## Tuesday, April 19, 2011

### Count the number of occurrences of a given key in an array

This was asked in on-site Microsoft interview (see original post).  It seems the answers to this question are not correct.  This question can be solved in O(logN) with binary search.  We know how to find the first occurrence of a given key in a sorted array (see my previous post).  It is easy to modify the algorithm to find the last occurrence of the given key in a sorted array:

/**
* find the last occurrence of the target in a sorted array
* @param input a sorted array
* @param target
* @return the last index of target or -1
*/
public static int findLastOccurence(int[] input, int target) {
int lower = -1;
int upper = input.length;

while (lower + 1 != upper) {
int  middle = (lower + upper) >>> 1;

if (input[middle] > target) {
upper = middle;
} else {
lower = middle;
}
}

if (lower != -1 && input[lower] == target) {
return lower;
} else {
return -1;
}
}

After knowing indexes of the first and last occurrences, it is easy to know the number of occurrences (or the range).  The run time for finding the number of occurrences is O(logN) since finding first and last occurrences is O(logN).

## Monday, April 18, 2011

### Find the first occurrence of an integer in sorted array

Normal binary search will return the target randomly if multiple targets are in the given array.  Below is the modified version of binary search, and it is faster than normal binary search since it requires only one comparison with each loop:

/**
* find the index of first occurrence of target
* @param input is a sorted array
* @param target
* @return the first index of target or -1
*/
public static int findFirstOccurence(int[] input, int target) {
int lower = -1;
int upper = input.length;

while (lower + 1 != upper) {
//avoid overflow.  See here
int  middle = (lower + upper) >>> 1;

if (input[middle] < target) {
lower = middle;
} else {
upper = middle;
}
}

if (upper < input.length && input[upper] == target) {
return upper;
} else {
return -1;
}
}

The run time of above method is still O(logn).

## Wednesday, April 13, 2011

### Implementation of the procedure RANDOM(a, b) that only makes calls to RANDOM(0, 1).

Implementation of the procedure RANDOM(a, b) that only makes calls to RANDOM(0, 1).  This is the question asked in book "Introduction to Algorithm."  I spent some time figuring out the following solution implemented in Java.  The assumption is that we can get random number 0 and 1 which I use Java class Random to generate.

import java.util.Random;
public class RandomGenerator {
private static Random rand = new Random();

/**
* generate a random nubmer between a (inclusive) and b (exclusive)
*/
public static int random(int a, int b) {
if (a >= b) {
throw new UnsupportedOperationException("2nd param must be greater than 1st param");
}
return a + generate(b - a);
}
/**
* generate the random between 0 to a (exclusive)
*/
private static int generate(int a) {
int run = leastPowerOfTwo(a);

while (true)  {
int power = 1;
int sum = 0;
for (int i = 0; i < run; i++) {
sum += random01() * power;
power *= 2;
}
if (sum < a) {
return sum;
}
}
}

/**
* find the least power of 2 which is greater than or equal to the given input a
*/
private static int leastPowerOfTwo(int a) {
int power = 0;
int temp = a;
while (temp > 0) {
temp >>>= 1;
power++;
}
//if a is a power of 2
if ((a & (a - 1)) == 0) {
power--;
}
return power;
}

/**
* a method to randomly generate 0 or 1
* @return 0 or 1
*/
private static int random01() {
return rand.nextInt(2);
}
}

The trick is to use the binary representation for the generated random number.  For example, to generate a random within [0, 5),  we need to call random(0, 1) 3 times since 2 ^ 3 = 8.  If call random(0, 1) 2 times, we only have 2 ^ 2 = 4 random numbers.  We only need to keep random number less than 5, and if we get a number is greater than 4, we just retry until we get the number is less than 5.   Below is the binary representation for 3 runs of random(0, 1)

#run #3   #2   #1
0    0     0        0
0    0     1        1
0    1     0        2
0    1     1        3
1    0     0        4
1    0     1        5
1    1     0        6
1    1     1        7

The informal proof of the above algorithm is that the chance to generate each number is same (1/8).  Since we only keep the number from 0 to 4, the chance to generate the number from 0 to 4 is still same.  Then the above algorithm does generate the correct random numbers.  I ran this program on my machine to generate random number from 7 to 12 (exclusive) for 10 million times, and here is the result:

7:   1997855
8:   2000311
9:   2000140
10: 2000369
11: 2001325

## Tuesday, April 12, 2011

### Maximum subarray

This is an interview question:

Given an integer array, find the max sum in any contiguous sub-array.  If all elements are negative, then the sum is 0. The following code just finds the max sum without returning the start and end indexes:
public class MaxSubArray {
public static int maxSum(int[] input) {
int currentSum = 0;
int maxSum = 0;
for (int i = 0; i < input.length; i++) {
currentSum = Math.max(currentSum + input[i], 0);
maxSum = Math.max(maxSum, currentSum);
}
return maxSum;
}
}
The following code returns max sum with starting and ending indexes. If all elements are negative, return start and end indexes as 0 and -1, respectively.
public class MaxSubArray {
public static MaxSubArrayResult compute(int[] input) {

MaxSubArrayResult result = new MaxSubArrayResult(0, -1, 0);
int currentSum = 0;
int currentStart = 0;
for (int currentEnd = 0; currentEnd < input.length; currentEnd++) {
currentSum += input[currentEnd];
if (currentSum > result.maxSum) {
result.start = currentStart;
result.end = currentEnd;
result.maxSum = currentSum;
} else if (currentSum < 0){
currentStart = currentEnd+ 1;
currentSum = 0;
}
}
return result;
}

public static class MaxSubArrayResult {
public int start;
public int end;
public int maxSum;

public MaxSubArrayResult(int start, int end, int sum) {
super();
this.start = start;
this.end = end;
this.maxSum = sum;
}

@Override
public String toString() {
return String.format("start = %d, end = %d, maxSum = %d", start, end, maxSum);
}
}
}
The run time for both method is linear (O(n)).