Showing posts with label LeetCode - Extended. Show all posts
Showing posts with label LeetCode - Extended. Show all posts

String Replace - Linkedin


Implement 28 - Implement strStr()
http://wxx5433.github.io/replace.html
function replace(string orig, string find, string repl)
-- replaces every instances of string find with string repl
-- returned the modified string
  1. After one replacing, if the suffix of replaced string is the prefix of the find string, should we replace again? For instance, replace("babcc", "abc", "cab"). We first change it to "bcabc", should we change it to "bccab" or not?

Analysis

We need to first count how many times the pattern have appeared in the original string. Otherwise, we need to copy the char array many times each time we replace a string. (Since len(find) can be different from len(repl)).

Complexity

Time: Suppose len(orig) = m, len(find) = n, then we need O(mn) time to count and O(m + count*n) time to replace.
Space: O(m), we need extra space to store the start index of a found pattern in the original string.
    public static String replace(String orig, String find, String repl) {
        List<Integer> index = findPattern(orig, find);
        int count = index.size();
        // compute length after replacement
        int newLen = orig.length() + count * (repl.length() - find.length());
        char[] result = new char[newLen];
        // replace
        int i = 0, k = 0;
        int resultIndex = 0;
        while( i < orig.length()) {
            if (k < count && i == index.get(k)) {
                for (int j = 0; j < repl.length(); ++j) {
                    result[resultIndex++] = repl.charAt(j);
                }
                ++k;
                i += find.length();
            } else {
                result[resultIndex++] = orig.charAt(i++);
            }
        }
        return new String(result);
    }

    /**
     * find how many times the pattern string have appeared in the orig string
     * @param orig original string
     * @param find pattern string to find
     * @return a list of the start index of a fuond pattern in the original string
     */
    private static List<Integer> findPattern(String orig, String find) {
        List<Integer> index = new ArrayList<Integer>();
        int i = 0, j = 0;
        // count how many times "find" have appeared
        while (i + find.length() <= orig.length()) {
            int k = i;
            j = 0;
            while (j < find.length()) {
                if (orig.charAt(k) != find.charAt(j)) {
                    break;
                }
                ++k;
                ++j;
            }
            if (j == find.length()) {
                index.add(i);
                i += find.length();
            } else {
                ++i;
            }
        }
        return index;
    }





Find Depth - Linkedin


http://wxx5433.github.io/find-depth.html
Consider this string representation for binary trees. Each node is of the form (lr), where l represents the left child and r represents the right child. If l is the character 0, then there is no left child. Similarly, if r is the character 0, then there is no right child. Otherwise, the child can be a node of the form (lr), and the representation continues recursively. For example: (00) is a tree that consists of one node. ((00)0) is a two-node tree in which the root has a left child, and the left child is a leaf. And ((00)(00)) is a three-node tree, with a root, a left and a right child.
Write a function that takes as input such a string, and returns -1 if the string is malformed, and the depth of the tree if the string is well-formed.
For instance:
  find_depth('(00)') -> 0 
  find_depth('((00)0)') -> 1 
  find_depth('((00)(00))') -> 1 //2?
  find_depth('((00)(0(00)))') -> 2 
  find_depth('((00)(0(0(00))))') -> 3 
  find_depth('x') -> -1 
  find_depth('0') -> -1 
  find_depth('()') -> -1 
  find_depth('(0)') -> -1 
  find_depth('(00)x') -> -1 
  find_depth('(0p)') -> -1

Analysis

We can substitute all occurence "(00)" to "0" in the string, this decrease the tree height by one. We repeatedly do this unitl no change during two iteration.
If the final result is "0", then it is a valid tree, otherwise, invalid.

