Minimum Increment or Bitwise OR operations to make two integers equal
Last Updated :
05 Mar, 2024
Given two integers X and Y (X < Y), we can perform the following operations any number of times:
- Increment X by 1, X = X + 1
- Increment Y by 1, Y = Y + 1
- Change X to X | Y, X = X | Y
The task is to determine the minimum number of operations required to make X = Y.
Input: X = 1, Y = 3
Output: 1
Explanation: It takes only one operation to make X = Y,
- Apply type 3 operation (X = X | Y), so X = 1 | 3 = 3 and Y = 3.
Input: X = 5, Y = 8
Output: 3
Explanation: It takes three operations to make X = Y,
- Perform type 1 operation (X = X + 1), so X = 6, Y = 8
- Perform type 1 operation (X = X + 1), so X = 7, Y = 8
- Perform type 1 operation (X = X + 1), so X = 8, Y = 8
Approach: To solve the problem, follow the below idea:
It can be observed that its optimal to apply type 3 operation at most once as if we apply X = X | Y, the value of X will become greater than or equal to Y. So, if it becomes equal to Y then we don’t need to apply any operation otherwise if it becomes greater than Y, then we need to apply only type 2 operations.
Now, we know that the max number of operations needed are (Y – X). Let’s say we apply some type 1 operations and some type 2 operations before applying type 3 operation, so the new value of X becomes X1 and the new value of Y becomes Y1 before performing the type 3 operation. So, now the total operations needed will be: F = (X1 – X) + (Y1 – Y) + ((X1 | Y1) – Y1) + 1, which can be simplified as F = X1 + (X1 | Y1) – (X + Y – 1). Now, we need minimize F which is same as minimizing X1 + (X1 | Y1), since (X + Y – 1) is constant.
We can iterate X1 from X to Y and for a fixed X1, we need to minimize the value of X1 | Y1, so we can construct Y1 bit by bit as follows:
- If ith bit of X1 = 0 and Y = 0, so ith bit of Y1 will be 0.
- If ith bit of X1 = 0 and Y = 1, so ith bit of Y1 will be 1.
- If ith bit of X1 = 1 and Y = 0, so ith bit of Y1 will be 1 and break.
- If ith bit of X1 = 1 and Y = 1, so ith bit of Y1 will be 1.
Step-by-step algorithm:
- Initialize minimum operations = Y – X.
- Iterate X1 from X to Y, and for every X1,
- Construct Y1 bit by bit.
- Calculate total number of operations as currentOperations.
- Update min operations if currentOperations < min operations.
- Return min operations as the answer.
Below is the implementation of the algorithm:
C++
#include <bits/stdc++.h>
using namespace std;
int getMinOperations(int X, int Y) {
int minOperations = Y - X;
for(int X1 = X; X1 <= Y; X1 ++) {
int currentOperations = 0;
currentOperations += (X1 - X);
int Y1 = 0;
for(int j = 30; j >= 0; j --) {
int bitX1 = (X1 >> j) & 1;
int bitY = (Y >> j) & 1;
if((bitX1 == 0 && bitY == 1) || (bitX1 == 1 && bitY == 1)) {
Y1 |= (1 << j);
}
else if(bitX1 == 1 && bitY == 0) {
Y1 |= (1 << j);
break;
}
}
currentOperations += (Y1 - Y);
currentOperations += 1;
currentOperations += ((X1 | Y1) - Y1);
minOperations = min(minOperations, currentOperations);
}
return minOperations;
}
int main() {
cout << getMinOperations(1, 3) << "\n";
cout << getMinOperations(5, 8) << "\n";
return 0;
}
|
Java
public class Main {
public static int getMinOperations(int X, int Y) {
int minOperations = Y - X;
for (int X1 = X; X1 <= Y; X1++) {
int currentOperations = 0;
currentOperations += (X1 - X);
int Y1 = 0;
for (int j = 30; j >= 0; j--) {
int bitX1 = (X1 >> j) & 1;
int bitY = (Y >> j) & 1;
if ((bitX1 == 0 && bitY == 1) || (bitX1 == 1 && bitY == 1)) {
Y1 |= (1 << j);
} else if (bitX1 == 1 && bitY == 0) {
Y1 |= (1 << j);
break;
}
}
currentOperations += (Y1 - Y);
currentOperations += 1;
currentOperations += ((X1 | Y1) - Y1);
minOperations = Math.min(minOperations, currentOperations);
}
return minOperations;
}
public static void main(String[] args) {
System.out.println(getMinOperations(1, 3));
System.out.println(getMinOperations(5, 8));
}
}
|
C#
using System;
class GFG
{
static int GetMinOperations(int X, int Y)
{
int minOperations = Y - X;
for (int X1 = X; X1 <= Y; X1++)
{
int currentOperations = 0;
currentOperations += (X1 - X);
int Y1 = 0;
for (int j = 30; j >= 0; j--)
{
int bitX1 = (X1 >> j) & 1;
int bitY = (Y >> j) & 1;
if ((bitX1 == 0 && bitY == 1) || (bitX1 == 1 && bitY == 1))
{
Y1 |= (1 << j);
}
else if (bitX1 == 1 && bitY == 0)
{
Y1 |= (1 << j);
break;
}
}
currentOperations += (Y1 - Y);
currentOperations += 1;
currentOperations += ((X1 | Y1) - Y1);
minOperations = Math.Min(minOperations, currentOperations);
}
return minOperations;
}
static void Main()
{
Console.WriteLine(GetMinOperations(1, 3));
Console.WriteLine(GetMinOperations(5, 8));
}
}
|
Javascript
function getMinOperations(X, Y) {
let minOperations = Y - X;
for (let X1 = X; X1 <= Y; X1++) {
let currentOperations = 0;
currentOperations += (X1 - X);
let Y1 = 0;
for (let j = 30; j >= 0; j--) {
let bitX1 = (X1 >> j) & 1;
let bitY = (Y >> j) & 1;
if ((bitX1 === 0 && bitY === 1) || (bitX1 === 1 && bitY === 1)) {
Y1 |= (1 << j);
} else if (bitX1 === 1 && bitY === 0) {
Y1 |= (1 << j);
break;
}
}
currentOperations += (Y1 - Y);
currentOperations += 1;
currentOperations += ((X1 | Y1) - Y1);
minOperations = Math.min(minOperations, currentOperations);
}
return minOperations;
}
console.log(getMinOperations(1, 3));
console.log(getMinOperations(5, 8));
|
Python3
def get_min_operations(X, Y):
min_operations = Y - X
for X1 in range(X, Y + 1):
current_operations = 0
current_operations += (X1 - X)
Y1 = 0
for j in range(30, -1, -1):
bit_X1 = (X1 >> j) & 1
bit_Y = (Y >> j) & 1
if (bit_X1 == 0 and bit_Y == 1) or (bit_X1 == 1 and bit_Y == 1):
Y1 |= (1 << j)
elif bit_X1 == 1 and bit_Y == 0:
Y1 |= (1 << j)
break
current_operations += (Y1 - Y)
current_operations += 1
current_operations += ((X1 | Y1) - Y1)
min_operations = min(min_operations, current_operations)
return min_operations
if __name__ == "__main__":
print(get_min_operations(1, 3))
print(get_min_operations(5, 8))
|
Time Complexity: O(X log Y), where X and Y are the two input numbers.
Auxiliary Space: O(1)
Share your thoughts in the comments
Please Login to comment...