Write a function which returns 1 that 2 is passed and return 2 when 1 is passed.
Source: Adobe Interview Experience | Set 19 (For MTS)
A simple solution is to compare the passed value with 1 and 2.
C
int invert(int x)
{
if (x == 1) return 2;
else return 1;
}
|
Java
static int invert(int x)
{
if (x == 1) return 2;
else return 1;
}
|
Python3
def invert(x):
if (x == 1):
return 2
else:
return 1
|
C#
static int invert(int x)
{
if (x == 1)
return 2;
else
return 1;
}
|
Javascript
function invert(x)
{
if (x === 1)
return 2;
else
return 1;
}
|
C++
int invert(int x) {
if (x == 1) {
return 2;
}
else {
return 1;
}
}
|
Another solution is to use subtraction
C
int invertSub(int x)
{
return (3-x);
}
|
Java
static int invertSub(int x)
{
return (3-x);
}
|
Python3
def invertSub(x):
return (3-x)
|
C#
static int invertSub(int x)
{
return (3-x);
}
|
Javascript
function invertSub(int x)
{
return (3-x);
}
|
C++
int invertSub(int x) {
return (3 - x);
}
|
We can also use bitwise xor operator.
C
int invertXOR(int x)
{
return (x ^ 1 ^ 2);
}
|
Java
import java.util.*;
class GFG
{
static int invertXOR(int x)
{
return (x ^ 1 ^ 2);
}
}
|
Python3
def invertXOR(x):
return x ^ 1 ^ 2
|
C#
using System;
class GFG {
static int invertXOR(int x) { return (x ^ 1 ^ 2); }
public static void Main(string[] args)
{
Console.WriteLine(invertXOR(1));
Console.WriteLine(invertXOR(2));
}
}
|
Javascript
function invertXOR(x)
{
return (x ^ 1 ^ 2);
}
|
C++
int invertXOR(int x) {
return (x ^ 1 ^ 2);
}
|
This article is contributed by Anuj. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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