One of the oldest trick question from programming interview is, How do you swap two integers without using temp variable? This was first asked to me on a C, C++ interview and then several times on various Java interviews. Beauty of this question lies both on trick to think about how you can swap two numbers with out third variable, but also problems associated with each approach. If a programmer can think about integer overflow and consider that in its solution then it creates a very good impression in the eye of interviewers. Ok, so let's come to the point, suppose you have tow integers i = 10 and j = 20, how will you swap them without using any variable so that j = 10 and i = 20? Though this is journal problem and solution work more or less in every other programming language, you need to do this using Java programming constructs. You can swap numbers by performing some mathematical operations e.g. addition and subtraction and multiplication and division, but they face problem of integer overflow.

There is a neat trick of swapping number using XOR bitwise operator which proves to be ultimate solution. We will explore and understand each of these three solutions in detail here, and if you are hungry for more programming questions and solutions, you can always take a look at Cracking the Coding Interview: 150 Programming Questions and Solutions, one of the best book to prepare for programming job interviews.

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You can use + and - operator in Java to swap two integers as shown below :

a = a + b;

b = a - b; // actually (a + b) - (b), so now b is equal to a

a = a - b; // (a + b) -(a), now a is equal to b

You can see that its really nice trick and first time it took sometime to think about this approach. I was really happy to even think about this solution because its neat, but happiness was short lived because interviewer said that it will not work in all conditions. He said that integer will overflow if addition is more than maximum value of int primitive as defined by Integer.MAX_VALUE and if subtraction is less than minimum value i.e. Integer.MIN_VALUE.

It confused me and I conceded only to find that the code is absolutely fine and work perfectly in Java, because overflows are clearly defined. Ofcourse same solution will not work in C or C++ but it will work absolutely fine in Java. Don't believe me? here is the proof :

a = Integer.MAX_VALUE;

b = 10;

System.out.printf("Before swap 'a': %d, 'b': %d %n", a, b);

a = (a + b) - (b = a);

System.out.printf("After swapping, 'a': %d, 'b': %d %n", a, b);

Output :

Before swap 'a': 2147483647, 'b': 10

After swapping, 'a': 10, 'b': 2147483647

Here addition of two numbers will overflow, Integer.MAX_VALUE + 10 = -2147483639, but we are also doing subtraction, which will compensate this value, because b is now Integer.MAX_VALUE and -2147483639 - 2147483647 will again overflow to give you 10 as output.

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Second solution to swap two integer numbers in Java without using temp variable is widely recognized as the best solution, as it will also work in language which doesn't handle integer overflow like Java e.g. C and C++. Java provides several bitwise and bitshift operator, one of them is XOR which is denoted by ^.

If you remember XOR truth table then you know that A XOR B will return 1 if A and B are different and 0 if A and B are same. This property gives birth to following code, popularly know as XOR trick:

x = x ^ y;

y = x ^ y; // now y = x

x = x ^ y; // now x = y

Let's see these examples in action by writing a simple Java program, as shown below.

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There is a neat trick of swapping number using XOR bitwise operator which proves to be ultimate solution. We will explore and understand each of these three solutions in detail here, and if you are hungry for more programming questions and solutions, you can always take a look at Cracking the Coding Interview: 150 Programming Questions and Solutions, one of the best book to prepare for programming job interviews.

###

**Solution 1 - Using Addition and Subtraction**

You can use + and - operator in Java to swap two integers as shown below :

a = a + b;

b = a - b; // actually (a + b) - (b), so now b is equal to a

a = a - b; // (a + b) -(a), now a is equal to b

You can see that its really nice trick and first time it took sometime to think about this approach. I was really happy to even think about this solution because its neat, but happiness was short lived because interviewer said that it will not work in all conditions. He said that integer will overflow if addition is more than maximum value of int primitive as defined by Integer.MAX_VALUE and if subtraction is less than minimum value i.e. Integer.MIN_VALUE.

