Archive

Tag Archives: position

Question

Write a program to right-rotate a string by m characters. Right-rotating a string means moving m characters at the left of string to the right. Is it required the time complexity is O(n) and helper memory size is O(1). For example, right-rotate “abcdefghi” by 3 characters gives “defghiabc”.

Solution

There are 2 ways to do it. The first one is that we divide the string into 2 parts (XY). X is the part to move to the right and Y is the one moving on to the left. Reverse them in place 2 times.

XY: "abcdefghi"
R(X) R(T): "cbaihgfed"
YX: "defghiabc"

Secondly we can swap character one by one. Let p0 at position 0, p1 at position m. Swap their characters repeatedly with moving both pointer right by 1.

abcdefghij
dbcaefghij
decabfghij
defabcghij
defgbcahij
defghcabij
defghiabcj
defghijbca  <- p1 doesn't move since it reaches the end
defghijacb
defghijabc
#include <iostream>

// reverse the string between start and end
void reverse_string(char* start_pointer, char* end_pointer)
{
	while (start_pointer < end_pointer) {
		char temp_char = *start_pointer;
		*start_pointer = *end_pointer;
		*end_pointer = temp_char;
		start_pointer++;
		end_pointer--;
	}
}

// solution 1
void left_rotate_string_1(char* string_pointer, int n)
{
	if (string_pointer == NULL) return;

	int length = strlen(string_pointer);
	if (n > length || n <= 0) return;

	char* first_start_pointer = string_pointer;
	char* first_end_pointer = string_pointer + n - 1;
	char* second_start_pointer = string_pointer + n;
	char* second_end_pointer = string_pointer + length - 1;
	reverse_string(first_start_pointer, first_end_pointer);
	reverse_string(second_start_pointer, second_end_pointer);
	reverse_string(first_start_pointer, second_end_pointer);
}

// solution 2
void left_rotate_string_2(char* string_pointer, int n)
{
	if (string_pointer == NULL) return;

	int length = strlen(string_pointer);
	if (n > length || n <= 0) return;

	char* pointer_1 = string_pointer;
	char* pointer_2 = string_pointer + n;
	char* pointer_end = string_pointer + length - 1;
	while (pointer_1 != pointer_2) {
		char temp_char = *pointer_1;
		*pointer_1 = *pointer_2;
		*pointer_2 = temp_char;
		if (pointer_1 != pointer_end) pointer_1++;
		if (pointer_2 != pointer_end) pointer_2++;
	}
}

// main
int main()
{
	char string_1[] = "abcdefghij";
	left_rotate_string_1(string_1, 3);
	printf("%sn", string_1); // defghijabc

	char string_2[] = "abcdefghij";
	left_rotate_string_2(string_2, 3);
	printf("%sn", string_2); // defghijabc
}

Question

Given a function prototype: int continumax(char *output_string,char *input_string). Implement it to find the longest consecutive digits. This function must return the length of the longest digits. The found longest digits should be written to the memory location that output_string is pointing. For example, if input_string is “abcd12345ed125ss123456789”, function returns 9 and output_string becomes “123456789”.

Solution

This question must be implemented in C/C++. Just skip letters, count the length and save the head position to a temp pointer.

Sample

#include <iostream>

int continumax(char *output_string, char *input_string)
{
	int max_length = 0;
	char *max_string = new char[sizeof(input_string)];

	while (true) {

		// skip all non-digit characters
		while (*input_string != 0 && (*input_string < '0' || *input_string > '9')) input_string++;
		if (*input_string == 0) break;

		// current char is digit
		int length = 0;
		char *temp_string = input_string;
		while (*input_string!=0 && *input_string >= '0' && *input_string <= '9') {
			length++;
			input_string++;
		}

		// check if this is longer
		if (length > max_length) {
			max_length = length;
			max_string = temp_string;
		}
	}

	// write maximum string to output and add null to end.
	int i;
	for (i = 0; i < max_length; i++) output_string[i] = max_string[i];
	output_string[i] = 0;

	return max_length;
}

int main()
{
	char input[] = "abcd12345ed125ss123456789";
	char *output = new char[sizeof(input)];
	int length = continumax(output, input);
	printf("Length: %dn", length);
	printf("Output: %sn", output);
}

Question

Simple task, reverse words in a sentence.

Solution

In Python, this can be simple because of the build-in functions. We can just split the sentence by spaces, reverse the list, and join them with spaces again.

