Google Interview Question


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def is_reset( cur_char, sub_index, sub_string ){
  sub_index > 0 && 
  ( cur_char @  sub_sequence_string[0:sub_index-1]  ) && // is in 
  ( cur_char !@  sub_sequence_string[sub_index-1:-1]  ) // not in 
}

def find_all_subsequences_indices( host_string, sub_sequence_string ){
  host_size = size(host_string)
  sub_size = size(sub_sequence_string)
  hi = 0 // host index 
  si = 0 // sub seq index 
  collector = list()
  tmp_list = list()
  while ( hi < host_size ){
    // this is for that ... god knows what. 
    // it is not a typical subsequence problem .. so reset when ... 
    if ( is_reset( host_string[hi] , si, sub_sequence_string ) ){
      si = 0 
      tmp_list.clear() 
    }
    if ( host_string[hi] == sub_sequence_string[si] ){
        tmp_list += hi 
        si += 1
        if ( si == sub_size ){
          collector.add( tmp_list )
          si = 0 
          tmp_list = list()
        }
    }
    hi += 1
  }
  collector // return 
}
/*
Given for example S = BCXXBXXCXDXBCD, then the result should be [[4,7,9],[11,12,13] 
find BCBC in String S = BCXXBXCXBCBC 
the result should be [[0,1,4,6],[8,9,10,11]]
*/

println( jstr( find_all_subsequences_indices( "BCXXBXXCXDXBCD" , "BCD" ) ) )
println( jstr( find_all_subsequences_indices( "BCXXBXCXBCBC" , "BCBC" ) ) )

- undefined June 01, 2019 | Flag Reply
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public static List<int[]> findLocation(String subString, String str) {
        if (subString == null || str == null) return null;
        List<int[]> result = new ArrayList<>();
        int[] location = new int[subString.length()];
        Map<Character, Integer> countChar = new HashMap<>();
        Map<Character, Integer> foundChar = new HashMap<>();
        int i = 0;
        int j = 0;

        while (j < str.length()) {

            countChar.put(str.charAt(j), countChar.containsKey(str.charAt(j)) ? countChar.get(str.charAt(j)) + 1 : 1);

            if (str.charAt(j) == subString.charAt(i)){
                foundChar.put(str.charAt(j), foundChar.containsKey(str.charAt(j)) ? foundChar.get(str.charAt(j)) + 1 : 1);
                location[i] = j;
                i++;
            }

            if(foundChar.containsKey(str.charAt(j)) && countChar.get(str.charAt(j)) > foundChar.get(str.charAt(j))) {
                countChar.clear();
                foundChar.clear();
                location = new int[subString.length()];
                if(str.charAt(j) == subString.charAt(0)) {
                    location[0] = j;
                    i = 1;
                }
            }

            if (i == subString.length()){
                result.add(location);
                location = new int[subString.length()];
                i = 0;
            }
            j++;
        }

        return result;
    }

- Danny June 01, 2019 | Flag Reply
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of 0 vote

Hello

- Hello World June 02, 2019 | Flag Reply
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This could be solved with regex, O(nm) or using modified KMP, i.e preprocess string first and then run KMP 2O(n).

- nooooob June 02, 2019 | Flag Reply
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of 0 votes

This wont work. KMP is only for substring not subsequence.

- code reviewer June 03, 2019 | Flag
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modified KMP, i.e preprocess string first and then run KMP 2O(n)?
Do you mean by removing the X and then use the KMP to do string matching?

- bottleneck56 June 04, 2019 | Flag
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Yes, it could be done in two ways.

1) Create an array of valid indices, i.e if you ever encounter X at index (k) it should point you to the valid character (B, C , D etc) i < k. And this could be factored in the KMP algorithm while matching the pattern with the string.

2) Remove Xs, and while doing it create an array that will give you indices of B, C, and D in the original input. In this case, run KMP and for valid outputs, you can get the indices information from the array while printing the results.

Let me know if you need an example.

- nooooob June 08, 2019 | Flag
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of 0 votes

yes please, example would be good here. Thanks in advance.

- code reviewer June 08, 2019 | Flag
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of 0 vote

def patSearch(txt,pat):
	txtpos=0
	patpos=0
	txtlen=len(txt)
	patlen=len(pat)
	result=[[]]
	while txtpos!=txtlen:
		if(txt[txtpos]==pat[patpos]):
			result[-1].append(txtpos)
			patpos+=1
			if(patpos==patlen):
				result.append([])
				patpos=0
		elif(txt[txtpos]!='X'):
			result[-1]=[]
			patpos=0
			if(txt[txtpos]==pat[patpos]):
				result[-1].append(txtpos)
				patpos+=1
		txtpos+=1
	return result
print(patSearch("BCXXBXCXBCBC" , "BCBC"))
print(patSearch("BCXXBXXCXDXBCD" , "BCD" ))

- Anonymous June 04, 2019 | Flag Reply
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of 0 vote

def patSearch(txt,pat):
	txtpos=0
	patpos=0
	txtlen=len(txt)
	patlen=len(pat)
	result=[[]]
	while txtpos!=txtlen:
		if(txt[txtpos]==pat[patpos]):
			result[-1].append(txtpos)
			patpos+=1
			if(patpos==patlen):
				result.append([])
				patpos=0
		elif(txt[txtpos]!='X'):
			result[-1]=[]
			patpos=0
			if(txt[txtpos]==pat[patpos]):
				result[-1].append(txtpos)
				patpos+=1
		txtpos+=1
	return result
print(patSearch("BCXXBXCXBCBC" , "BCBC"))
print(patSearch("BCXXBXXCXDXBCD" , "BCD" ))

- Anonymous June 04, 2019 | Flag Reply
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0
of 0 vote

def patSearch(txt,pat):
	txtpos=0
	patpos=0
	txtlen=len(txt)
	patlen=len(pat)
	result=[[]]
	while txtpos!=txtlen:
		if(txt[txtpos]==pat[patpos]):
			result[-1].append(txtpos)
			patpos+=1
			if(patpos==patlen):
				result.append([])
				patpos=0
		elif(txt[txtpos]!='X'):
			result[-1]=[]
			patpos=0
			if(txt[txtpos]==pat[patpos]):
				result[-1].append(txtpos)
				patpos+=1
		txtpos+=1
	return result
print(patSearch("BCXXBXCXBCBC" , "BCBC"))
print(patSearch("BCXXBXXCXDXBCD" , "BCD" ))

- Anonymous June 04, 2019 | Flag Reply
Comment hidden because of low score. Click to expand.
-1
of 1 vote

class Code{

}

- Anonymous June 01, 2019 | Flag Reply


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