It confused me and I conceded only to find that the code is absolutely fine and work perfectly in Java, because overflows are clearly defined. Ofcourse same solution will not work in C or C++ but it will work absolutely fine in Java. Don't believe me? here is the proof :

a = Integer.MAX_VALUE;

b = 10;

System.out.printf("Before swap 'a': %d, 'b': %d %n", a, b);

a = (a + b) - (b = a);

System.out.printf("After swapping, 'a': %d, 'b': %d %n", a, b);

Output :

Before swap 'a': 2147483647, 'b': 10

After swapping, 'a': 10, 'b': 2147483647

Here addition of two numbers will overflow, Integer.MAX_VALUE + 10 = -2147483639, but we are also doing subtraction, which will compensate this value, because b is now Integer.MAX_VALUE and -2147483639 - 2147483647 will again overflow to give you 10 as output.

###

**Solution 2 - Using XOR trick**

Second solution to swap two integer numbers in Java without using temp variable is widely recognized as the best solution, as it will also work in language which doesn't handle integer overflow like Java e.g. C and C++. Java provides several bitwise and bitshift operator, one of them is XOR which is denoted by ^.

If you remember XOR truth table then you know that A XOR B will return 1 if A and B are different and 0 if A and B are same. This property gives birth to following code, popularly know as XOR trick:

x = x ^ y;

y = x ^ y; // now y = x

x = x ^ y; // now x = y

Let's see these examples in action by writing a simple Java program, as shown below.

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**Progarm to Swap Two Integers without tempoarry variable in Java**

In this Java program, I will show you couple of ways to swap two integers without using any temporary or third variable, and what are problems comes with each approach and which one will work in all conditions. Actually the two popular solution works perfectly fine in Java but candidates, especially those coming from C and C++ background often think that first solution is broken and will fail on integer boundaries, but that's not true. Integer overflows are clearly define in Java e.g. Integer. MAX_VALUE + 1 will result in Integer. MIN_VALUE, which means if you do both addition and subtraction in with same set of numbers, your result will be fine, as seen in above example.

/**

* Java program to swap two integers without using temp variable

*

* @author java67

*/

public class Problem {

public static void main(String args[]) {

int a = 10;

int b = 20;

// one way using arithmetic operator e.g. + or -

// won't work if sum overflows

System.out.println("One way to swap two numbers without temp variable");

System.out.printf("Before swap 'a': %d, 'b': %d %n", a, b);

a = a + b;

b = a - b; // actually (a + b) - (b), so now b is equal to a

a = a - b; // (a + b) -(a), now a is equal to b

System.out.printf("After swapping, 'a': %d, 'b': %d %n", a, b);

// another example

a = Integer.MIN_VALUE;

b = Integer.MAX_VALUE;

System.out.printf("Before swap 'a': %d, 'b': %d %n", a, b);

a = (a + b) - (b = a);

System.out.printf("After swapping, 'a': %d, 'b': %d %n", a, b);

// Another way to swap integers without using temp variable is

// by using XOR bitwise operator

// Known as XOR trick

System.out.println("Swap two integers without third variable

using XOR bitwise Operator");

int x = 30;

int y = 60;

System.out.printf("Before swap 'x': %d, 'y': %d %n", x, y);

x = x ^ y;

y = x ^ y;

x = x ^ y;

System.out.printf("After swapping, 'x': %d, 'y': %d %n", x, y);

}

}

**Output**

One way to swap two numbers without temp variable

Before swap 'a': 10, 'b': 20

After swapping, 'a': 20, 'b': 10

Before swap 'a': -2147483648, 'b': 2147483647

After swapping, 'a': 2147483647, 'b': -2147483648

Swap two integers without third variable using XOR bitwise Operator

Before swap 'x': 30, 'y': 60

After swapping, 'x': 60, 'y': 30

That's all about how to swap two integers without using temporary variable in Java. Unlike popular belief that first solution will not work on integer boundaries i.e. around maximum and minimum values, which is also true for languages like C and C++, it work perfectly fine in Java. Why? because overflow is clearly define and addition and subtraction together negate the effect of overflow. There is no doubt about second solution as it is the best solution and not subject to any sign or overflow problem. It is also believed to be fasted way to swap two numbers without using temp variable in Java.