What is Logarithmic Time Complexity? A Complete Tutorial - GeeksforGeeks (2024)

// C++ code for reducing a number to its logarithm#include <bits/stdc++.h>using namespace std;int main(){ int N = 16; int number_of_operations = 0; cout << "Logarithmic reduction of N: "; for (int i = N; i > 1; i = i / 2) { cout << i << " "; number_of_operations++; } cout << '\n' << "Algorithm Runtime for reducing N to 1: " << number_of_operations;}

/*package whatever //do not write package name here */import java.io.*;class GFG { public static void main (String[] args) { int N = 16; int number_of_operations = 0; System.out.print("Logarithmic reduction of N: "); for (int i = N; i > 1; i = i / 2) { System.out.print(i + " "); number_of_operations++; } System.out.println(); System.out.print("Algorithm Runtime for reducing N to 1: " + number_of_operations); }}

 # python3 code for the above approach# Driver Codeif __name__ == "__main__": N = 16 number_of_operations = 0 print("Logarithmic reduction of N: ", end = "") i = N while(i>1) : print( i , end = " ") number_of_operations += 1 i = i // 2 print() print("Algorithm Runtime for reducing N to 1:", number_of_operations) # This code is contributed by sanjoy_62.

// C# implementation of above approachusing System;using System.Collections.Generic;class GFG {// Driver Codepublic static void Main(){ int N = 16; int number_of_operations = 0;  Console.Write("Logarithmic reduction of N: "); for (int i = N; i > 1; i = i / 2) { Console.Write(i + " "); number_of_operations++; } Console.WriteLine(); Console.WriteLine("Algorithm Runtime for reducing N to 1: " + number_of_operations);}}

let number_of_operations = 0;let n= 16;for(let i=n; i>1; i=i/2) { console.log(i); number_of_operations++;}console.log(number_of_operations);

// C++ program for finding the index of 8#include <iostream>using namespace std;int find_position(int val, int Arr[], int n, int& steps){ int l = 0, r = n - 1; while (l <= r) { steps++; int m = l + (r - l) / 2; if (Arr[m] == val) return m; else if (Arr[m] < val) l = m + 1; else r = m - 1; } return -1;}// Driver Codeint main(){ int Arr[8] = { 2, 4, 6, 8, 10, 12, 14, 16 }; int steps = 0; // Function Call int idx = find_position(8, Arr, 8, steps); cout << "8 was present on index: "<<idx << endl; // Since the worst case runtime of Binary search is // log(N) so the count of steps must be less than log(N) cout << "Algorithm Runtime: " << steps << endl; return 0;}

// Java program for finding the index of 8import java.io.*;class GFG { static int steps = 0; static int find_position(int val, int Arr[], int n) { int l = 0, r = n - 1; while (l <= r) { steps++; int m = l + (r - l) / 2; if (Arr[m] == val) return m; else if (Arr[m] < val) l = m + 1; else r = m - 1; } return -1; } // Driver Code public static void main (String[] args) { int Arr[] = { 2, 4, 6, 8, 10, 12, 14, 16 }; steps = 0; // Function Call int idx = find_position(8, Arr, 8); System.out.println("8 was present on index: "+idx); // Since the worst case runtime of Binary search is // log(N) so the count of steps must be less than log(N) System.out.println("Algorithm Runtime: " + steps); }}// This code is contributed by Aman Kumar

# Python program for finding the index of 8def find_position(val,Arr,n): global steps l=0 r=n-1 while(l<=r): steps+=1 m=l+(r-l)//2 if(Arr[m] == val): return m elif(Arr[m] < val): l=m+1 else: l=m-1 return -1 # Driver codeArr=[2, 4, 6, 8, 10, 12, 14, 16]steps=0# Function Callidx = find_position(8, Arr, 8)print("8 was present on index: {0}".format(idx))# Since the worst case runtime of Binary search is# log(N) so the count of steps must be less than log(N)print("Algorithm Runtime: {0}".format(steps))# This code is contributed by Pushpesh Raj.

using System;namespace GFG { class Program { static int steps = 0; static int FindPosition(int val, int[] arr, int n) { int l = 0, r = n - 1; while (l <= r) { steps++; int m = l + (r - l) / 2; if (arr[m] == val) { return m; } else if (arr[m] < val) { l = m + 1; } else { r = m - 1; } } return -1; } static void Main(string[] args) { int[] arr = { 2, 4, 6, 8, 10, 12, 14, 16 }; steps = 0; int idx = FindPosition(8, arr, 8); Console.WriteLine("8 was present on index: " + idx); Console.WriteLine("Algorithm runtime: " + steps); } }}//This code is contributed by Edula Vinay Kumar Reddy

