Although the practice is the only way that ensures increased performance in programming contests but having some tricks up your sleeve ensures an upper edge and fast debugging.**1) Checking if the number is even or odd without using the % operator:**

Although this trick is not much better than using a % operator but is sometimes efficient (with large numbers).

**Use & operator:**

System.out.println((a & 1) == 0 ? "EVEN" : "ODD" );

**Example:**

num = 5 Binary: "101 & 1" will be 001, so false num = 4 Binary: "100 & 1" will be 000, so true.

**2) Fast Multiplication or Division by 2**

Multiplying by 2 means shifting all the bits to left and dividing by 2 means shifting to the right.

**Example:** 2 (Binary 10): shifting left 4 (Binary 100) and right 1 (Binary 1)

n = n << 1; // Multiply n with 2 n = n >> 1; // Divide n by 2

**3) Swapping of 2 numbers using XOR: **

This method is fast and doesnâ€™t require the use of the 3rd variable.

// A quick way to swap a and b a ^= b; b ^= a; a ^= b;

**4) Faster I/O: **

Refer here for Fast I/O in java

**5) For String manipulations: **

Use StringBuffer for string manipulations, as String in java is immutable. Refer here.

**6) Calculating the most significant digit: **

To calculate the most significant digit of any number log can be directly used to calculate it.

Suppose the number is N then Let double K = Math.log10(N); now K = K - Math.floor(K); int X = (int)Math.pow(10, K); X will be the most significant digit.

**7) Calculating the number of digits directly:**

To calculate the number of digits in a number, instead of looping we can efficiently use log :

No. of digits in N = Math.floor(Math.log10(N)) + 1;

**8) Inbuilt GCD Method:**

Java has inbuilt GCD method in BigInteger class. It returns a BigInteger whose value is the greatest common divisor of abs(this) and abs(val). Returns 0 if this==0 && val==0.

**Syntax : **

public BigInteger gcd(BigInteger val)Parameters :val - value with which the GCD is to be computed.Returns :GCD(abs(this), abs(val))

Below is the implementation of GCD:

## Java

`// Java program to demonstrate how` `// to use gcd method of BigInteger class` `import` `java.math.BigInteger;` `class` `Test {` ` ` `public` `static` `int` `gcd(` `int` `a, ` `int` `b)` ` ` `{` ` ` `BigInteger b1 = BigInteger.valueOf(a);` ` ` `BigInteger b2 = BigInteger.valueOf(b);` ` ` `BigInteger gcd = b1.gcd(b2);` ` ` `return` `gcd.intValue();` ` ` `}` ` ` `public` `static` `long` `gcd(` `long` `a, ` `long` `b)` ` ` `{` ` ` `BigInteger b1 = BigInteger.valueOf(a);` ` ` `BigInteger b2 = BigInteger.valueOf(b);` ` ` `BigInteger gcd = b1.gcd(b2);` ` ` `return` `gcd.longValue();` ` ` `}` ` ` `// Driver method` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `System.out.println(gcd(` `3` `, ` `5` `));` ` ` `System.out.println(gcd(10000000000L, 600000000L));` ` ` `}` `}` |

**Output**

1 200000000

**9) checking for a prime number:**

Java has an inbuilt isProbablePrime() method in BigInteger class. It returns true if this BigInteger is probably prime(with some certainty), false if it’s definitely composite.

BigInteger.valueOf(1235).isProbablePrime(1)

**10) ****Efficient trick to know if a number is a power of 2****:**

The normal technique of division the complexity comes out to be O(logN), but it can be solved using O(v) where v is the number of digits of the number in binary form.

## Java

`/* Method to check if x is power of 2*/` `static` `boolean` `isPowerOfTwo (` `int` `x)` `{` ` ` `/* First x in the below expression is` ` ` `for the case when x is 0 */` ` ` `return` `x!=` `0` `&& ((x&(x-` `1` `)) == ` `0` `); ` `}` |

**11) Sorting Algorithm:**

- Arrays.sort() used to sort elements of an array.
- Collections.sort() used to sort elements of a collection.

For primitives, Arrays.sort() uses dual pivot quicksort algorithms.

**12) Searching Algorithm:**

- Arrays.binarySearch()(SET 1 | SET2) used to apply binary search on a sorted array.
- Collections.binarySearch() used to apply binary search on a collection based on comparators.

**13) Copy Algorithm: **

- Arrays.copyOf() and copyOfRange() copy the specified array.
- Collections.copy() copies specified collection.

**14) Rotation and Frequency **

We can use Collections.rotate() to rotate a collection or an array by a specified distance. You can also use Collections.frequency() method to get the frequency of a specified element in a collection or an array.

**15) Most data structures are already implemented in the ****Collections Framework**.

For example Stack, LinkedList, HashSet, HashMaps, Heaps etc.

**16) Use ****Wrapper class**** functions for getting radix conversions of a number** Sometimes you require radix conversion of a number. For this, you can use wrapper classes.

## Java

`// Java program to demonstrate use of wrapper` `// classes for radix conversion` `class` `Test {` ` ` `// Driver method` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `int` `a = ` `525` `;` ` ` `long` `b = 12456545645L;` ` ` `String binaryA = Integer.toString(a, ` `2` `);` ` ` `System.out.println(` `"Binary representation"` ` ` `+ ` `" of A : "` `+ binaryA);` ` ` `String binaryB = Long.toString(b, ` `2` `);` ` ` `System.out.println(` `"Binary representation"` ` ` `+ ` `" of B : "` `+ binaryB);` ` ` `String octalA = Integer.toString(a, ` `8` `);` ` ` `System.out.println(` `"Octal representation"` ` ` `+ ` `" of A : "` `+ octalA);` ` ` `String octalB = Long.toString(b, ` `8` `);` ` ` `System.out.println(` `"Octal representation"` ` ` `+ ` `" of B : "` `+ octalB);` ` ` `}` `}` |

**Output**

Binary representation of A : 1000001101 Binary representation of B : 1011100110011101111100110101101101 Octal representation of A : 1015 Octal representation of B : 134635746555

**17) NullPointerException(Why ?)** Refer here and here to avoid it.

This article is contributed by **Gaurav Miglani**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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