Given a positive integer N, the task is to count the number of integers upto N which can be represented as the sum of two or more consecutive numbers.
Examples:
Input: N = 7
Output: 4
Explanation: In the range [1, 7]: {3, 5, 6, 7} can be represented as sum of consecutive numbers.
- 3 = 1 + 2
- 5 = 2 + 3
- 6 = 1 + 2 + 3
- 7 = 3 + 4
Input: N = 15
Output: 11
Naive Approach: Follow the steps below to solve the problem:
- Iterate over all integers in the range [1, N] and for each integer, check if it can be represented as the sum of two or more consecutive integers or not.
- To check whether a number can be expressed as the sum of two or more consecutive numbers or not, refer to this article.
Below is the implementation of the above approach:
C++
// C++ Program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to check if the number N// can be expressed as sum of 2 or// more consecutive numbers or notbool isPossible(int N){ return ((N & (N - 1)) && N);}// Function to count integers in the range// [1, N] that can be expressed as sum of// 2 or more consecutive numbersvoid countElements(int N){ // Stores the required count int count = 0; for (int i = 1; i <= N; i++) { if (isPossible(i)) count++; } cout << count;}// Driver Codeint main(){ int N = 15; countElements(N); return 0;} |
Java
// Java program of the above approachimport java.io.*;class GFG{// Function to check if the number N// can be expressed as sum of 2 or// more consecutive numbers or notstatic int isPossible(int N){ return (((N & (N - 1)) & N));}// Function to count integers in the range// [1, N] that can be expressed as sum of// 2 or more consecutive numbersstatic void countElements(int N){ // Stores the required count int count = 0; for (int i = 1; i <= N; i++) { if (isPossible(i) != 0) count++; } System.out.println(count);}// Driver Code public static void main(String[] args){ int N = 15; countElements(N);}}// This code is contributed by jana_sayantan. |
Python3
# Python3 Program to implement# the above approach# Function to check if the number N# can be expressed as sum of 2 or# more consecutive numbers or notdef isPossible(N): return ((N & (N - 1)) and N)# Function to count integers in the range# [1, N] that can be expressed as sum of# 2 or more consecutive numbersdef countElements(N): # Stores the required count count = 0 for i in range(1, N + 1): if (isPossible(i)): count += 1 print (count) # Driver Codeif __name__ == '__main__': N = 15 countElements(N) # This code is contributed by mohit kumar 29 |
C#
// C# program for the above approachusing System;class GFG{// Function to check if the number N// can be expressed as sum of 2 or// more consecutive numbers or notstatic int isPossible(int N){ return (((N & (N - 1)) & N));}// Function to count integers in the range// [1, N] that can be expressed as sum of// 2 or more consecutive numbersstatic void countElements(int N){ // Stores the required count int count = 0; for (int i = 1; i <= N; i++) { if (isPossible(i) != 0) count++; } Console.Write(count);}// Driver Codestatic public void Main(){ int N = 15; countElements(N);}}// This code is contributed by code_hunt. |
Javascript
<script>// JavaScript program of the above approach // Function to check if the number N // can be expressed as sum of 2 or // more consecutive numbers or not function isPossible(N) { return (((N & (N - 1)) & N)); } // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers function countElements(N) { // Stores the required count var count = 0; for (i = 1; i <= N; i++) { if (isPossible(i) != 0) count++; } document.write(count); } // Driver Code var N = 15; countElements(N);// This code contributed by Rajput-Ji </script> |
Output:
11
Time Complexity: O(N)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach, follow the steps below to solve the problem
- All numbers except the ones which are powers of 2 can be expressed as a sum of consecutive numbers.
- Count the number of powers of 2 ? N and store it in a variable, k say Count
- Print (N – Count) as the required answer.
Below is the implementation of the above approach:
C++
// C++ implementation// of the above approach#include <bits/stdc++.h>using namespace std;// Function to count integers in the range// [1, N] that can be expressed as sum of// 2 or more consecutive numbersvoid countElements(int N){ int Cur_Ele = 1; int Count = 0; // Count powers of 2 up to N while (Cur_Ele <= N) { // Increment count Count++; // Update current power of 2 Cur_Ele = Cur_Ele * 2; } cout << N - Count;}// Driver Codeint main(){ int N = 15; countElements(N); return 0;} |
Java
// Java implementation// of the above approachimport java.util.*;class GFG{ // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers static void countElements(int N) { int Cur_Ele = 1; int Count = 0; // Count powers of 2 up to N while (Cur_Ele <= N) { // Increment count Count++; // Update current power of 2 Cur_Ele = Cur_Ele * 2; } System.out.print(N - Count); } // Driver Code public static void main(String[] args) { int N = 15; countElements(N); }}// This code is contributed by shikhasingrajput |
Python3
# python 3 implementation# of the above approach# Function to count integers in the range# [1, N] that can be expressed as sum of# 2 or more consecutive numbersdef countElements(N): Cur_Ele = 1 Count = 0 # Count powers of 2 up to N while (Cur_Ele <= N): # Increment count Count += 1 # Update current power of 2 Cur_Ele = Cur_Ele * 2 print(N - Count)# Driver Codeif __name__ == '__main__': N = 15 countElements(N) |
C#
// C# implementation// of the above approachusing System;public class GFG{ // Function to count integers in the range // [1, N] that can be expressed as sum of // 2 or more consecutive numbers static void countElements(int N) { int Cur_Ele = 1; int Count = 0; // Count powers of 2 up to N while (Cur_Ele <= N) { // Increment count Count++; // Update current power of 2 Cur_Ele = Cur_Ele * 2; } Console.Write(N - Count); } // Driver Code public static void Main(String[] args) { int N = 15; countElements(N); }}// This code is contributed by 29AjayKumar |
Javascript
<script>// Javascript implementation// of the above approach// Function to count integers in the range// [1, N] that can be expressed as sum of// 2 or more consecutive numbersfunction countElements(N){ var Cur_Ele = 1; var Count = 0; // Count powers of 2 up to N while (Cur_Ele <= N) { // Increment count Count++; // Update current power of 2 Cur_Ele = Cur_Ele * 2; } document.write(N - Count);}// Driver Codevar N = 15;countElements(N);// This code is contributed by todaysgaurav </script> |
Output:
11
Time Complexity: O(log(N))
Auxiliary Space: O(1)
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