// JavaScript program for finding the index of 8let steps = 0;function find_position(val, Arr, n) { let l = 0; let r = n - 1; while (l <= r) { steps += 1; let m = Math.floor(l + (r - l) / 2); if (Arr[m] === val) { return m; } else if (Arr[m] < val) { l = m + 1; } else { r = m - 1; } } return -1;}// Driver codelet Arr = [2, 4, 6, 8, 10, 12, 14, 16];// Function Calllet idx = find_position(8, Arr, 8);console.log(`8 was present on index: ${idx}`);// Since the worst case runtime of Binary search is// log(N) so the count of steps must be less than log(N)console.log(`Algorithm Runtime: ${steps}`);

#include <bits/stdc++.h>using namespace std;const long long MAX_SIZE = 1000001;// isPrime[] : isPrime[i] is true if number is prime// prime[] : stores all prime number less than N// SPF[] that store smallest prime factor of number// [for Exp : smallest prime factor of '8' and '16'// is '2' so we put SPF[8] = 2 , SPF[16] = 2 ]vector<long long> isprime(MAX_SIZE, true);vector<long long> prime;vector<long long> SPF(MAX_SIZE);// Function generate all prime number less than N in O(n)void manipulated_seive(int N){ // 0 and 1 are not prime isprime[0] = isprime[1] = false; // Fill rest of the entries for (long long int i = 2; i < N; i++) { // If isPrime[i] == True then i is // prime number if (isprime[i]) { // put i into prime[] vector prime.push_back(i); // A prime number is its own smallest // prime factor SPF[i] = i; } // Remove all multiples of i*prime[j] which are // not prime by making isPrime[i*prime[j]] = false // and put smallest prime factor of i*Prime[j] as // prime[j] [ for exp :let i = 5 , j = 0 , prime[j] // = 2 [ i*prime[j] = 10 ] so smallest prime factor // of '10' is '2' that is prime[j] ] this loop run // only one time for number which are not prime for (long long int j = 0; j < (int)prime.size() && i * prime[j] < N && prime[j] <= SPF[i]; j++) { isprime[i * prime[j]] = false; // put smallest prime factor of i*prime[j] SPF[i * prime[j]] = prime[j]; } }}// Driver program to test above functionint main(){ int N = 13; // Must be less than MAX_SIZE manipulated_seive(N); // Print all prime number less than N for (int i = 0; i < prime.size() && prime[i] <= N; i++) cout << prime[i] << " "; return 0;}

import java.util.*;public class Main { static final int MAX_SIZE = 1000001; // isprime[] : isprime[i] is true if number is prime // prime[] : stores all prime numbers less than N // SPF[] that store smallest prime factor of number // [for Exp : smallest prime factor of '8' and '16' // is '2' so we put SPF[8] = 2 , SPF[16] = 2 ] static boolean[] isprime = new boolean[MAX_SIZE]; static List<Integer> prime = new ArrayList<Integer>(); static int[] SPF = new int[MAX_SIZE]; // Function generate all prime numbers less than N in // O(n) static void manipulated_seive(int N) { Arrays.fill(isprime, true); // 0 and 1 are not prime isprime[0] = isprime[1] = false; // Fill rest of the entries for (int i = 2; i < N; i++) { // If isprime[i] is true then i is prime number if (isprime[i]) { // put i into prime[] list prime.add(i); // A prime number is its own smallest prime // factor SPF[i] = i; } // Remove all multiples of i*prime[j] which are // not prime by making isprime[i*prime[j]] = // false and put the smallest prime factor of // i*Prime[j] as prime[j] [for example: let i = // 5, j = 0, prime[j] = 2 [i*prime[j] = 10] so // the smallest prime factor of '10' is '2' that // is prime[j]] this loop runs only one time for // numbers which are not prime for (int j = 0; j < prime.size() && i * prime.get(j) < N && prime.get(j) <= SPF[i]; j++) { isprime[i * prime.get(j)] = false; // put the smallest prime factor of // i*prime[j] SPF[i * prime.get(j)] = prime.get(j); } } } // Driver program to test above function public static void main(String[] args) { int N = 13; // Must be less than MAX_SIZE manipulated_seive(N); // Print all prime numbers less than N for (int i = 0; i < prime.size() && prime.get(i) <= N; i++) { System.out.print(prime.get(i) + " "); } }}// This code is contributed by divyansh2212