Complexity

Time: O(n * depth)
Space: O(n)

public static int getLength(String str) {
    int depth = -1;
    while (!str.equals("0")) {
        String newStr = str.replace("(00)", "0");
        if (newStr.equals(str)) {
            return -1;
        }
        str = newStr;
        ++depth;
    }
    return depth;
}

public static int getLength2(String str) {
    int depth = -1;
    char[] pattern = {'(', '0', '0', ')'};
    char[] strArr = str.toCharArray();
    while (strArr.length != 1) {
        List<Integer> points = new ArrayList<Integer>();
        int start = 0;
        while (start + 3 < strArr.length) { // find all pattern
            int i = start;
            while (i < start + 4) {
                if (strArr[i] != pattern[i - start]) {
                    break;
                }
                ++i;
            }
            if (i - start == 4) {
                points.add(start);
                start += 4;
            } else {
                ++start;
            }
        }
        if (points.size() == 0) {  // cannot change anymore
            break;
        }
        // substitute all pattern
        char[] newArr = new char[strArr.length - points.size()*3];
        int j = 0;
        int k = 0;
        while (j < strArr.length) {
            if (k == points.size()) {
                newArr[j - k * 3] = strArr[j++];
            } else if (j == points.get(k)) {
                newArr[j - k * 3] = '0';
                ++k;
                j += 4;
            } else {
                newArr[j - k * 3] = strArr[j++];
            }
        }
        strArr = newArr;
        ++depth;
    }
    if (depth == -1 || strArr.length != 1 || strArr[0] != '0') {
        return -1;
    }
    return depth;
}
https://github.com/prasanthj/algorithms/blob/master/src/main/java/edu/osu/cse/StringBinaryTreeDepth.java
public static int getTreeDepth(String inp) {
  if(inp == null) {
    return -1;
  }
 
  if(inp.length() < 4) {
    return -1;
  }
 
  if(inp.charAt(0) != '(' || inp.charAt(inp.length() -1) != ')') {
    return -1;
  }
 
  int depth = -1;
  int zeroCount = 0;
  Stack<Character> stack = new Stack<Character>();
  for(int i=0; i<inp.length(); i++) {
    char c = inp.charAt(i);
    if(c == '(') {
      stack.push(c);
      if(stack.size() > depth) {
        depth = stack.size();
      }
      zeroCount = 0;
    } else if(c == ')') {
      stack.pop();
      zeroCount = 0;
    } else {
      if( c != '0') {
        return -1;
      } else {
        zeroCount += 1;
        if(zeroCount > 2) {
          return -1;
        }
      }
    }
  }
 
  // invalid pattern
  if(stack.size() > 0) {
    return -1;
  }
 
  return depth-1;
}

Generate number - Linkedin


http://wxx5433.github.io/generate-number.html
There is a particular sequence that only uses numbers 1, 2, 3, 4 and no two adjacent numbers are the same.
Write a program that given n1 1s, n2 2s, n3 3s, n4 4s will output the number of such sequences using all these numbers.
Output your answer modulo 1000000007 (10^9 + 7).

Analysis

We can use DFS to find the result. Each time we try to append a number, we check if it is the same as previous number. If it is, then it's invalid.

Complexity

Time: O(4^len), len is the totally lengt
Space: O(len), the length of the path DFS will search.
There is a particular sequence only uses the numbers 1, 2, 3, 4 and no two adjacent numbers are the same.
Write a program that given n1 1s, n2 2s, n3 3s, n4 4s will output the number of such sequences using all these numbers.
Output your answer modulo 1000000007 (10^9 + 7).
Solution 1: the easy way to do it would be using recursion and back trace.  if count of 1 is greater than 0 and the string’s end is not 1, try 1 and back trace, so do 2, 3, 4.
- DP + cache? Map: key is string: "n1-n2-n3-n4"
        private int count = 0;
        public int Sequence(int n1, int n2, int n3, int n4)
        {
            this.SequenceRecursion(n1, n2, n3, n4, string.Empty);

            return count;
        }

        private bool SequenceRecursion(int n1, int n2, int n3, int n4, string str)
        {
            if (n1 == 0 && n2 == 0 && n3 == 0 && n4 == 0)
            {
                count++;
                return true;
            }

            if (str.Length == 0 || (str.Length > 0 && n1 > 0 && !str[str.Length - 1].Equals('1')))
            {
                str += "1";
                this.SequenceRecursion(n1 - 1, n2, n3, n4, str);
                str = str.Substring(0, str.Length - 1);
            }

            if (str.Length == 0 || (str.Length > 0 && n2 > 0 && !str[str.Length - 1].Equals('2')))
            {
                str += "2";
                this.SequenceRecursion(n1, n2 - 1, n3, n4, str);
                str = str.Substring(0, str.Length - 1);
            }

