# Comparing Integers in Array

Sam Benry

Ranch Hand

Posts: 89

posted 7 years ago

lets say we have

int[] c = new int[5];

this is later filled with 5 different values from 2 to 14 (repetition is allowed)

I need a way to find out how many numbers are repeated and how many times this number is repeated.

lets say the array was { 3 , 12 , 3 , 3 , 12 }

3 is repeated 3 times

12 is repeated 2 times

what is a good method for doing this?

int[] c = new int[5];

this is later filled with 5 different values from 2 to 14 (repetition is allowed)

I need a way to find out how many numbers are repeated and how many times this number is repeated.

lets say the array was { 3 , 12 , 3 , 3 , 12 }

3 is repeated 3 times

12 is repeated 2 times

what is a good method for doing this?

posted 7 years ago

'good' depends on many things... how big will the real array be? what is the range of the expected values?

one way might be a map that uses the value as the key. each time you read a value, like '3', you check to see if that key/value is defined. if so, you increment the value. If not, you create it with a value of 1.

when you're done, you get a list of all the keys, and use that to get the counts.

if your range is sufficiently small, you could just define an int array of say (from your example) 15. Then, as you read each value, increment the int at that index by 1.

'good' is a balance between speed, readability, maintainability, and ease of programming.

one way might be a map that uses the value as the key. each time you read a value, like '3', you check to see if that key/value is defined. if so, you increment the value. If not, you create it with a value of 1.

when you're done, you get a list of all the keys, and use that to get the counts.

if your range is sufficiently small, you could just define an int array of say (from your example) 15. Then, as you read each value, increment the int at that index by 1.

'good' is a balance between speed, readability, maintainability, and ease of programming.

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Guido Sautter

Ranch Hand

Posts: 142

posted 7 years ago

this sounds like a programming exercise ... hints:

The trivial solution uses two nested for loops and runs in o(n�) - avoiding dupliacte output will take an additional data structure, eg a second array or a set.

The half-trivial solution involves sorting and runs in o(n*log(n)).

The nifty solution involves a HashMap and runs in linear time.

The trivial solution uses two nested for loops and runs in o(n�) - avoiding dupliacte output will take an additional data structure, eg a second array or a set.

The half-trivial solution involves sorting and runs in o(n*log(n)).

The nifty solution involves a HashMap and runs in linear time.