import mathMAX_SIZE = 1000001# isprime[]: isprime[i] is True if number is prime# prime[]: stores all prime numbers less than N# SPF[] that store smallest prime factor of number# [for Exp: smallest prime factor of '8' and '16'# is '2' so we put SPF[8] = 2, SPF[16] = 2]isprime = [True] * MAX_SIZEprime = []SPF = [0] * MAX_SIZE# Function generate all prime numbers less than N in O(n)def manipulated_seive(N): global isprime, prime, SPF # 0 and 1 are not prime isprime[0] = isprime[1] = False # Fill rest of the entries for i in range(2, N): # If isprime[i] is True then i is prime number if isprime[i]: # put i into prime[] list prime.append(i) # A prime number is its own smallest prime factor SPF[i] = i # Remove all multiples of i*prime[j] which are # not prime by making isprime[i*prime[j]] = False # and put the smallest prime factor of i*Prime[j] as # prime[j] [for example: let i = 5, j = 0, prime[j] # = 2 [i*prime[j] = 10] so the smallest prime factor # of '10' is '2' that is prime[j]] this loop runs # only one time for numbers which are not prime j = 0 while j < len(prime) and i * prime[j] < N and prime[j] <= SPF[i]: isprime[i * prime[j]] = False # put the smallest prime factor of i*prime[j] SPF[i * prime[j]] = prime[j] j += 1# Driver program to test above functionif __name__ == "__main__": N = 13 # Must be less than MAX_SIZE manipulated_seive(N) # Print all prime numbers less than N for i in range(len(prime)): if prime[i] <= N: print(prime[i], end=" ") else: break

using System;using System.Collections.Generic;class MainClass{ static readonly int MAX_SIZE = 1000001; // isprime[] : isprime[i] is true if number is prime // prime[] : stores all prime numbers less than N // SPF[] that store smallest prime factor of number // [for Exp : smallest prime factor of '8' and '16' // is '2' so we put SPF[8] = 2 , SPF[16] = 2 ] static bool[] isprime = new bool[MAX_SIZE]; static List<int> prime = new List<int>(); static int[] SPF = new int[MAX_SIZE]; // Function generate all prime numbers less than N in // O(n) static void manipulated_seive(int N) { Array.Fill(isprime, true); // 0 and 1 are not prime isprime[0] = isprime[1] = false; // Fill rest of the entries for (int i = 2; i < N; i++) { // If isprime[i] is true then i is prime number if (isprime[i]) { // put i into prime[] list prime.Add(i); // A prime number is its own smallest prime // factor SPF[i] = i; } // Remove all multiples of i*prime[j] which are // not prime by making isprime[i*prime[j]] = // false and put the smallest prime factor of // i*Prime[j] as prime[j] [for example: let i = // 5, j = 0, prime[j] = 2 [i*prime[j] = 10] so // the smallest prime factor of '10' is '2' that // is prime[j]] this loop runs only one time for // numbers which are not prime for (int j = 0; j < prime.Count && i * prime[j] < N && prime[j] <= SPF[i]; j++) { isprime[i * prime[j]] = false; // put the smallest prime factor of // i*prime[j] SPF[i * prime[j]] = prime[j]; } } } // Driver program to test above function public static void Main(string[] args) { int N = 13; // Must be less than MAX_SIZE manipulated_seive(N); // Print all prime numbers less than N for (int i = 0; i < prime.Count && prime[i] <= N; i++) { Console.Write(prime[i] + " "); } }}

const MAX_SIZE = 1000001;// isprime[]: isprime[i] is true if number is prime// prime[]: stores all prime numbers less than N// SPF[] that store smallest prime factor of number// [for Exp: smallest prime factor of '8' and '16'// is '2' so we put SPF[8] = 2, SPF[16] = 2]let isprime = Array(MAX_SIZE).fill(true);let prime = [];let SPF = Array(MAX_SIZE).fill(0);// Function generate all prime numbers less than N in O(n)function manipulated_seive(N) { // 0 and 1 are not prime isprime[0] = isprime[1] = false; // Fill rest of the entries for (let i = 2; i < N; i++) { // If isprime[i] is true then i is prime number if (isprime[i]) { // put i into prime[] list prime.push(i); // A prime number is its own smallest prime factor SPF[i] = i; } // Remove all multiples of i*prime[j] which are // not prime by making isprime[i*prime[j]] = false // and put the smallest prime factor of i*Prime[j] as // prime[j] [for example: let i = 5, j = 0, prime[j] // = 2 [i*prime[j] = 10] so the smallest prime factor // of '10' is '2' that is prime[j]] this loop runs // only one time for numbers which are not prime let j = 0; while (j < prime.length && i * prime[j] < N && prime[j] <= SPF[i]) { isprime[i * prime[j]] = false; // put the smallest prime factor of i*prime[j] SPF[i * prime[j]] = prime[j]; j++; } }}// Driver program to test above functionconst N = 13; // Must be less than MAX_SIZEmanipulated_seive(N);// Print all prime numbers less than Nconsole.log(prime.join(' '));// Contributed by adityasha4x71