            if (str.Length == 0 || (str.Length > 0 && n3 > 0 && !str[str.Length - 1].Equals('3')))
            {
                str += "3";
                this.SequenceRecursion(n1, n2, n3 - 1, n4, str);
                str = str.Substring(0, str.Length - 1);
            }

            if (str.Length == 0 || (str.Length > 0 && n4 > 0 && !str[str.Length - 1].Equals('4')))
            {
                str += "4";
                this.SequenceRecursion(n1, n2, n3, n4 - 1, str);
                str = str.Substring(0, str.Length - 1);
            }

            return false;
        }
Solution 2: if we use 4 dimension arrays to store how many number of particular sequence without adjacent. It would save a lot of time.
dp1[i][j][k][l] : how many number of particular sequence without adjacent end with 1 and has i 1s, j 2s, k 3s, l 4s.
dp2[i][j][k][l] : how many number of particular sequence without adjacent end with 2 and has i 1s, j 2s, k 3s, l 4s.
So dp1[i][j][k][l] = dp2[i – 1][j][k][l] + dp3[i – 1][j][k][l] +dp4[i – 1][j][k][l]; dp1 would be the sum of i – 1 1s end with 2(dp2), 3(dp3), 4(dp4);
        public int Sequence2(int n1, int n2, int n3, int n4)
        {
            var dp1 = new int[n1 + 1, n2 + 1, n3 + 1, n4 + 1];
            var dp2 = new int[n1 + 1, n2 + 1, n3 + 1, n4 + 1];
            var dp3 = new int[n1 + 1, n2 + 1, n3 + 1, n4 + 1];
            var dp4 = new int[n1 + 1, n2 + 1, n3 + 1, n4 + 1];

            const int MOD = 1000000007;
            dp1[1, 0, 0, 0] = 1;
            dp2[0, 1, 0, 0] = 1;
            dp3[0, 0, 1, 0] = 1;
            dp4[0, 0, 0, 1] = 1;

            for (int i = 0; i <= n1; i++)
            {
                for (int j = 0; j <= n2; j++)
                {
                    for (int k = 0; k <= n3; k++)
                    {
                        for (int l = 0; l <= n4; l++)                                        {                             
                            if (i + j + k + l > 1)
                            {
                                if (i > 0) dp1[i, j, k, l] = dp2[i - 1, j, k, l] + dp3[i - 1, j, k, l] + dp4[i - 1, j, k, l] % MOD;
                                if (j > 0) dp2[i, j, k, l] = dp1[i, j - 1, k, l] + dp3[i, j - 1, k, l] + dp4[i, j - 1, k, l] % MOD;
                                if (k > 0) dp3[i, j, k, l] = dp2[i, j, k - 1, l] + dp1[i, j, k - 1, l] + dp4[i, j, k - 1, l] % MOD;
                                if (l > 0) dp4[i, j, k, l] = dp2[i, j, k, l - 1] + dp3[i, j, k, l - 1] + dp1[i, j, k, l - 1] % MOD;
                            }
                        }
                    }

                }
            }
            return dp1[n1, n2, n3, n4] + dp2[n1, n2, n3, n4] + dp3[n1, n2, n3, n4] + dp4[n1, n2, n3, n4] % MOD;
        }

public static void generateNumber(List<Integer> result, String number, int[] count, int len) {
    if (number.length() == len) {
        BigInteger num = new BigInteger(number).mod(new BigInteger("1000000007"));
        result.add(new Integer(num.toString()));
        return ;
    }

    for (int i = 0; i < 4; ++i) {
        int num = i + 1;
        if (count[i] > 0 && (number.length() == 0 
                || number.charAt(number.length() - 1) - '0' != num)) {
            --count[i];
            generateNumber(result, number + num, count, len);
            ++count[i];
        }
    }
}
https://www.careercup.com/question?id=5653460783464448

Find an integer array corresponding to the string specifying increase-decrease transitions


http://www.ideserve.co.in/learn/integer-array-corresponding-to-increase-decrease-sequence
Given a string of size 'n' where each character can be either 'd' or 'i' and nothing else. If character 'd' denotes decrease in value and character 'i' denotes increase in value then how can we build an integer array of size 'n+1' created by using numbers from 1 to 'n+1' such that this array holds one to one correspondence with the input string.

