Program to find the Sum of numbers in a given range in C
Find the Sum of Numbers in a Given Range in C
Given two integer inputs num1 and num2, the objective is to write a code to Find the Sum of Numbers in a Given Range in C. To do so we’ll keep iterating and adding the numbers from num1 until num2 to the Sum variable.
Example Input : 2 4 Output : 2 + 3 + 4 = 9
Find the Sum of Numbers in a Given Interval in C
Given two integer inputs the objective is to Find the Sum of all the integers that lay within the given interval input by writing a code in C Language. Some Methods for solving the above problem are mentioned below,
- Method 1 : Using Brute Force
- Method 2 : Using the Formula
- Method 3 : Using Recursion
We’ll discuss about the above mentioned methods in the sections below in detail.
Method 1 : Using Brute Force
Algorithm and Working
The algorithm and working of the above mentioned problem is mentioned below,
- Include the required Header files using include keyword.
- Initialize the required variables.
- Create a variable sum = 0
- Run for loop in iterations of (I) in between [a, b]
- In each iteration add I to sum value
- Print sum value
Now let’s try and implement the algorithm in C language.
C Code
#include <stdio.h>
int main()
{
int a = 5;
int b = 10;
int sum = 0;
for (int i = a; i <= b; i++)
sum = sum + i;
printf("%d",sum);
return 0;
}
// Time complexity : O(N)
// Space complexity : O(1)
Output
45
Method 2 : Using the Formula
Algorithm and Working
This method uses a concept that the sum of first n natural numbers can be found using direct formulae – n(n+1)/2
For sum between [a, b] we can simply –
- Create a variable sum = 0
- sum = b*(b+1)/2 – a*(a+1)/2 + a
- Print the sum
Note – An extra ‘a’ is added at the end for offset (check last part of the formula)
Example – Assume you had to calculate sum b/w [3, 5]. Doing b*(b+1)/2 – a*(a+1)/2 what we have done is subtracted (1, 2, 3) from (1, 2, 3, 4, 5). Thus only got the sum of (4, 5). So extra 3 i.e. a is added in the formula
C Code
#include <stdio.h>
int main()
{
int a = 5;
int b = 10;
int sum = b*(b+1)/2 - a*(a+1)/2 + a;
printf("%d",sum);
return 0;
}
// Time complexity : O(1)
// Space complexity : O(1)
Output
45
Method 3 : Using Recursion
Working and Algorithm
Uses recursion starting with a (lower number) but only going till b (upper number)
For sum between [a, b] we can simply –
- Include the Required header files using include keyword.
- Initialize the required variables.
- Create a variable sum = 0
- Call function getSum(0, a, b)
- In each recursion, call return i + getSum(sum, i + 1, b)
- In base Case i.e. when recursion call goes beyond upper limit return sum
Let’s try and implement the above algorithm in C Language.
C Code
#include <stdio.h>
int getSum(int sum, int i, int b){
// stop when any recursion call tries to go over b (larger number)
if (i > b)
return sum;
return i + getSum(sum, i + 1, b);
}
int main()
{
int a = 5;
int b = 10;
int sum = getSum(0, a, b);
printf("%d",sum);
return 0;
}
// Time complexity : O(N)
// Space complexity : O(1)
// Auxilary Space complexity : O(1)
// Because of function call stack
Output
45
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#python program
sum=0
for i in range(30,40+1):
sum=sum+i
print(sum)