Calculate sum of all integers from 1 to N, excluding perfect power of 2

0 0 3 minutes read

Given a positive integer N, the task is to calculate the sum of all integers from 1 to N but excluding the number which is a perfect power of 2.
Examples:

Input: N = 2
Output: 0
Input: N = 1000000000
Output: 499999998352516354

Naive Approach:
The naive approach is to iterate every number from 1 to N and compute the sum in the variable by excluding the number which is a perfect power of 2. But to compute the sum to the number 10^9, the above approach will give Time Limit Error.
Time Complexity: O(N)
Efficient Approach:
To find desired sum, below are the steps:

Find the sum of all the number till N using the formula discussed in this article in O(1) time.
Since sum of all perfect power of 2 forms a Geometric Progression. Hence the sum of all powers of 2 less than N is calculated by the below formula:

The number of element with perfect power of 2 less than N is given by log₂N,
Let r = log₂N
And the sum of all numbers which are perfect power of 2 is given by 2^r – 1.

Subtract the sum of all perfect powers of 2 calculated above from the sum of first N numbers to get the result.

Below is the implementation of the above approach:

C++

// C++ implementation of the
// approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the required
// summation
void findSum(int N)
{
    // Find the sum of first N
    // integers using the formula
    int sum = (N) * (N + 1) / 2;
     
    int r = log2(N) + 1;
     
    // Find the sum of numbers 
    // which are exact power of
    // 2 by using the formula
    int expSum = pow(2, r) - 1;    
 
    // Print the final Sum
    cout << sum - expSum << endl;
}
 
// Driver's Code
int main()
{
    int N = 2;
 
    // Function to find the
    // sum
    findSum(N);
    return 0;
}

Java

// Java implementation of the above approach 
import java.lang.Math;
 
class GFG{
     
// Function to find the required 
// summation 
public static void findSum(int N) 
{ 
 
    // Find the sum of first N 
    // integers using the formula 
    int sum = (N) * (N + 1) / 2; 
         
    int r = (int)(Math.log(N) / 
                  Math.log(2)) + 1; 
         
    // Find the sum of numbers 
    // which are exact power of 
    // 2 by using the formula 
    int expSum = (int)(Math.pow(2, r)) - 1;     
     
    // Print the final Sum 
    System.out.println(sum - expSum); 
} 
 
// Driver Code
public static void main(String[] args)
{
    int N = 2; 
 
    // Function to find the sum 
    findSum(N); 
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python 3 implementation of the
# approach
from math import log2,pow
 
# Function to find the required
# summation
def findSum(N):
    # Find the sum of first N
    # integers using the formula
    sum = (N) * (N + 1) // 2
     
    r = log2(N) + 1
     
    # Find the sum of numbers 
    # which are exact power of
    # 2 by using the formula
    expSum = pow(2, r) - 1
 
    # Print the final Sum
    print(int(sum - expSum))
 
# Driver's Code
if __name__ == '__main__':
    N = 2
 
    # Function to find the
    # sum
    findSum(N)
     
# This code is contributed by Surendra_Gangwar

C#

// C# implementation of the above approach 
using System;
 
class GFG{ 
     
// Function to find the required 
// summation 
public static void findSum(int N) 
{ 
 
    // Find the sum of first N 
    // integers using the formula 
    int sum = (N) * (N + 1) / 2; 
         
    int r = (int)(Math.Log(N) / 
                  Math.Log(2)) + 1; 
         
    // Find the sum of numbers 
    // which are exact power of 
    // 2 by using the formula 
    int expSum = (int)(Math.Pow(2, r)) - 1; 
     
    // Print the final Sum 
    Console.Write(sum - expSum); 
} 
 
// Driver Code 
public static void Main(string[] args) 
{ 
    int N = 2; 
 
    // Function to find the sum 
    findSum(N); 
} 
} 
 
// This code is contributed by rutvik_56

Javascript

<script>
 
// Javascript implementation of the above approach 
 
// Function to find the required 
// summation 
function findSum(N) 
{ 
     
    // Find the sum of first N 
    // integers using the formula 
    var sum = (N) * (N + 1) / 2; 
         
    var r = (Math.log(N) / 
             Math.log(2)) + 1; 
         
    // Find the sum of numbers 
    // which are exact power of 
    // 2 by using the formula 
    var expSum = (Math.pow(2, r)) - 1;     
     
    // Print the final Sum 
    document.write(sum - expSum); 
} 
     
// Driver code
var N = 2; 
 
// Function to find the sum 
findSum(N); 
 
// This code is contributed by Kirti
 
</script>

Output:

Time Complexity: O(1)

Auxiliary Space: O(1)

Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 zambiatek!