Maximize a given unsigned number by swapping bits at it’s extreme positions.
Last Updated :
15 Nov, 2022
Given a number maximize it by swapping bits at its extreme positions i.e. at the first and last position, second and second last position, and so on.
Examples:
Input : 4 (0000...0100)
Output : 536870912 (0010...0000)
In the above example swapped 3rd and 3rd last
bit to maximize the given unsigned number.
Input : 12 (00000...1100)
Output : 805306368 (0011000...0000)
In the above example 3rd and 3rd last bit and
4th and 4th last bit are swapped to maximize
the given unsigned number.
Naive Approach:
1. Convert the number into its bit representation and store its bit representation in an array.
2. Traverse the array from both ends, if the less significant bit of the bit representation is greater than the more significant bit i.e. if the less significant bit is 1 and the more significant bit is 0 then swap them else take no action.
3. Convert the obtained binary representation back to the number.
Efficient Approach:
1. Create a copy of the original number because the original number would be modified, iteratively obtaining the bits at the extreme positions.
2. If less significant bit is 1 and the more significant bit is 0 then swap the bits in the bit from only, and continue the process until less significant bit’s position is less than more significant bit’s position.
3. Display the maximized number.
C++
#include <bits/stdc++.h>
using namespace std;
#define ull unsigned long long int
ull findMax(ull num)
{
ull num_copy = num;
int j = sizeof (unsigned long long int ) * 8 - 1;
int i = 0;
while (i < j) {
int m = (num_copy >> i) & 1;
int n = (num_copy >> j) & 1;
if (m > n) {
int x = (1 << i | 1 << j);
num = num ^ x;
}
i++;
j--;
}
return num;
}
int main()
{
ull num = 4;
cout << findMax(num);
return 0;
}
|
Java
class GFG {
static int findMax( int num) {
byte size_of_int = 4 ;
int num_copy = num;
int j = size_of_int * 8 - 1 ;
int i = 0 ;
while (i < j) {
int m = (num_copy >> i) & 1 ;
int n = (num_copy >> j) & 1 ;
if (m > n) {
int x = ( 1 << i | 1 << j);
num = num ^ x;
}
i++;
j--;
}
return num;
}
static public void main(String[] args) {
int num = 4 ;
System.out.println(findMax(num));
}
}
|
Python 3
def findMax( num):
num_copy = num
j = 4 * 8 - 1 ;
i = 0
while (i < j) :
m = (num_copy >> i) & 1
n = (num_copy >> j) & 1
if (m > n) :
x = ( 1 << i | 1 << j)
num = num ^ x
i + = 1
j - = 1
return num
if __name__ = = "__main__" :
num = 4
print (findMax(num))
|
C#
using System;
public class GFG {
static int findMax( int num) {
byte size_of_int = 4;
int num_copy = num;
int j = size_of_int * 8 - 1;
int i = 0;
while (i < j) {
int m = (num_copy >> i) & 1;
int n = (num_copy >> j) & 1;
if (m > n) {
int x = (1 << i | 1 << j);
num = num ^ x;
}
i++;
j--;
}
return num;
}
static public void Main() {
int num = 4;
Console.Write(findMax(num));
}
}
|
Javascript
<script>
function findMax(num)
{
let num_copy = num;
let j = 4 * 8 - 1;
let i = 0;
while (i < j) {
let m = (num_copy >> i) & 1;
let n = (num_copy >> j) & 1;
if (m > n) {
let x = (1 << i | 1 << j);
num = num ^ x;
}
i++;
j--;
}
return num;
}
let num = 4;
document.write(findMax(num));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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