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// summationvoid 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 Codeint main(){ int N = 2; // Function to find the // sum findSum(N); return 0;} |
Java
// Java implementation of the above approachimport java.lang.Math;class GFG{ // Function to find the required// summationpublic 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 Codepublic 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# approachfrom math import log2,pow# Function to find the required# summationdef 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 Codeif __name__ == '__main__': N = 2 # Function to find the # sum findSum(N) # This code is contributed by Surendra_Gangwar |
C#
// C# implementation of the above approachusing System;class GFG{ // Function to find the required// summationpublic 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 Codepublic 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// summationfunction 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 codevar N = 2;// Function to find the sumfindSum(N);// This code is contributed by Kirti</script> |
0
Time Complexity: O(1)
Auxiliary Space: O(1)
.png)
