Sum of range in a series of first odd then even natural numbers

The sequence first consists of all the odd numbers starting from 1 to n and then remaining even numbers starting 2 up to n. Let’s suppose we have n as 1000. Then the sequence becomes 1 3 5 7….999 2 4 6….1000
We are given a range (L, R), we need to find sum of numbers of this sequence in a given range.

Note: Here the range is given as (L, R) L and R are included in the range

Examples: 

Input  : n = 10
         Range 1 6
Output : 27
Explanation:
Sequence is 1 3 5 7 9 2 4 6 8 10
Sum in range (2, 6) 
= 1 + 3 + 5 + 7 + 9 + 2 
= 27

Input  : n = 5
         Range 1 2
Output : 4
Explanation:
sequence is 1 3 5 2 4
sum = 1 + 3 = 4

The idea is to first find sum of numbers before left(excluding left), then find sum of numbers before right (including right). We get result as second sum minus first sum.

How to find sum till a limit?

We first count how many odd numbers are there, then we use formulas for sum of odd natural numbers and sum of even natural numbers to find the result.
How to find count of odd numbers?  

• If n is odd then the number of odd numbers are ((n/2) + 1)
• If n is even then number of odd numbers are (n/2)

By simple observation, we get the number of odd numbers is ceil(n/2). So, the number of even numbers are n – ceil(n/2). 

• Sum of first N odd numbers is (N^2)  
• Sum of first N even numbers is (N^2) + N 

For a given number x how will we find the sum in the sequence from 1 to x? Let us suppose x is less than the number of odd numbers.  

• Then we simply return (x*x)

If the x is greater than the number of odd numbers 

• var = x-odd;
• That means we need first var even numbers
• we return (odd*odd) + (var*var) + var;

C++

// CPP program to find sum in the given range in
// the sequence 1 3 5 7.....N 2 4 6...N-1
#include <bits/stdc++.h>
using namespace std;

// For our convenience
#define ll long long

// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
ll sumTillX(ll x, ll n)
{
    // number of odd numbers
    ll odd = ceil(n / 2.0);

    if (x <= odd)
       return x * x;

    // number of extra even
    // numbers required
    ll even = x - odd;

    return ((odd * odd) + (even * even) + even);
}

int rangeSum(int N, int L, int R)
{
   return sumTillX(R, N) - sumTillX(L-1, N);
}

// Driver code
int main()
{
    ll N = 10, L = 1, R = 6;
    cout << rangeSum(N, L, R);
    return 0;
}

C

// C program to find sum in the given range in
// the sequence 1 3 5 7.....N 2 4 6...N-1
#include <stdio.h>
#include <math.h>

// For our convenience
#define ll long long

// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
ll sumTillX(ll x, ll n)
{
    // number of odd numbers
    ll odd = ceil(n / 2.0);

    if (x <= odd)
       return x * x;

    // number of extra even
    // numbers required
    ll even = x - odd;

    return ((odd * odd) + (even * even) + even);
}

int rangeSum(int N, int L, int R)
{
   return sumTillX(R, N) - sumTillX(L-1, N);
}

// Driver code
int main()
{
    ll N = 10, L = 1, R = 6;
    printf("%d",rangeSum(N, L, R));
    return 0;
}

// This code is contributed by kothavvsaakash.

Java

// Java program to find
// sum in the given
// range in the sequence
// 1 3 5 7.....N
// 2 4 6...N-1

class GFG {

    // Function that returns sum
    // in the range 1 to x in the
    // sequence 1 3 5 7.....N 2 4 6...N-1
    static double sumTillX(double x,
                           double n)
    {

        // number of odd numbers
        double odd = Math.ceil(n / 2.0);

        if (x <= odd)
            return x * x;

        // number of extra even
        // numbers required
        double even = x - odd;

        return ((odd * odd) + (even *
                       even) + even);
    }

    static double rangeSum(double N,
                           double L,
                           double R)
    {
        return sumTillX(R, N) -
               sumTillX(L-1, N);
    }

    // Driver Code
    public static void main(String args[])
    {
        long N = 10, L = 1, R = 6;
        int n = 101;
        System.out.println((int)rangeSum(N, L, R));

    }
}

// This code is contributed by Sam007

Python 3

# Python 3 program to find sum in the
# given range in the sequence 1 3 5 7
# .....N 2 4 6...N-1
import math

# For our convenience
#define ll long long

# Function that returns sum in the
# range 1 to x in the sequence
# 1 3 5 7.....N 2 4 6...N-1
def sumTillX(x, n):

    # number of odd numbers
    odd = math.ceil(n / 2.0)

    if (x <= odd):
        return x * x;

    # number of extra even
    # numbers required
    even = x - odd;

    return ((odd * odd) +
            (even * even) + even);


def rangeSum(N, L, R):

    return (sumTillX(R, N) -
                sumTillX(L-1, N));

# Driver code
N = 10
L = 1
R = 6
print(rangeSum(N, L, R))

# This code is contributed by
# Smitha

C#

// C# program to find sum in the given
// range in the sequence 1 3 5 7.....N
// 2 4 6...N-1
using System;

public class GFG {

    // Function that returns sum
    // in the range 1 to x in the
    // sequence 1 3 5 7.....N 2 4 6...N-1
    static double sumTillX(double x, double n)
    {

        // number of odd numbers
        double odd = Math.Ceiling(n / 2.0);

        if (x <= odd)
            return x * x;

        // number of extra even
        // numbers required
        double even = x - odd;

        return ((odd * odd) + (even * even)
                                    + even);
    }

    static double rangeSum(double N, double L,
                                       double R)
    {
        return sumTillX(R, N) - sumTillX(L-1, N);
    }

    // Driver code
    public static void Main()
    {
        long N = 10, L = 1, R = 6;
        Console.Write(rangeSum(N, L, R));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP program to find sum
// in the given range in the
// sequence 1 3 5 7.....
// N 2 4 6...N-1

// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....
// N 2 4 6...N-1
function sumTillX($x, $n)
{

    // number of odd numbers
    $odd = ceil($n / 2.0);

    if ($x <= $odd)
    return $x * $x;

    // number of extra even
    // numbers required
    $even = $x - $odd;

    return (($odd * $odd) +
            ($even * $even) +
             $even);
}

function rangeSum($N, $L, $R)
{
    return sumTillX($R, $N) -
           sumTillX($L-1, $N);
}

// Driver code
$N = 10; $L = 1; $R = 6;
echo(rangeSum($N, $L, $R));

// This code is contributed by Ajit.
?>

Javascript

<script>
// javascript program to find sum in the given range in
// the sequence 1 3 5 7.....N 2 4 6...N-1

// Function that returns sum
// in the range 1 to x in the
// sequence 1 3 5 7.....N 2 4 6...N-1
function sumTillX( x,  n)
{

    // number of odd numbers
    let odd = Math.ceil(n / 2.0);

    if (x <= odd)
       return x * x;

    // number of extra even
    // numbers required
    let even = x - odd;

    return ((odd * odd) + (even * even) + even);
}

function rangeSum( N,  L,  R)
{
   return sumTillX(R, N) - sumTillX(L-1, N);
}

// Driver code
    let N = 10, L = 1, R = 6;
     document.write(rangeSum(N, L, R));

// This code is contributed by Rajput-Ji

</script>

Output:

27

Time Complexity : O(1)
Auxiliary Space: O(1)