When working on leetcode 70 climbing stairs: You are climbing a stair case. It takes n steps to reach to the top.Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top?
Here is my first solution:
class Solution {
public int fib (int n){
if (n <= 2) return n;
return fib(n-1) + fib(n-2);
}
public int climbStairs(int n) {
return fib (n+1);
}
}
when n <44, it works, but n >=44, it doesn't work.because of this, it leads to the failure in submission in leetcode.
but when use the 2nd solution, shows below
class Solution {
public int climbStairs(int n) {
if (n <= 2) return n;
int[] allWays = new int[n];
allWays[0] = 1;
allWays[1] = 2;
for (int i = 2; i < n; i++){
allWays[i] = allWays[i-1] + allWays[i-2];
}
return allWays[n-1];
}
}
the second solution is accepted by leetcode. however, when n >=46, it gives a negative number.
Can anyone give me some explanation why the first solution fails? what's the difference between the two solutions? Thanks.
Your intuition is correct. The number of ways to reach the top indeed follows the fibonacci sequence.
The first solution computes the fibonacci numbers recursively (fib(n) = fib(n - 1) + fib(n-2)). To compute any value, the function needs to recursively call itself twice. Every function call takes up space in a region of memory called the stack. Whats probably happening is when n is too large, too many recursive calls are happening and the program runs out of space to execute more calls.
The second solution uses Dynamic programming and memoization. This effectively saves space and computation time. If you don't know these topics, I would encourage you to read into them.
You are getting negative values since the 47th Fibonacci number is greater than the maximum value that can be represented by type int. You can try using long or the BigInteger class to represent larger values.
To understand the solution you need to understand the concept of Dynamic Programming and Recursion
In the first solution to calculate the n-th Fibonacci number, your algorithm is
fib(n)= fib(n-1)+fib(n-2)
But the second solution is more optimized
This approach stores values in an array so that you don't have to calculate fib(n) every time.
Example:
fib(5) = fib(4) + fib(3)
= (fib(3) + fib(2)) + (fib(2) + fib(1))
By first solution, you are calculating fib(2) twice for n = 4.
By second solution you are storing fibonacci values in an array
Example:
for n =4,
first you calculate fib(2) = fib(1)+fib(0) = 1
then you calculate f(3) = f(2)+f(1)
We don't have to calculate the fib(2) since it is already stored in the array.
Check this link for more details
for n = 44 no of ways = 1134903170
and for n = 45 no of ways = 1836311903
so for n = 46 number of ways will be n=44 + n=45 i.e 2971215073
sign INTEGER can only store upto 2147483647 i.e. 2^31 - 1
Because of with for n=46 it is showing -ve number
I am trying to compute hamming distances between each node in a graph of n nodes. Each node in this graph has a label of the same length (k) and the alphabet used for labels is {0, 1, *}. The '*' operates as a don't care symbol. For example, hamming distances between labels 101*01 and 1001*1 is equal to 1 (we say they only differ at the 3rd index).
What I need to do is to find all 1-hamming-distance neighbors of each node and report exactly at which index those two labels differ.
I am comparing each nodes label with all others character by character as follows:
// Given two strings s1, s2
// returns the index of the change if hd(s1,s2)=1, -1 otherwise.
int count = 0;
char c1, c2;
int index = -1;
for (int i = 0; i < k; i++)
{
// do not compute anything for *
c1 = s1.charAt(i);
if (c1 == '*')
continue;
c2 = s2.charAt(i);
if (c2 == '*')
continue;
if (c1 != c2)
{
index = i;
count++;
// if hamming distance is greater than 1, immediately stop
if (count > 1)
{
index = -1;
break;
}
}
}
return index;
I may have a couple of millions nodes. k is usually around 50. I am using JAVA, this comparison takes n*n*k time and operates slow. I considered making use of tries and VP-trees but could not figure out which data structure works for this case. I also studied the Simmetrics library but nothing flashed into my mind. I would really appreciate any suggestions.
Try this approach:
Convert the keys into ternary numbers (base 3). i.e. 0=0, 1=1, *=2
10 digits ternary give you a range of 0..59049 which fits in 16 bits.
That means two of those would form a 32 bit word. Create a lookup table with 4 billion entries that return the distance between those two 10 digit ternary words.
