Given an integer N, the task is to find the smallest integer X such that it has no odd position set and X ? N.
Note: The positioning of bits is assumed from the right side and the first bit is assumed to be the 0th bit.
Examples:
Input: N = 9
Output: 16
16’s binary representation is 10000, which has its 4th bit
set which is the smallest number possible satisfying the given condition.
Input: N = 5
Output: 5
Input: N = 19
Output: 20
Approach: The problem can be solved using a greedy approach and some bit properties. The property that if smaller powers of two are taken exactly once and added up they can never exceed a higher power of two (e.g., (1 + 2 + 4) < 8). The following greedy approach is used to solve the above problem:
- Initially count the number of bits.
- Get the leftmost index of the set bit.
- If the leftmost set bit is at an odd index, then answer will always be (1 << (leftmost_bit_index + 1)).
- Else, greedily form a number by setting all the even bits from 0 to leftmost_bit_index. Now greedily remove a power of two from right to check if we get a number that satisfies the given conditions.
- If the above condition does not give us our number, then we simply set the next leftmost even bit and return the 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 the total bitsint countBits(int n){ int count = 0; // Iterate and find the // number of set bits while (n) { count++; // Right shift the number by 1 n >>= 1; } return count;}// Function to find the nearest numberint findNearestNumber(int n){ // Count the total number of bits int cnt = countBits(n); // To get the position cnt -= 1; // If the last set bit is // at odd position then // answer will always be a number // with the left bit set if (cnt % 2) { return 1 << (cnt + 1); } else { int tempnum = 0; // Set all the even bits which // are possible for (int i = 0; i <= cnt; i += 2) tempnum += 1 << i; // If the number still is less than N if (tempnum < n) { // Return the number by setting the // next even set bit return (1 << (cnt + 2)); } else if (tempnum == n) return n; // If we have reached this position // it means tempsum > n // hence turn off even bits to get the // first possible number for (int i = 0; i <= cnt; i += 2) { // Turn off the bit tempnum -= (1 << i); // If it gets lower than N // then set it and return that number if (tempnum < n) return tempnum += (1 << i); } }}// Driver codeint main(){ int n = 19; cout << findNearestNumber(n);} |
Java
// Java implementation of the approachimport java.util.*;class GFG { // Function to count the total bits static int countBits(int n) { int count = 0; // Iterate and find the // number of set bits while (n > 0) { count++; // Right shift the number by 1 n >>= 1; } return count; } // Function to find the nearest number static int findNearestNumber(int n) { // Count the total number of bits int cnt = countBits(n); // To get the position cnt -= 1; // If the last set bit is // at odd position then // answer will always be a number // with the left bit set if (cnt % 2 == 1) { return 1 << (cnt + 1); } else { int tempnum = 0; // Set all the even bits which // are possible for (int i = 0; i <= cnt; i += 2) { tempnum += 1 << i; } // If the number still is less than N if (tempnum < n) { // Return the number by setting the // next even set bit return (1 << (cnt + 2)); } else if (tempnum == n) { return n; } // If we have reached this position // it means tempsum > n // hence turn off even bits to get the // first possible number for (int i = 0; i <= cnt; i += 2) { // Turn off the bit tempnum -= (1 << i); // If it gets lower than N // then set it and return that number if (tempnum < n) { return tempnum += (1 << i); } } } return Integer.MIN_VALUE; } // Driver code public static void main(String[] args) { int n = 19; System.out.println(findNearestNumber(n)); }}// This code is contributed by 29AjayKumar |
C#
// C# implementation of the approachusing System;class GFG { // Function to count the total bits static int countBits(int n) { int count = 0; // Iterate and find the // number of set bits while (n > 0) { count++; // Right shift the number by 1 n >>= 1; } return count; } // Function to find the nearest number static int findNearestNumber(int n) { // Count the total number of bits int cnt = countBits(n); // To get the position cnt -= 1; // If the last set bit is // at odd position then // answer will always be a number // with the left bit set if (cnt % 2 == 1) { return 1 << (cnt + 1); } else { int tempnum = 0; // Set all the even bits which // are possible for (int i = 0; i <= cnt; i += 2) { tempnum += 1 << i; } // If the number still is less than N if (tempnum < n) { // Return the number by setting the // next even set bit return (1 << (cnt + 2)); } else if (tempnum == n) { return n; } // If we have reached this position // it means tempsum > n // hence turn off even bits to get the // first possible number for (int i = 0; i <= cnt; i += 2) { // Turn off the bit tempnum -= (1 << i); // If it gets lower than N // then set it and return that number if (tempnum < n) { return tempnum += (1 << i); } } } return int.MinValue; } // Driver code public static void Main() { int n = 19; Console.WriteLine(findNearestNumber(n)); }}// This code is contributed by anuj_67.. |
Python3
