## Friday, November 29, 2013

### No. 49 - Longest Substring without Duplication

Problem: Given a string, please get the length of the longest substring which does not have duplicated characters. Supposing all characters in the string are in the range from ‘a’ to ‘z’.

Analysis: It’s not difficult to get all substrings of a string, and to check whether a substring has duplicated characters. The only concern about this brute-force strategy is performance. A string with n characters has O(n2) substrings, and it costs O(n) time to check whether a substring has duplication. Therefore, the overall cost is O(n3).

We may improve the efficiency with dynamic programming. Let’s denote the length of longest substring ending with the ith character by L(i).

We scan the string one character after another. When the ith character is scanned, L(i-1) is already know. If the ith character has not appeared before, L(i) should be L(i-1)+1. It’s more complex when the ith character is duplicated. Firstly we get the distance between the ith character and its previous occurrence. If the distance is greater than L(i-1), the character is not in longest substring without duplication ending with the (i-1)th character, so L(i) should also be L(i-1)+1. If the distance is less than L(i-1), L(i) is the distance, and it means between the two occurrence of the ith character there are no other duplicated characters.

This solution can be implemented in Java as the following code:

public static int longestSubstringWithoutDuplication(String str) {
int curLength = 0;
int maxLength = 0;

int position[] = new int[26];
for(int i = 0; i < 26; ++i) {
position[i] = -1;
}

for(int i = 0; i < str.length(); ++i) {
int prevIndex = position[str.charAt(i) - 'a'];
if(prevIndex < 0 || i - prevIndex > curLength) {
++curLength;
}
else {
if(curLength > maxLength) {
maxLength = curLength;
}

curLength = i - prevIndex;
}
position[str.charAt(i) - 'a'] = i;
}

if(curLength > maxLength) {
maxLength = curLength;
}

return maxLength;
}

L(i) is implemented as curLength in the code above. An integer array is used to store the positions of each character.

Code with unit tests is shared at http://ideone.com/CmY3xN.

More coding interview questions are discussed in my book <Coding Interviews: Questions, Analysis & Solutions>. You may find the details of this book on Amazon.com, or Apress.

The author Harry He owns all the rights of this post. If you are going to use part of or the whole of this ariticle in your blog or webpages, please add a reference to http://codercareer.blogspot.com/. If you are going to use it in your books, please contact him via zhedahht@gmail.com . Thanks.

## Thursday, November 28, 2013

### No. 48 - Least Number after Deleting Digits

Problem: Please get the least number after deleting k digits from the input number. For example, if the input number is 24635, the least number is 23 after deleting 3 digits.

Analysis: Let’s delete a digit from the number at each step. What’s the first digit to be deleted from the number 24635, in order to get the least number with the remaining digits? We may list all the remaining numbers after deleting a digit, in the following table:

 Deleted Digit Remaining Number 2 4635 4 2635 6 2435 3 2465 5 2463

The number 2435 is the least one in all remaining numbers, by deleting the digit 6. Notice that the digit 6 is the first digit in the number 24635 which is greater than the next digit.

Let’s delete another digit from the number 2435, the remaining least number after the first step. We may summarize the remaining numbers after delete every digit from it in the following table:

 Deleted Digit Remaining Number 2 435 4 235 3 245 5 243

The number 235 is the least one in all remaining numbers, by deleting the digit 4. Notice that the digit 4 is the first digit in the number 2435 which is greater than the next digit.

The remaining three digits in the number 235 are increasingly sorted. What is the next digit to be deleted to get the least remaining number? Again, we may list the remaining numbers after deleting each digit in a table:

 Deleted Digit Remaining Number 2 35 3 25 5 23

The number 23 is the least one in all remaining numbers, by deleting the last digit 5.

If we are going to deleting more digits from a number whose digits are increasingly sorted to get the least number, the last digit is deleted at each step.

Now we get the rules to delete digits to get the least remaining number: If there are digits who are greater than the next one, delete the first digit. If all digits in the number are increasingly sorted, delete the last digit gets deleted. The process repeats until the required k digits are deleted.

The code can be implemented in Java as the following:

public static String getLeastNumberDeletingDigits_1(String number, int k) {
String leastNumber = number;
while(k > 0 && leastNumber.length() > 0) {
int firstDecreasingDigit = getFirstDecreasing(leastNumber);
if(firstDecreasingDigit >= 0) {
leastNumber = removeDigit(leastNumber, firstDecreasingDigit);
}
else {
leastNumber = removeDigit(leastNumber, leastNumber.length() - 1);
}

--k;
}

return leastNumber;
}

private static int getFirstDecreasing(String number) {
for(int i = 0; i < number.length() - 1; ++i) {
int curDigit = number.charAt(i) - '0';
int nextDigit = number.charAt(i + 1) - '0';
if(curDigit > nextDigit) {
return i;
}
}

return -1;
}

private static String removeDigit(String number, int digitIndex) {
String result = "";
if(digitIndex > 0) {
result = number.substring(0, digitIndex);
}
if(digitIndex < number.length() - 1) {
result += number.substring(digitIndex + 1);
}

return result;
}

Optimization: Save the start index for the next round of search for the first decreasing digit

In the method getFirstDecreasing above to get the first digit which is greater than the next one, we always start from the first digit. Is it necessary to start over in every round of search?

The answer is no. If the ith digit is the first digit which is greater than the next one, all digits before the ith digit are increasingly sorted. The (i-1)th digit might be less than the (i+1)th digit, the next digit of the (i-1)th digit after the ith digit is deleted. Therefore, it is safe to start from the (i-1)th digit in the next round of search.

With this optimization strategy, the efficiency gets improved from O(n*k) to O(n), if the length of the input number has n digits and k digits are deleted.

The optimized solution can be implemented as:

class DecreasingResult {
public int firstDecreasing;
public int nextStart;
}

public static String getLeastNumberDeletingDigits_2(String number, int k) {
String leastNumber = number;
int start = 0;
while(k > 0 && leastNumber.length() > 0) {
DecreasingResult result = getNextDecreasing(leastNumber, start);
if(result.firstDecreasing >= 0) {
leastNumber = removeDigit(leastNumber, result.firstDecreasing);
}
else {
leastNumber = removeDigit(leastNumber, leastNumber.length() - 1);
}

start = result.nextStart;
--k;
}

return leastNumber;
}

private static DecreasingResult getNextDecreasing(String number, int start) {
int firstDecreasing = -1;
int nextStart;

for(int i = start; i < number.length() - 1; ++i) {
int curDigit = number.charAt(i) - '0';
int nextDigit = number.charAt(i + 1) - '0';
if(curDigit > nextDigit) {
firstDecreasing = i;
break;
}
}

if(firstDecreasing == 0) {
nextStart = 0;
}
else if (firstDecreasing > 0) {
nextStart = firstDecreasing - 1;
}
else {
nextStart = number.length();
}

DecreasingResult result = new DecreasingResult();
result.firstDecreasing = firstDecreasing;
result.nextStart = nextStart;

return result;
}

Code with unit tests is shared at http://ideone.com/0Mdfcf.

More coding interview questions are discussed in my book <Coding Interviews: Questions, Analysis & Solutions>. You may find the details of this book on Amazon.com, or Apress.

The author Harry He owns all the rights of this post. If you are going to use part of or the whole of this ariticle in your blog or webpages, please add a reference to http://codercareer.blogspot.com/. If you are going to use it in your books, please contact him via zhedahht@gmail.com . Thanks.