You can now use the lookup table to check 10 characters of the key with one lookup. If you use 5 characters, then 3^5 gives you 243 values which would fit into one byte, so the lookup table would only be 64 KB.
By using shift operations, you can create lookup tables of different sizes to balance memory and speed.
That way, you can optimize the loop to abort much more quickly.
To get the position of the first difference, you can use a second lookup table which contains the index of the first difference for two key substrings.
If you have millions of nodes, then you will have many that start with the same substring. Try to sort them into buckets where one bucket contains nodes that start with the same key. The goal here is to make the buckets as small as possible (to reduce the n*n).
Instead of / additional to the string, store a mask for 1 bits and a mask for * bits. One could use BitSet, but let's try without.
static int mask(String value, char digit) {
int mask = 0;
int bit = 2; // Start with bits[1] as per specification.
for (int i = 0; i < value.length(); ++i) {
if (value.charAt(i) == digit) {
mask |= bit;
}
bit <<= 1;
}
return mask;
}
class Cell {
int ones;
int stars;
}
int difference(Cell x, Cell y) {
int distance = 0;
return (x.ones & ~y.stars) ^ (y.ones & ~x.stars);
}
int hammingDistance(Cell x, Cell y) {
return Integer.bitCount(difference(x, y));
}
boolean differsBy1(Cell x, Cell y) {
int diff = difference(x, y);
return diff == 0 ? false : (diff & (diff - 1)) == 0;
}
int bitPosition(int diff) {
return Integer.numberOfTrailingZeroes(diff);
}
Interesting problem. It would be easy it weren't for the wild card symbol.
If the wildcard was a regular character in the alphabet, then for a given string you could enumerate all k hamming distance 1 strings. Then look these strings up in a multi-map. So for example for 101 you look up 001,111 and 100.
The don't care symbol makes it so that you can't do that lookup. However if the multi-map is build such that each node is stored by all its possible keys you can do that lookup again. So for example 1*1 is stored as 111 and 101. So when you do the look up for 10* you look up 000,010,011,001,111 which would find 1*1 which was stored by 111.
The upside of this is also that you can store all labels as integers rather then trinary structures so with an int[3] as the key value you can use any k < 96.
Performance would depend on the backing implementation of the multi-map. Ideally you'd use a hash implementation for key sizes < 32 and a tree-implementation for anything above. With the tree-implementation all nodes be connected to their distance-1 neighbors in O(n*k*log(n)). Building the multi-map takes O(n * 2 ^ z) where z is maximum number of wildcard characters for any string. If the average number of wildcards is low this should be an acceptable performance penalty.
edit: You improve look up performance for all nodes to O(n*log(n)) by also inserting the hamming distance 1 neighbors into the multi-map but that might just explode its size.
Note: I'm typing this in a lunch break. I haven't checked the details yet.
This is a question regarding a piece of coursework so would rather you didn't fully answer the question but rather give tips to improve the run time complexity of my current algorithm.
I have been given the following information:
A function g(n) is given by g(n) = f(n,n) where f may be defined recursively by
I have implemented this algorithm recursively with the following code:
public static double f(int i, int j)
{
if (i == 0 && j == 0) {
return 0;
}
if (i ==0 || j == 0) {
return 1;
}
return ((f(i-1, j)) + (f(i-1, j-1)) + (f(i, j-1)))/3;
}
This algorithm gives the results I am looking for, but it is extremely inefficient and I am now tasked to improve the run time complexity.
I wrote an algorithm to create an n*n matrix and it then computes every element up to the [n][n] element in which it then returns the [n][n] element, for example f(1,1) would return 0.6 recurring. The [n][n] element is 0.6 recurring because it is the result of (1+0+1)/3.
I have also created a spreadsheet of the result from f(0,0) to f(7,7) which can be seen below:
Now although this is much faster than my recursive algorithm, it has a huge overhead of creating a n*n matrix.
Any suggestions to how I can improve this algorithm will be greatly appreciated!
I can now see that is it possible to make the algorithm O(n) complexity, but is it possible to work out the result without creating a [n][n] 2D array?
I have created a solution in Java that runs in O(n) time and O(n) space and will post the solution after I have handed in my coursework to stop any plagiarism.
This is another one of those questions where it's better to examine it, before diving in and writing code.
The first thing i'd say you should do is look at a grid of the numbers, and to not represent them as decimals, but fractions instead.