In C++, this can be complicated. The following might be one of the possible answers. We have to scan from the head to tail, and find the position that a word starts and ends. Then reverse the letters in this word. At the end, reverse all the letters in the whole sentence.

Examples

# function to reverse a sentence
def reverse_sentence(sentence):
    return ' '.join(reversed(sentence.split()))

# main
print reverse_sentence("I am a student.")
print reverse_sentence("Testing 1 2 3")
#include <iostream>

using namespace std;

// function to reverse a word, in place
void reverse_word(string* word, int begin, int end)
{
	while (begin < end) {
		char letter = word->at(begin);
		word->at(begin) = word->at(end);
		word->at(end) = letter;
		begin++;
		end--;
	}
}

// function to reverse a sentence, in place
void reverse_sentence(string* sentence)
{
	int length = sentence->size();
	int pointer = 0;
	// for each character
	for (int i = 0; i < sentence->size(); i++) {
		// find word head
		if (pointer == -1) {
			if (sentence->at(i) != ' ') pointer = i;
			continue;
		}
		// find word tail
		if (sentence->at(i) == ' ') {
			reverse_word(sentence, pointer, i - 1);
			pointer = -1;
		}
		// end of string
		if (i == length - 1) {
			reverse_word(sentence, pointer, i - 1);
		}
	}
	// reverse letters for the whole sentence
	reverse_word(sentence, 0, sentence->size() - 1);
}

// main part
int main()
{
	string text = "I am a student.";
	reverse_sentence(&text);
	cout << text << endl;
}
public class InterviewPractice10
{
	private static String reverse_sentence(String sentence)
	{
		String[] splited = sentence.split(" ");
		StringBuilder builder = new StringBuilder();
		for (int i = splited.length - 1; i >= 0; i--) {
			builder.append(splited[i]);
			if (i > 0) builder.append(" ");
		}
		return builder.toString();
	}

	public static void main(String[] args)
	{
		System.out.println(reverse_sentence("I am a student."));
		System.out.println(reverse_sentence("Testing 1 2 3"));
	}
}

Question

Define a stack structure with “min” function — a function to get the minimum value within the stack. The time complexity of min, push and pop functions must be O(1).

Solution

結合鍊錶一起做。首先我做插入以下數字: 10, 7, 3, 3, 8, 5, 2, 6

0: 10 -> NULL (MIN=10, POS=0)
1: 7  -> [0]  (MIN=7,  POS=1) // 用數組表示堆棧,第0個元素表示棧底
2: 3  -> [1]  (MIN=3,  POS=2)
3: 3  -> [2]  (MIN=3,  POS=3)
4: 8  -> NULL (MIN=3,  POS=3) // 技巧在這裡,因為8比當前的MIN大,所以彈出8不會影響當前的MIN
5: 5  -> NULL (MIN=3,  POS=3)
6: 2  -> [2]  (MIN=2,  POS=6) // 如果2出棧了,那麼3就是MIN
7: 6  -> [6]

出棧的話採用類似方法修正。

所以,藉助輔助棧,保存最小值,且隨時更新輔助棧中的元素。push第一個元素進A,也把它push進B,當向Apush的元素比B中的元素小, 則也push進B,即更新B。否則,不動B,保存原值。向棧A push元素時,順序由下至上。輔助棧B中,始終保存著最小的元素。

然後,當pop A中的元素小於B中棧頂元素時,則也要pop B中棧頂元素。

Samples

The following is an optimized implementation.  Repeated minimum value will not be stored through comparing values while pushing or popping values.

#include <vector>
#include <cassert>

using namespace std;

/**
 * Class defining a new type of stack supporting any datatype.
 */
template <typename T>
class StackWithMin
{
private:
    vector<T> dataStack; // stacks to store data
    vector<size_t> minStack; // stack to store position of minimum value

public:

    /**
     * Push data to the stack at the same time update the minimum stack if the
     * new data is smaller then the current minimum value. Throw exception if
     *  stack is empty.
     */
    void push(T data) {
        dataStack.push_back(data);
        if (minStack.empty() || data < dataStack[minStack.back()])
            minStack.push_back(dataStack.size()-1);
    }

    /**
     * Pop data to the stack at the same time pop from the min stack if it is
     * popping the minimum value. Throw exception if stack is empty.
     */
    void pop() {
        assert(!dataStack.empty());
        if (dataStack.back() == dataStack[minStack.back()])
            minStack.pop_back();
        dataStack.pop_back();
    }

