通过递增它们或对它们进行按位或来最小化使两个整数相等的步骤

给定两个正整数A和B 。任务是使用最小操作使它们相等，例如：

A = A + 1（将 a 增加 1）。
B = B + 1（将 b 增加 1）。
A = A | B（将 A 替换为 A 和 B 的按位或）。

例子：

Input: A = 5, B = 9
Output: 4
Explanation: It is better to use first operation i.e increase A by 1 four times,
Input: A = 2, B = 5
Output: 2
Explanation: It is better to apply second operation then third operation,

编程需要懂一点英语

贪心方法：问题可以在位操作的帮助下使用贪心技术来解决。

直觉：

Try to increase A or B by 1, the steps will be the maximum steps possible.
Now to reduce these steps,
- We need to find an intermediate number X such that X OR B = B, because only then we can jump more than 1 number in a single step.
- Once we have found possible values for X, we can check that which value among them is reachable for A in least steps.
- Those least steps + 1 step (for doing bitwise OR of X with B) will be one of the lesser number of steps for A to reach B.
Another way to reduce these steps:
- Consider the case when instead of making X OR B = B, we find possible values of Y such that A OR Y = Y, as B can also be moved as per given problem.
- So we can find least step needed to move B to Y and then add 1 more step to do bitwise OR of A with B.
Now try to find the minimum among the both possible lesser steps as the required number of steps to change A to B.

编程需要懂一点英语

插图：

Suppose A = 2, B = 5

Case 1: Possible value of X such that (X OR B = B) => [0, 1, 4, 5]
Now the steps required to convert A to B if we convert A to each possible value of X first, are:
Convert A to 0 => not possible as we cannot decrement A
Convert A to 1 => not possible as we cannot decrement A
Convert A to 4 => 2 increment operation, and then 1 operation for 4 OR 5 to make A as 5. Hence total operation = 3
Convert A to 5 => 3 increment operation to make A as 5. Hence total operation = 3
Case 2: Possible value of Y such that (A OR Y = Y) => [2, 6, 7, …]
Now the steps required to convert A to B if we convert B to each possible value of Y first, are:
Convert B to 2 => not possible as we cannot decrement B
Convert B to 6 => 1 increment operation, and then 1 operation for 2 OR 6 to make A as 6. Hence total operation = 2
Convert B to 7 => 2 increment operation, and then 1 operation for 2 OR 7 to make A as 7. Hence total operation = 3
Similarly for any conversion of B to value greater than 7 will take more steps.

Therefore the least steps required to convert A to B using given operations = min(3, 2) = 2 steps.

编程需要懂一点英语

按照下面提到的步骤来实施该方法：

从i = A 迭代到 B并检查 ( i | B ) 是否与B相同以及所需的步骤。
找到以这种方式使 A 和 B 相等所需的最小步骤（例如x ）。
现在迭代j = B 到 B+x ：
- 检查j是否满足上面提到的情况 2。
- 更新使 A 和 B 相等所需的最小步骤。
返回最小步数。

下面是上述方法的实现：

Java

// Java code to implement the approach
 
import java.io.*;
import java.lang.*;
class GFG {
 
    // Function to find min steps
    // to convert A to B
    public static int AToB(int a, int b)
    {
        int ans = Integer.MAX_VALUE;
 
        // If a is greater than b, swap them
        if (a > b) {
            int temp = b;
            b = a;
            a = temp;
        }
 
        // Check for every i from 0 to b
        for (int i = a; i <= b; i++) {
 
            // If i or b equals to b then
            // update the answer
            if ((i | b) == b) {
                int j = Math.abs(a - i);
                if (i != b)
                    j++;
                ans = Math.min(ans, j);
            }
        }
 
        for (int i = b + 1; i <= b + ans; i++) {
            if ((i | a) == i) {
                ans = Math.min(ans, i - b + 1);
            }
        }
 
        return ans;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int A = 2;
        int B = 5;
        System.out.println(AToB(A, B));
    }
}

输出

时间复杂度： O(B * log B)
辅助空间： O(1)