The first thing that should be obvious is that the total number of you have is just a measure of the distance from the origin, .
If you look at a grid in this way, you can get all of the denominators:
Note that the first row and column are not all 1s - they've been chosen to follow the pattern, and the general formula which works for all of the other squares.
The numerators are a little bit more tricky, but still doable. As with most problems like this, the answer is related to combinations, factorials, and then some more complicated things. Typical entries here include Catalan numbers, Stirling's numbers, Pascal's triangle, and you will nearly always see Hypergeometric functions used.
Unless you do a lot of maths, it's unlikely you're familiar with all of these, and there is a hell of a lot of literature. So I have an easier way to find out the relations you need, which nearly always works. It goes like this:
Write a naive, inefficient algorithm to get the sequence you want.
Copy a reasonably large amount of the numbers into google.
Hope that a result from the Online Encyclopedia of Integer Sequences pops up.
3.b. If one doesn't, then look at some differences in your sequence, or some other sequence related to your data.
Use the information you find to implement said sequence.
So, following this logic, here are the numerators:
Now, unfortunately, googling those yielded nothing. However, there are a few things you can notice about them, the main being that the first row/column are just powers of 3, and that the second row/column are one less than powers of three. This kind boundary is exactly the same as Pascal's triangle, and a lot of related sequences.
Here is the matrix of differences between the numerators and denominators:
Where we've decided that the f(0,0) element shall just follow the same pattern. These numbers already look much simpler. Also note though - rather interestingly, that these numbers follow the same rules as the initial numbers - except the that the first number is one (and they are offset by a column and a row). T(i,j) = T(i-1,j) + T(i,j-1) + 3*T(i-1,j-1):
1
1 1
1 5 1
1 9 9 1
1 13 33 13 1
1 17 73 73 17 1
1 21 129 245 192 21 1
1 25 201 593 593 201 25 1
This looks more like the sequences you see a lot in combinatorics.
If you google numbers from this matrix, you do get a hit.
And then if you cut off the link to the raw data, you get sequence A081578, which is described as a "Pascal-(1,3,1) array", which exactly makes sense - if you rotate the matrix, so that the 0,0 element is at the top, and the elements form a triangle, then you take 1* the left element, 3* the above element, and 1* the right element.
The question now is implementing the formulae used to generate the numbers.
Unfortunately, this is often easier said than done. For example, the formula given on the page:
T(n,k)=sum{j=0..n, C(k,j-k)*C(n+k-j,k)*3^(j-k)}
is wrong, and it takes a fair bit of reading the paper (linked on the page) to work out the correct formula. The sections you want are proposition 26, corollary 28. The sequence is mentioned in Table 2 after proposition 13. Note that r=4
The correct formula is given in proposition 26, but there is also a typo there :/. The k=0 in the sum should be a j=0:
Where T is the triangular matrix containing the coefficients.
The OEIS page does give a couple of implementations to calculate the numbers, but neither of them are in java, and neither of them can be easily transcribed to java:
There is a mathematica example:
Table[ Hypergeometric2F1[-k, k-n, 1, 4], {n, 0, 10}, {k, 0, n}] // Flatten
which, as always, is ridiculously succinct. And there is also a Haskell version, which is equally terse:
a081578 n k = a081578_tabl !! n !! k
a081578_row n = a081578_tabl !! n
a081578_tabl = map fst $ iterate
(\(us, vs) -> (vs, zipWith (+) (map (* 3) ([0] ++ us ++ [0])) $
zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([1], [1, 1])
I know you're doing this in java, but i could not be bothered to transcribe my answer to java (sorry). Here's a python implementation:
from __future__ import division
import math
#
# Helper functions
#
def cache(function):
cachedResults = {}
def wrapper(*args):
if args in cachedResults:
return cachedResults[args]
else:
result = function(*args)
cachedResults[args] = result
return result
return wrapper
#cache
def fact(n):
return math.factorial(n)
#cache
def binomial(n,k):
if n < k: return 0
return fact(n) / ( fact(k) * fact(n-k) )
def numerator(i,j):
"""
Naive way to calculate numerator
"""
if i == j == 0:
return 0
elif i == 0 or j == 0:
return 3**(max(i,j)-1)
else:
return numerator(i-1,j) + numerator(i,j-1) + 3*numerator(i-1,j-1)
def denominator(i,j):
return 3**(i+j-1)
def A081578(n,k):
"""
http://oeis.org/A081578
"""
total = 0
for j in range(n-k+1):
total += binomial(k, j) * binomial(n-k, j) * 4**(j)
return int(total)
def diff(i,j):
"""
Difference between the numerator, and the denominator.