# Python implementation of the above approach # Function to count the total bitsdef countBits(n): count = 0; # Iterate and find the # number of set bits while (n>0): count+=1; # Right shift the number by 1 n >>= 1; return count; # Function to find the nearest numberdef findNearestNumber(n): # Count the total number of bits cnt = countBits(n); # To get the position cnt -= 1; # If the last set bit is # at odd position then # answer will always be a number # with the left bit set if (cnt % 2): return 1 << (cnt + 1); else: tempnum = 0; # Set all the even bits which # are possible for i in range(0,cnt+1,2): tempnum += 1 << i; # If the number still is less than N if (tempnum < n): # Return the number by setting the # next even set bit return (1 << (cnt + 2)); elif (tempnum == n): return n; # If we have reached this position # it means tempsum > n # hence turn off even bits to get the # first possible number for i in range(0,cnt+1,2): # Turn off the bit tempnum -= (1 << i); # If it gets lower than N # then set it and return that number if (tempnum < n): tempnum += (1 << i); return tempnum;# Driver coden = 19;print(findNearestNumber(n));# This code contributed by PrinciRaj1992 |
Javascript
<script>// Javascript implementation of the above approach// Function to count the total bitsfunction countBits(n){ let count = 0; // Iterate and find the // number of set bits while (n) { count++; // Right shift the number by 1 n >>= 1; } return count;}// Function to find the nearest numberfunction findNearestNumber(n){ // Count the total number of bits let cnt = countBits(n); // To get the position cnt -= 1; // If the last set bit is // at odd position then // answer will always be a number // with the left bit set if (cnt % 2) { return 1 << (cnt + 1); } else { let tempnum = 0; // Set all the even bits which // are possible for (let i = 0; i <= cnt; i += 2) tempnum += 1 << i; // If the number still is less than N if (tempnum < n) { // Return the number by setting the // next even set bit return (1 << (cnt + 2)); } else if (tempnum == n) return n; // If we have reached this position // it means tempsum > n // hence turn off even bits to get the // first possible number for (let i = 0; i <= cnt; i += 2) { // Turn off the bit tempnum -= (1 << i); // If it gets lower than N // then set it and return that number if (tempnum < n) return tempnum += (1 << i); } }}// Driver code let n = 19; document.write(findNearestNumber(n));</script> |
Output
20
Time Complexity: O(log N)
Auxiliary Space: O(1)
Another Approach: Another approach to solving this problem is to check each number starting from N until we find a number that satisfies the given conditions. Bitwise operators can be used to check if the number has odd-positioned bits set or not.
Below is the implementation of the above approach:
C++
// C++ implementation of the above approach#include <bits/stdc++.h>using namespace std;// Function to find the nearest numberint findNearestNumber(int n){ while (true) { bool hasOddBitSet = false; int temp = n; // Check if the number has odd positioned bits set while (temp) { if (temp & 0x2) hasOddBitSet = true; temp >>= 2; } // If the number has no odd positioned bits set, return it if (!hasOddBitSet) return n; // Otherwise, increment the number and try again n++; }}// Driver codeint main(){ int n = 19; cout << findNearestNumber(n);} |
C#
using System;public class Program{ public static int FindNearestNumber(int n) { while (true) { bool hasOddBitSet = false; int temp = n; // Check if the number has odd positioned bits set while (temp > 0) { if ((temp & 0x2) != 0) { hasOddBitSet = true; } temp >>= 2; } // If the number has no odd positioned bits set, return it if (!hasOddBitSet) { return n; } // Otherwise, increment the number and try again n++; } } public static void Main() { int n = 19; Console.WriteLine(FindNearestNumber(n)); }} |
Java
import java.util.*;public class Main { // Function to find the nearest number public static int findNearestNumber(int n) { while (true) { boolean hasOddBitSet = false; int temp = n; // Check if the number has odd positioned bits set while (temp != 0) { if ((temp & 0x2) != 0) hasOddBitSet = true; temp >>= 2; } // If the number has no odd positioned bits set, return it if (!hasOddBitSet) return n; // Otherwise, increment the number and try again n++; } } // Driver code public static void main(String[] args) { int n = 19; System.out.println(findNearestNumber(n)); }} |
Javascript
// Function to find the nearest numberfunction findNearestNumber(n) { while (true) { let hasOddBitSet = false; let temp = n; // Check if the number has odd positioned bits set while (temp) { if (temp & 0x2) hasOddBitSet = true; temp >>= 2; } // If the number has no odd positioned bits set, return it if (!hasOddBitSet) return n; // Otherwise, increment the number and try again n++; }}// Driver codelet n = 19;console.log(findNearestNumber(n)); |
Python3
# Python3 implementation of the above approach# Function to find the nearest numberdef findNearestNumber(n): while True: hasOddBitSet = False temp = n # Check if the number has odd positioned bits set while temp != 0: if temp & 0x2: hasOddBitSet = True temp >>= 2 # If the number has no odd positioned bits set, return it if not hasOddBitSet: return n # Otherwise, increment the number and try again n += 1# Driver coden = 19print(findNearestNumber(n)) |
Output
20
Time Complexity: The time complexity of the given approach is O(log N) in the worst-case scenario. This is because we only need to iterate through the bits of the number once to count the total number of bits, and then perform constant time operations on the bits. The worst-case scenario would occur if we need to iterate through all the bits of the number to find the leftmost set bit.
Auxiliary Space: The auxiliary space used by the given approach is O(1) as we are not using any additional data structures, and all the operations are performed on the given number itself. Therefore, the space complexity is constant.
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