For example, if the given string is "di" then because string size is 2, we need to use numbers 1,2,3 and build integer array [3,1,2] where first pair formed by first two elements 3,1 corresponds to character 'd' since there is decrease in value from 3 to 1 and then second pair 1,2 corresponds to character 'i' - increase in value from 1 to 2. Another way to build this could have been [2,1,3]. Now pair (2,1) corresponds to 'd' and second pair (1,3) corresponds to character 'i'.

When we consider 'n' such pairs formed out of adjacent elements from 'n+1' elements, 'n' pairs should correspond to 'n' characters of input string in the same sequence.

Another example could be for input string 'ddddi', one of the outputs could be [6,4,3,2,1,5].

Write a program to create any one of the correct output integer array given an input string having characters 'd' and 'i'.
The idea is to use the higher numbers for satisfying 'i' transitions and remaining numbers for 'd' transitions.

Let's look at an example. Say we have a string consisting of only 2 'i's and 3 'd's - 'ddidi'
Then out of numbers 1,2,3,4,5,6 we reserve two higher numbers 6 and 5 for 'i' transitions because no matter what the current number we can always satisfy 'i' transitions using these higher numbers. The number next to reserved numbers for 'i' transitions is put at the 0th position. In this case, number next to 6 and 5 would be 4 hence we make output[0] = 4. Note that out of 1,2,3,4,5,6 we have marked top 2 numbers that is 6 and 5 for 'i' transitions and next number to them that is 4 as a start value for output array - therefore remaining numbers(that is 3,2,1) would be used for 'd' transitions. Now we mark lowest number from the numbers reserved for 'i' transitions as increaseValue - in this case increaseValue would be 5. Similarly, we mark highest number from numbers that are going to used for 'd' transitions as decreaseValue - in this case decreaseValue would be 3.

Now the algorithm is very simple. Whenever we see a 'd' at string[i], we just put decreaseValue at output[i+1] and decrement decreaseValue by 1. And whenever we see a 'i' at string[i], we just put increaseValue at output[i+1] and increment increaseValue by 1. Output[0] is already initilaized to the number next to numbers reserved for 'i' and 'd' transitions.

In this case, for string 'ddidi' we will be using numbers 1,2,3,4,5,6
Numbers reserved for 'i' - 6,5
StartValue to be put at output[0] - 4
Numbers reserved for 'd' - 3,2,1
increaseValue marked at - 5
decreaseValue marked at - 3

output[0] = 4
output[1] = 3, decreaseValue updated to 2 (since 0th character for input is 'd')
output[2] = 2, decreaseValue updated to 1 (first character is 'd')
output[3] = 5, increaseValue updated to 6 (second character is 'i')
output[4] = 1, decreaseValue updated to 0 (third character is 'd')
output[5] = 6, increaseValue updated to 7 (fourth character is 'i')

Time taken by this algorithm is O(n).
    public void createSequence(String input, int[] output)
    {
        if (input.length() == 0)
        {
            return;
        }
         
        int iCount = 0;
        // count the number of increases required
        for (int i = 0;  i < input.length(); i++)
        {
            if (input.charAt(i) == 'i')
            {
                iCount += 1;
            }
        }
         
        // now in numbers 1 to n+1 reserve 'iCount' higher numbers to be used for 'i'
        // for example if there are 3 'i's in 6 character string,
        // then reserve numbers 7,6 and 5 for 'i'
         
        int n = input.length();
         
        // if we see a 'i', put 'increaseValue' in the output array and increment 'increaseValue' by 1
        int increaseValue = n + 2 - iCount;
         
        // keep startValue fixed
        int startValue = increaseValue - 1;
         
        // if we see a 'd', put 'decreaseValue' in the output array and decrement 'decreaseValue' by 1
        int decreaseValue = startValue - 1;
         
        output[0] = startValue;
        for (int i = 0;  i < input.length(); i++)
        {
            if (input.charAt(i) == 'i')
            {
                output[i+1] = increaseValue;
                increaseValue += 1;
            }
             
            if (input.charAt(i) == 'd')
            {
                output[i+1] = decreaseValue;
                decreaseValue -= 1;
            }
        }
    }

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