Answer will then be 1-diff/denom.
"""
if i == j == 0:
return 1/3
elif i==0 or j==0:
return 0
else:
return A081578(j+i-2,i-1)
def answer(i,j):
return 1 - diff(i,j) / denominator(i,j)
# And a little bit at the end to demonstrate it works.
N, M = 10,10
for i in range(N):
row = "%10.5f"*M % tuple([numerator(i,j)/denominator(i,j) for j in range(M)])
print row
print ""
for i in range(N):
row = "%10.5f"*M % tuple([answer(i,j) for j in range(M)])
print row
So, for a closed form:
Where the are just binomial coefficients.
Here's the result:
One final addition, if you are looking to do this for large numbers, then you're going to need to compute the binomial coefficients a different way, as you'll overflow the integers. Your answers are lal floating point though, and since you're apparently interested in large f(n) = T(n,n) then I guess you could use Stirling's approximation or something.
Well for starters here are some things to keep in mind:
This condition can only occur once, yet you test it every time through every loop.
if (x == 0 && y == 0) {
matrix[x][y] = 0;
}
You should instead: matrix[0][0] = 0; right before you enter your first loop and set x to 1. Since you know x will never be 0 you can remove the first part of your second condition x == 0 :
for(int x = 1; x <= i; x++)
{
for(int y = 0; y <= j; y++)
{
if (y == 0) {
matrix[x][y] = 1;
}
else
matrix[x][y] = (matrix[x-1][y] + matrix[x-1][y-1] + matrix[x][y-1])/3;
}
}
No point in declaring row and column since you only use it once. double[][] matrix = new double[i+1][j+1];
This algorithm has a minimum complexity of Ω(n) because you just need to multiply the values in the first column and row of the matrix with some factors and then add them up. The factors stem from unwinding the recursion n times.
However you therefore need to do the unwinding of the recursion. That itself has a complexity of O(n^2). But by balancing unwinding and evaluation of recursion, you should be able to reduce complexity to O(n^x) where 1 <= x <= 2. This is some kind of similiar to algorithms for matrix-matrix multiplication, where the naive case has a complexity of O(n^3) but Strassens's algorithm is for example O(n^2.807).
Another point is the fact that the original formula uses a factor of 1/3. Since this is not accurately representable by fixed point numbers or ieee 754 floating points, the error increases when evaluating the recursion successively. Therefore unwinding the recursion could give you higher accuracy as a nice side effect.
For example when you unwind the recursion sqr(n) times then you have complexity O((sqr(n))^2+(n/sqr(n))^2). The first part is for unwinding and the second part is for evaluating a new matrix of size n/sqr(n). That new complexity actually can be simplified to O(n).
To describe time complexity we usually use a big O notation. It is important to remember that it only describes the growth given the input. O(n) is linear time complexity, but it doesn't say how quickly (or slowly) the time grows when we increase input. For example:
n=3 -> 30 seconds
n=4 -> 40 seconds
n=5 -> 50 seconds
This is O(n), we can clearly see that every increase of n increases the time by 10 seconds.
n=3 -> 60 seconds
n=4 -> 80 seconds
n=5 -> 100 seconds
This is also O(n), even though for every n we need twice that much time, and the raise is 20 seconds for every increase of n, the time complexity grows linearly.
So if you have O(n*n) time complexity and you will half the number of operations you perform, you will get O(0.5*n*n) which is equal to O(n*n) - i.e. your time complexity won't change.
This is theory, in practice the number of operations sometimes makes a difference. Because you have a grid n by n, you need to fill n*n cells, so the best time complexity you can achieve is O(n*n), but there are a few optimizations you can do:
Cells on the edges of the grid could be filled in separate loops. Currently in majority of the cases you have two unnecessary conditions for i and j equal to 0.
