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 i^th character by L(i).

We scan the string one character after another. When the i^th character is scanned, L(i-1) is already know. If the i^th character has not appeared before, L(i) should be L(i-1)+1. It’s more complex when the i^th character is duplicated. Firstly we get the distance between the i^th 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 i^th 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.

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 i^th digit is the first digit which is greater than the next one, all digits before the i^th 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 i^th 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.

No. 47 - Search in a Rotation of an Array

Question: When some elements at the beginning of an array are moved to the end, it gets a rotation of the original array. Please implement a function to search a number in a rotation of an increasingly sorted array. Assume there are no duplicated numbers in the array.

For example, array {3, 4, 5, 1, 2} is a rotation of array {1, 2, 3, 4, 5}. If the target number to be searched is 4, the index of the number 4 in the rotation 1 should be returned. If the target number to be searched is 6, -1 should be returned because the number does not exist in the rotated array.

Analysis: Binary search is suitable for sorted arrays. Let us try to utilize it on a rotation of a sorted array. Notice that a rotation of a sorted array can be partitioned into two sorted sub-arrays, and numbers in the first sub-array are greater than numbers in the second one.

Two pointers P1 and P2 are utilized. P1 references to the first element in the array, and P2 references to the last element. According to the rotation rule, the first element should be greater than the last one.

The algorithm always compares the number in middle with numbers pointed by P1 and P2 during binary search. If the middle number is in the first increasingly sorted sub-array, it is greater than the number pointed by P1.

If the value of target number to be search is between the number pointed by P1 and the middle number, we then search the target number in the first half sub-array. In such a case the first half sub-array is in the first increasing sub-array, we could utilize the binary search algorithm. For example, if we search the number 4 in a rotation {3, 4, 5, 1, 2}, we could search the target number 4 in the sub-array {3, 4, 5} because 4 is between the first number 3 and the middle number 5.

If the value of target number is not between the number pointed by P1 and the middle number, we search the target in the second half sub-array. Notice that the second half sub-array also contains two increasing sub-array and itself is also a rotation, so we could search recursively with the same strategy. For example, if we search the number 1 in a rotation {3, 4, 5, 1, 2}, we could search the target number 1 in the sub-array {5, 1, 2} recursively.

The analysis above is for two cases when the middle number is in the first increasing sub-array. Please analyze the other two cases when the middle number is in the second increasing sub-array yourself, when the middle number is less than the number pointed by P2.

The code implementing this algorithm is listed below, in C/C++:

int searchInRotation(int numbers[], int length, int k)

{

    if(numbers == NULL || length <= 0)

        return -1;

   

    return searchInRotation(numbers, k, 0, length - 1);

}

int searchInRotation(int numbers[], int k, int start, int end)

{

    if(start > end)

        return -1;

       

    int middle = start + (end - start) / 2;

    if(numbers[middle] == k)

        return middle;

   

    // the middle number is in the first increasing sub-array

    if(numbers[middle] >= numbers[start])

    {

        if(k >= numbers[start] && k < numbers[middle])

            return binarySearch(numbers, k, start, middle - 1);

        return searchInRotation(numbers, k, middle + 1, end);

    }

    // the middle number is in the second increasing sub-array

    else if(numbers[middle] <= numbers[end])

    {

        if(k > numbers[middle] && k <= numbers[end])

            return binarySearch(numbers, k, middle + 1, end);

        return searchInRotation(numbers, k, start, middle - 1);

    }

   

    // It should never reach here if the input is valid

    assert(false);

}

Since the function binarySearch is for the classic binary search algorithm, it is not listed here. You might implement your own binary search code if you are interested.

In each round of search, half of the array is excluded for the next round, so the time complexity is O(logn).

You may wonder why we assume there are no duplications in the input array. We determine whether the middle number is in the first or second sub-array by comparing the middle number and the numbers pointed by P1 or P2. When the middle number, the number pointed by P1 and P2 are identical, we don’t know whether the middle number is in the first or second increasing sub-array.

Let’s look at some examples. Two arrays {1, 0, 1, 1, 1} and {1, 1, 1, 0, 1} are both rotations of an increasingly sorted array {0, 1, 1, 1, 1}, which are visualized in Figure 1.

Figure 1: Two rotations of an increasingly sorted array {0, 1, 1, 1, 1}: {1, 0, 1, 1, 1} and {1, 1, 1, 0, 1}. Elements with gray background are in the second increasing sub-array.

In Figure 1, the elements pointed by P1 and P2, as well as the middle element are all 1. The middle element with index 2 is in the second sub-array in Figure 1 (a), while the middle element is in the first sub-array in Figure 1 (b).

Since we can’t determine whether the middle number in the first or second increasing sub-array, we have to search sequentially for such cases, and our code listed above should be revised.

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.