# Convert the Integer to maximum possible Integer under the constrained operations

Problem Statement: Given an input Integer Num,the task is to determine the maximum possible integer that can be formed from input number by performing below operations.

• Convert the integer to its Binary number
• Swap only unlike bits of the Binary number
• The Binary number will be N-Bit Binary number where N is length of Binary String

Determine the maximum possible Number.

Example:

input:11

output:14

explanation:11 => 1011 replace 0th bit with 2nd bit =>1110 1110=>14

input:

output:12

input:7

output:7

Approach: If we can swap all set bits of the input binary number with unset bits such that all set bits comes to left of the N-bit binary number then we can always get the maximum possible value.We can only make all set bits to left of N-bit binary number when there is atleast one '1' and one '0' such that  swaps will become possible.

Below is the Java implementation of the above approach:

Code

// Java programme to convert the Integer to Maximum Possible Integer
import java.io.*;

class Solution {
// function to find the max possible number
public static int findMaxNum(int num)
{
// converting input integer to binary string
String binaryNumber = Integer.toBinaryString(num);
// maximum possible binary number after conversion
String maxBinaryNumber = "";
// keeping the count of 0's and 1's
int count0 = 0, count1 = 0;
// find the total number of bits
int N = binaryNumber.length();
// counting the 1's and 0's
for (int i = 0; i < N; i++) {
if (binaryNumber.charAt(i) == '1') {
count1++;
}
else {
count0++;
}
}
// if 1's and 0's count is greater than 1 then swaps
// are possible
if (count1 >= 1 && count0 >= 1) {
// shifting all set bits to left of maximized
// N-bit number
for (int i = 0; i < count1; i++) {
maxBinaryNumber += '1';
}
// shifting all unset bits to right of maximized
// N-bit number
for (int i = 0; i < count0; i++) {
maxBinaryNumber += '0';
}
// returing the maximized possible number
return Integer.parseInt(maxBinaryNumber, 2);
}
// if  any of 1's and 0's count is less than 1 then
// swaps are not possible then output will be the
// input number
return num;
}
// driver code
public static void main(String[] args)
{
// input integer
int num = 11;
// function call
System.out.println(findMaxNum(num));
}
}

output:

14

Time Complexity:O(N),if N=length of Binary Representation of Input Number

Auxiliary Space:O(1)

Feel free to comment if you have any doubt in comments section

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