You grid has a line of symmetry, you could utilize it to calculate only half of it and then copy the results onto the other half. For every i and j grid[i][j] = grid[j][i]
On final note, the clarity and readability of the code is much more important than performance - if you can read and understand the code, you can change it, but if the code is so ugly that you cannot understand it, you cannot optimize it. That's why I would do only first optimization (it also increases readability), but wouldn't do the second one - it would make the code much more difficult to understand.
As a rule of thumb, don't optimize the code, unless the performance is really causing problems. As William Wulf said:
More computing sins are committed in the name of efficiency (without necessarily achieving it) than for any other single reason - including blind stupidity.
EDIT:
I think it may be possible to implement this function with O(1) complexity. Although it gives no benefits when you need to fill entire grid, with O(1) time complexity you can instantly get any value without having a grid at all.
A few observations:
denominator is equal to 3 ^ (i + j - 1)
if i = 2 or j = 2, numerator is one less than denominator
EDIT 2:
The numerator can be expressed with the following function:
public static int n(int i, int j) {
if (i == 1 || j == 1) {
return 1;
} else {
return 3 * n(i - 1, j - 1) + n(i - 1, j) + n(i, j - 1);
}
}
Very similar to original problem, but no division and all numbers are integers.
If the question is about how to output all values of the function for 0<=i<N, 0<=j<N, here is a solution in time O(N²) and space O(N). The time behavior is optimal.
Use a temporary array T of N numbers and set it to all ones, except for the first element.
Then row by row,
use a temporary element TT and set it to 1,
then column by column, assign simultaneously T[I-1], TT = TT, (TT + T[I-1] + T[I])/3.
Thanks to will's (first) answer, I had this idea:
Consider that any positive solution comes only from the 1's along the x and y axes. Each of the recursive calls to f divides each component of the solution by 3, which means we can sum, combinatorially, how many ways each 1 features as a component of the solution, and consider it's "distance" (measured as how many calls of f it is from the target) as a negative power of 3.
JavaScript code:
function f(n){
var result = 0;
for (var d=n; d<2*n; d++){
var temp = 0;
for (var NE=0; NE<2*n-d; NE++){
temp += choose(n,NE);
}
result += choose(d - 1,d - n) * temp / Math.pow(3,d);
}
return 2 * result;
}
function choose(n,k){
if (k == 0 || n == k){
return 1;
}
var product = n;
for (var i=2; i<=k; i++){
product *= (n + 1 - i) / i
}
return product;
}
Output:
for (var i=1; i<8; i++){
console.log("F(" + i + "," + i + ") = " + f(i));
}
F(1,1) = 0.6666666666666666
F(2,2) = 0.8148148148148148
F(3,3) = 0.8641975308641975
F(4,4) = 0.8879743941472337
F(5,5) = 0.9024030889600163
F(6,6) = 0.9123609205913732
F(7,7) = 0.9197747256986194
I was reading through the solution to Project Euler Problem 12 on MathBlog and I have some trouble understanding the logic behind the codes. The program uses prime factorisation to find the number of divisors of a triangle number.
private int PrimeFactorisationNoD(int number, int[] primelist) {
int nod = 1;
int exponent;
int remain = number;
for (int i = 0; i < primelist.Length; i++) {
// In case there is a remainder this is a prime factor as well
// The exponent of that factor is 1
if (**primelist[i] * primelist[i] > number**) {
return nod * 2;
}
exponent = 1;
while (remain % primelist[i] == 0) {
exponent++;
remain = remain / primelist[i];
}
nod *= exponent;
//If there is no remainder, return the count
if (remain == 1) {
return nod;
}
}
return nod;
}
I understand most part of the program except for the highlighted portion "primelist[i] * primelist[i] > number". I have trouble understanding the necessity of the this line of code. I will use an example to illustrate my point. Let say I have a number 510 = 2*3*5*17. The highlighted code will only be true when Primelist goes to number 23. But by the time the list goes to number 17, the condition remain == 1 will be true and program would have exited the loop. Would it be better if I change the code to if(remain==primelist[i]) since the loop would end when primelist goes to number 17 instead of 21?
The if condition speeds up the code in certain situations (although it should have "remain" in place of "number"). Once primelist[i] is reached we know that remain is not divisible by primelist[0] through primelist[i-1]. If primelist[i]^2>remain then we can conclude that remain is some prime between primelist[i] and primelist[i]^2-1 (inclusive), as if remain=ab then both a,b would have to be at least primelist[i] so remain would be at least primelist[i]^2, a contradiction. Thus we can stop searching for primes dividing remain.
