sum = 0;
// If n is even
if (n % 2 == 0)
return sumeven(n, l, r);
// If n is odd
else
return sumodd(n, l, r);
}
// Driver Code
let n = 12;
let l = 1, r = 11;
document.write(rangesum(n, l, r));
// This code is contributed by rag2127
</script>
Time complexity: O(1)
Auxiliary Space: O(1)
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