long long sumFirstN(long long n) {
if(n==0)
return 0;
return sumFirstN(n-1)+n;
}
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Sum Of First N Numbers
Easy
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Average time to solve is 10m
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104 upvotes
Problem statement
You are given an integer ‘n’.
Your task is determining the sum of the first ‘n’ natural numbers and returning it.
Input: ‘n’ = 3
Output: 6
Explanation: The sum of the first 3 natural numbers is 1 + 2 + 3, equal to 6.
Detailed explanation ( Input/output format, Notes, Images )
Input Format:
The first line is an integer ‘n’, denoting the integer.
Return the sum of the first n natural numbers.
You don't need to print anything. Just implement the given function.
Sample Input 1:
3
Sample Output 1 :
6
Explanation Of Sample Input 1:
Input: ‘n’ = 3
Output: 6
Explanation: The sum of the first 3 natural numbers is 1 + 2 + 3, equal to 6.
Sample Input 2:
5
Sample Output 2:
15
Explanation of sample output 2:
Input: ‘n’ = 5
Output: 15
Explanation: The sum of the first 5 natural numbers is 1 + 2 + 3 + 4 + 5, equal to 15.
Expected Time Complexity:
The expected time complexity is O(1).
Expected Space Complexity:
The expected space complexity is O(1).
Constraints:
1 <= n <= 10^9
Time Limit: 1 sec
Hint
Hint 1
Can we use a formula to calculate the answer?
Approaches (1)
Approach 1
Maths
We know that the formula to calculate the sum of the first ‘n’ natural numbers is ‘(n) * (n + 1) / 2’. So, what we will do is we will put the given ‘n’ in this formula and store it in a variable. Finally, we will return the answer.
The steps are as follows:
function sumFirstN(int n):
- Initialize a variable ‘ans’.
- Assign ‘ans’ to ‘n * (n + 1) / 2’.
- Return ‘ans’.
Time Complexity
O(1).
We are calculating the answer in constant operation.
Hence, the time complexity is O(1).
Space Complexity
O(1).
We are using a variable to keep the answer.
Hence, the space complexity is O(1).
Code Solution
(100% EXP penalty)
Sum Of First N Numbers
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Interview problems
The most effective C++ solution
long long sumFirstN(long long n,long sum = 0) {
// Write your code here.
if(n==0){
return sum;
}
sum = n+sumFirstN(n-1,sum);
}
303 views
1 reply
1 upvote
Easy sol. in JAVA (one test case wasn't passed)
public class Solution {
public static long ans =0;
public static long sumFirstN(long n) {
// Write your code here.
if(n==1) {
return 1;
}
long sum_nm1= sumFirstN(n-1);
long sum_n = n + sum_nm1;
return sum_n;
}
}
137 views
0 replies
0 upvotes
Interview problems
Java soln
public class Solution {
public static long sumFirstN(long n) {
int sum=0;
return printNSum(n,sum);
}
public static long printNSum(long x, int sum1){
int p=1;
while(p<=x){
sum1=sum1+p;
p++;
}
return sum1;
}
}
134 views
1 reply
0 upvotes
Interview problems
java code with easy using (am)
import java.util.*;
public class Solution {
public static long sumFirstN(long n) {
long sum=n*(n+1)/2;
return sum;
}
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
System.out.println(sumFirstN(sc.nextLong()));
}
}
77 views
0 replies
0 upvotes
Interview problems
With the help of Arithmetic formula
"As we know, the formula for the sum of the first 'n' numbers is sum = n * (n + 1) / 2;
So, if we implement this formula in our code, it would be much more efficient unless it's provided that we do it by some other method. With this approach, the time and space complexity would be O(1)." long long sumFirstN(long long n) {
// Write your code here.
long long ans=0;
ans=n*(n+1)/2;
return ans;
}
114 views
0 replies
1 upvote
Interview problems
java answer
public static long sumFirstN(long n) {
// Write your code here.
if(n>0){
n=n+sumFirstN(n-1);
}
return n;
}
114 views
0 replies
0 upvotes
Interview problems
C++ (One line)
long long sumFirstN(long long n) {
// Write your code here.
return (n*(n +1))/2;
}
279 views
1 reply
2 upvotes
Interview problems
Easy Java Code
public class Solution {
public static long sumFirstN(long n) {
if(n<=0){
return n;
}
return n + sumFirstN(n-1);
}
}
107 views
0 replies
0 upvotes
Interview problems
Python Code - But getting TLE in last testcase
from typing import List
def sumFirstN(n: int) -> int:
if n == 0: # Base case: when n is 0, return 0
return 0
else:
return n + sumFirstN(n - 1) 71 views
0 replies
0 upvotes
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C++ (g++ 5.4)
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