Calculate sum of all integers from 1 to N, excluding perfect power of 2
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 2r – 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> |
0
Time Complexity: O(1)
Auxiliary Space: O(1)