Given a number N, the task is to find the longest range of integers [L, R] such that 1 ≤ L ≤ R ≤ N and the bitwise AND of all the numbers in that range is positive.
Examples:
Input: N = 7
Output: 4 7
Explanation: Check and from 1 to 7
Bitwise AND operations:
from 1 to 7 is 0
from 2 to 7 is 0
from 3 to 7 is 0
from 4 to 7 is 4
Therefore, maximum range comes out from L = 4 to R = 7.Input: K = 16
Output: 8 15
Approach: The problem can be solved based on the following mathematical observation. If 2K is the closest exponent of 2 greater than N then the maximum range will be either of the two:
- From 2(K – 2) to (2(K – 1) – 1) [both value inclusive] or,
- From 2(K – 1) to N
Because these ranges confirm that all the numbers in the range will have the most significant bit set for all of them. If the ranges vary for powers of 2 then the bitwise AND of the range will become 0.
Below is the implementation of the above approach.
C++
// C++ code to implement above approach#include <bits/stdc++.h>using namespace std;// Function to find the closest exponent of 2// which is greater than Kint minpoweroftwo(int K){ int count = 0; while (K > 0) { count++; K = K >> 1; } return count;}// Function to find the longest rangevoid findlongestrange(int N){ int K = minpoweroftwo(N); int y = N + 1 - pow(2, K - 1); int z = (pow(2, K - 1) - pow(2, K - 2)); if (y >= z) { cout << pow(2, K - 1) << " " << N; } else { cout << pow(2, K - 2) << " " << pow(2, K - 1) - 1; }}// Driver codeint main(){ int N = 16; findlongestrange(N); return 0;} |
C
// C code to implement above approach#include <math.h>#include <stdio.h>// Function to find the closest exponent of 2 // which is greater than Kint minpoweroftwo(int K){ int count = 0; while (K > 0) { count++; K = K >> 1; } return count;}// Function to find the longest rangevoid findlongestrange(int N){ int K = minpoweroftwo(N); int y = N + 1 - pow(2, K - 1); int z = (pow(2, K - 1) - pow(2, K - 2)); if (y >= z) { printf("%d %d", (int)pow(2, K - 1), N); } else { printf("%d %d", (int)pow(2, K - 2), (int)pow(2, K - 1)-1); }}// Driver codeint main(){ int N = 16; findlongestrange(N); return 0;} |
Java
// Java code to implement above approachclass GFG { // Function to find the closest exponent of 2 // which is greater than K static int minpoweroftwo(int K) { int count = 0; while (K > 0) { count++; K = K >> 1; } return count; } // Function to find the longest range static void findlongestrange(int N) { int K = minpoweroftwo(N); int y = (int) (N + 1 - Math.pow(2, K - 1)); int z = (int) (Math.pow(2, K - 1) - Math.pow(2, K - 2)); if (y >= z) { System.out.println(Math.pow(2, K - 1) + " " + N); } else { System.out.print((int) Math.pow(2, K - 2)); System.out.print(" "); System.out.print((int) Math.pow(2, K - 1) - 1); } } // Driver code public static void main(String args[]) { int N = 16; findlongestrange(N); }}// This code is contributed by gfgking. |
Python3
# Python code to implement above approach# Function to find the closest exponent of 2# which is greater than Kdef minpoweroftwo(K): count = 0; while (K > 0): count += 1; K = K >> 1; return count;# Function to find the longest rangedef findlongestrange(N): K = minpoweroftwo(N); y = int(N + 1 - pow(2, K - 1)); z = int(pow(2, K - 1) - pow(2, K - 2)); if (y >= z): print(pow(2, K - 1) , " " , N); else: print(pow(2, K - 2)); print(" "); print(pow(2, K - 1) - 1);# Driver codeif __name__ == '__main__': N = 16; findlongestrange(N);# This code is contributed by 29AjayKumar |
C#
// C# code to implement above approachusing System;class GFG { // Function to find the closest exponent of 2 // which is greater than K static int minpoweroftwo(int K) { int count = 0; while (K > 0) { count++; K = K >> 1; } return count; } // Function to find the longest range static void findlongestrange(int N) { int K = minpoweroftwo(N); int y = (int)(N + 1 - Math.Pow(2, K - 1)); int z = (int)(Math.Pow(2, K - 1) - Math.Pow(2, K - 2)); if (y >= z) { Console.Write(Math.Pow(2, K - 1) + " " + N); } else { Console.Write((int)Math.Pow(2, K - 2)); Console.Write(" "); Console.Write((int)Math.Pow(2, K - 1) - 1); } } // Driver code public static void Main() { int N = 16; findlongestrange(N); }}// This code is contributed by ukasp. |
Javascript
<script> // JavaScript code for the above approach // Function to find the closest exponent of 2 // which is greater than K function minpoweroftwo(K) { let count = 0; while (K > 0) { count++; K = K >> 1; } return count; } // Function to find the longest range function findlongestrange(N) { let K = minpoweroftwo(N); let y = N + 1 - Math.pow(2, K - 1); let z = (Math.pow(2, K - 1) - Math.pow(2, K - 2)); if (y >= z) { document.write(Math.pow(2, K - 1) + " " + N); } else { document.write(Math.pow(2, K - 2) + " " + (Math.pow(2, K - 1) - 1)); } } // Driver code let N = 16; findlongestrange(N); // This code is contributed by Potta Lokesh </script> |
Output
8 15
Time Complexity: O(logN)
Auxiliary Space: O(1)
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