For an example where this is faster, take number=7. Then the condition is triggered when we reach 3 (as 3^2=9>7), so we do not need to check all the primes up to 7.
First, it should better use remain:
primelist[i] * primelist[i] > remain
It is an optimization, as there can be no divisors between the square root of remain and remain, so you only have the factor remain left.
Also, the variable name exponent is lying, it really contains the exponent plus one. Better initialize it to zero and multiply by exponent + 1.
This question was asked in my interview.
random(0,1) is a function that generates integers 0 and 1 randomly.
Using this function how would you design a function that takes two integers a,b as input and generates random integers including a and b.
I have No idea how to solve this.
We can do this easily by bit logic (E,g, a=4 b=10)
Calculate difference b-a (for given e.g. 6)
Now calculate ceil(log(b-a+1)(Base 2)) i.e. no of bits required to represent all numbers b/w a and b
now call random(0,1) for each bit. (for given example range will be b/w 000 - 111)
do step 3 till the number(say num) is b/w 000 to 110(inclusive) i.e. we need only 7 levels since b-a+1 is 7.So there are 7 possible states a,a+1,a+2,... a+6 which is b.
return num + a.
I hate this kind of interview Question because there are some
answer fulfilling it but the interviewer will be pretty mad if you use them. For example,
Call random,
if you obtain 0, output a
if you obtain 1, output b
A more sophisticate answer, and probably what the interviewer wants is
init(a,b){
c = Max(a,b)
d = log2(c) //so we know how much bits we need to cover both a and b
}
Random(){
int r = 0;
for(int i = 0; i< d; i++)
r = (r<<1)| Random01();
return r;
}
You can generate random strings of 0 and 1 by successively calling the sub function.
So we have randomBit() returning 0 or 1 independently, uniformly at random and we want a function random(a, b) that returns a value in the range [a,b] uniformly at random. Let's actually make that the range [a, b) because half-open ranges are easier to work with and equivalent. In fact, it is easy to see that we can just consider the case where a == 0 (and b > 0), i.e. we just want to generate a random integer in the range [0, b).
Let's start with the simple answer suggested elsewhere. (Forgive me for using c++ syntax, the concept is the same in Java)
int random2n(int n) {
int ret = n ? randomBit() + (random2n(n - 1) << 1) : 0;
}
int random(int b) {
int n = ceil(log2(b)), v;
while ((v = random2n(n)) >= b);
return v;
}
That is-- it is easy to generate a value in the range [0, 2^n) given randomBit(). So to get a value in [0, b), we repeatedly generate something in the range [0, 2^ceil(log2(b))] until we get something in the correct range. It is rather trivial to show that this selects from the range [0, b) uniformly at random.
As stated before, the worst case expected number of calls to randomBit() for this is (1 + 1/2 + 1/4 + ...) ceil(log2(b)) = 2 ceil(log2(b)). Most of those calls are a waste, we really only need log2(n) bits of entropy and so we should try to get as close to that as possible. Even a clever implementation of this that calculates the high bits early and bails out as soon as it exits the wanted range has the same expected number of calls to randomBit() in the worst case.
We can devise a more efficient (in terms of calls to randomBit()) method quite easily. Let's say we want to generate a number in the range [0, b). With a single call to randomBit(), we should be able to approximately cut our target range in half. In fact, if b is even, we can do that. If b is odd, we will have a (very) small chance that we have to "re-roll". Consider the function:
int random(int b) {
if (b < 2) return 0;
int mid = (b + 1) / 2, ret = b;
while (ret == b) {
ret = (randomBit() ? mid : 0) + random(mid);
}
return ret;
}
This function essentially uses each random bit to select between two halves of the wanted range and then recursively generates a value in that half. While the function is fairly simple, the analysis of it is a bit more complex. By induction one can prove that this generates a value in the range [0, b) uniformly at random. Also, it can be shown that, in the worst case, this is expected to require ceil(log2(b)) + 2 calls to randomBit(). When randomBit() is slow, as may be the case for a true random generator, this is expected to waste only a constant number of calls rather than a linear amount as in the first solution.
function randomBetween(int a, int b){
int x = b-a;//assuming a is smaller than b
float rand = random();
return a+Math.ceil(rand*x);
}