    /**
     * return the current minimum value.
     */
    T min() {
        assert(!dataStack.empty() && !minStack.empty());
        return dataStack[minStack.back()];
    }
};

/**
 * Main program
 */
int main()
{
    StackWithMin<int> stack = StackWithMin<int>();
    stack.push(10);
    stack.push(7);
    stack.push(3);
    stack.push(3);
    stack.push(8);
    stack.push(5);
    stack.push(2);
    stack.push(6);
    printf("Minimum value: %dn", stack.min());
}
import java.util.Stack;

public class Pactice02 {

    public static void main(String[] args) {
        System.out.println("Hello World!");
        AdvancedStack stack = new AdvancedStack();
        stack.push(10);
        stack.push(7);
        stack.push(3);
        stack.push(3);
        stack.push(8);
        stack.push(5);
        stack.push(2);
        stack.push(6);
        System.out.println("Minimum value: "+ stack.getMinimum());
        stack.pop();
        System.out.println("Minimum value: "+ stack.getMinimum());
        stack.pop();
        System.out.println("Minimum value: "+ stack.getMinimum());
        stack.pop();
        System.out.println("Minimum value: "+ stack.getMinimum());
    }

    public static class AdvancedStack extends Stack {

        private Stack mMinimumStack = new Stack();

        @Override
        public E push(E item) {
            if (mMinimumStack.empty() || item.doubleValue()  0) {
                return mMinimumStack.peek();
            }
            return null;
        }
    }
}

While you are making a cube in unity which is transparent in gameplay, you would probably want to make it visible only in editor mode with gizmos. However, gizmos doesn’t rotate with the object transform. So you can use the following tricks to rotate the gizmos to fit your object. The code shown is to draw a cube which is totally the same as your object. By setting up the matrix of gizmos, we can transform gizmos to the local position, scale and rotation.

Gizmos.color = Color.green;
Gizmos.matrix = transform.localToWorldMatrix;
Gizmos.DrawWireCube(Vector3.zero,Vector3.one);

I have been working with Cocos2D for a while and I am still a newbie. The way to deal with animations in Cocos2D 1.0.0 – the stable version currently – is different from the way before using CCSpriteSheet. Now Cocos2D adapts animations as cache shared in the project using CCSpriteFrameCache. It also requires a plist file as well as a texture file.

First, put the following assets into the project “Resources” folder to let the program use.

  • grossini.plist – The file contain the animation information such as frame details.
  • grossini.pvr.gz – The texture file in pvr format which allow better performance in Cocos2D.

Second, add the following codes into the function init. Take a look at the comments for program explanation. =]

// Capture the window size
CGSize size = [[CCDirector sharedDirector] winSize];

// Load the plist into the cache
CCSpriteFrameCache *cache = [CCSpriteFrameCache sharedSpriteFrameCache];
[cache addSpriteFramesWithFile:@&quot;grossini.plist&quot;];

// Create sprite with the first frame
CCSprite *sprite = [CCSprite spriteWithSpriteFrameName:@&quot;grossini_dance_01.png&quot;];
sprite.position = ccp( size.width/2, size.height/2);
sprite.anchorPoint = ccp(0.5f, 0.5f);

// Create array for the animation frames
NSMutableArray *animFrames = [NSMutableArray array];
for(int i = 0; i &lt; 14; i++) {
    CCSpriteFrame *frame = [cache spriteFrameByName:[NSString stringWithFormat:@&quot;grossini_dance_%02d.png&quot;,(i+1)]];
    [animFrames addObject:frame];
}

// Convert the array into animation
CCAnimation *animation = [CCAnimation animationWithFrames:animFrames];

// Run the sprite with the created animation and display it by adding it to the scene
[sprite runAction:[CCRepeatForever actionWithAction: [CCAnimate actionWithDuration:2.8f animation:animation restoreOriginalFrame:NO] ]];
[self addChild:sprite z:0];

After all, don’t forget to deallocate the resources loaded in cache in function dealloc.

// Deallocate the resource in cache
CCSpriteFrameCache *cache = [CCSpriteFrameCache sharedSpriteFrameCache];
[cache removeSpriteFramesFromFile:@&quot;animations/grossini.plist&quot;];

Finally, this is a screenshot. Enjoy. XD

%d bloggers like this: