Google Interview Question for Principal Software Engineers


Country: United States




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0
of 2 vote

public class Solution {
    public static String[] shortestSubArray(String[] words, String[] keywords) {
        // Sliding window of words and their indices
        Map<String, Integer> wordAtIdx = new LinkedHashMap<>();

        Set<String> keywordSet = new HashSet<>(Arrays.asList(keywords));

        int solutionSize = Integer.MAX_VALUE; // can alo be words.length + 1
        int solutionStartIdx = -1;

        for (int i = 0; i < words.length; i++) {
            if (keywordSet.contains(words[i])) {
                // Remove (if necessary) and re-add word to end of sliding window
                wordAtIdx.remove(words[i]);
                wordAtIdx.put(words[i], i);

                // Check for new solution if our window matches the keyword size
                if (wordAtIdx.size() == keywordSet.size()) {
                    int windowStartIdx = wordAtIdx.values().iterator().next();
                    int candidateSize = i - windowStartIdx + 1;

                    if (candidateSize < solutionSize) {
                        solutionSize = candidateSize;
                        solutionStartIdx = windowStartIdx;
                    }

                    // Remove the leftmost element from the window
                    wordAtIdx.remove(words[windowStartIdx]);
                }
            }
        }

        return solutionStartIdx == -1 ? null
                : Arrays.copyOfRange(words, solutionStartIdx, solutionStartIdx + solutionSize);
    }
}

- jgriesser January 26, 2018 | Flag Reply
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0
of 0 vote

@jgriesser I think you would wanna use linkedHashSet, instead of a set since hashset doesn't guarantee order

- Shah January 26, 2018 | Flag Reply
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0
of 0 vote

Leetcode: 76. Minimum Window Substring

- UnknownUser January 30, 2018 | Flag Reply
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0
of 0 vote

76. Minimum Window Substring from LC

- unknownuser January 30, 2018 | Flag Reply
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0
of 0 vote

Leetcode: 76. Minimum Window Substring

- CodeNinja January 30, 2018 | Flag Reply
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0
of 0 vote

If the order has to be same, then its 727 LC, minimum window subsequence

- harrypottir January 31, 2018 | Flag Reply
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0
of 0 vote

def lastIndex(words, keywords):
    last = 0
    for keyword in keywords:
        try:
            i = words.index(keyword)
            words = words[i + 1:]
            last += i
        except ValueError:
            raise
    return last + 1

def minSubstring(words, keywords):
    first = keywords[0]
    startIndex = endIndex = 0

    for i, word in enumerate(words):
        if first != word:
            continue
        try:
            start = i 
            end = lastIndex(words[start + 1:], keywords[1:])
            if end - start > startIndex - endIndex:
                startIndex = start
                endIndex = start + end
        except ValueError:
            continue

    if startIndex != 0: 
        return words[startIndex:endIndex + 1]
    else:
        return ''

- Anonymous February 03, 2018 | Flag Reply
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0
of 0 vote

public class ShortestStringContainingKeywords {	
   public static void main(String[] args){
	   String[][] testcases = {
			   {"sky", "cloud", "google", "search", "sky", "work", "blue"},
			   {"sky", "blue"},
			   {"sky", "cloud", "google", "search", "sky", "work", "blue", "sky"},
			   {"sky", "blue"}};
	   
	   for(int i = 0; i < testcases.length/2; i++){
		   String doc[] = testcases[2*i];
		   String keywords[] = testcases[2*i+1];
		  String res = search(doc, keywords);
		   System.out.println(Arrays.toString(doc) + "\t" + res);
	   }
   }

   /**
    * keep track of list of occurrences of word, 
    * -- if all words covered have a feasible solution
    * -- if word occurring second time, move forward iterators 
    * 	until there is a word that is covered only once 
    */
   private static String search(String[] doc, String[] keywords) {
	   Map<String, Integer> keywordId = new HashMap<String, Integer>();
	   int numKeywords = 0;
	   for(String kw : keywords){
		   keywordId.put(kw, numKeywords);
		   numKeywords++;
	   }
	   
	   short n = (short) doc.length;
	   short optWindowStart = 0, currWindowStart = 0;
	   short optWindowLength = (short) (n + 1);
	   int[] numMatches = new int[numKeywords];
	   for(int i = 0; i < numKeywords; i++){
		   numMatches[i] = 0;
	   }
	   int numKeywordMatches = 0;
	   List<Integer> tokenIdToMatchingKw = new ArrayList<Integer>();
	   
	   for(short i = 0; i < n; i++){
		   if(keywordId.containsKey(doc[i])){
			   int kwIndex = keywordId.get(doc[i]);
			   tokenIdToMatchingKw.add(kwIndex);
			   if(numMatches[kwIndex] == 0){
				   numKeywordMatches++;
			   }
			   numMatches[kwIndex]++;
			   
			   if(numKeywordMatches == numKeywords){
				   short j = currWindowStart;
				   while(j < i){
					   int kwj = tokenIdToMatchingKw.get(j);
					   if(kwj != -1){
						   if(numMatches[kwj] > 1){
							   numMatches[kwj]--;
							   j++;
						   } else {
							   // local maximum
							   currWindowStart = j;
							   short currWindowLength = (short) (i - j + 1);
							   if(currWindowLength < optWindowLength){
								   optWindowLength = currWindowLength;
								   optWindowStart = currWindowStart;
							   }
							   break;
						   }
					   }
					   j++;
				   }
			   }
		   } else {
			   tokenIdToMatchingKw.add(-1);
		   }
	   }
	   
	   if(optWindowLength > n)
		   return null;
	   
	   String res = "";
	   for (short j = optWindowStart; j < optWindowStart + optWindowLength; j++) {
		   if(j > 0){
			   res += " " + doc[j];
		   } else {
			   res += doc[j];
		   }
	   }
	   return res;
   }
}

- just_do_it February 11, 2018 | Flag Reply
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0
of 0 vote

public class ShortestString {
	
	public static void main(String args[]){
		String input = "sky cloud google search sky work blue";
		String keywords = "sky blue";
		String[] keywordsList = keywords.split(" ");
		List<Integer> firstWordIndices = new ArrayList<>();
		List<Integer> lastWordIndices = new ArrayList<>();
		int cInd = -1;
		int lastIndex = input.lastIndexOf(keywordsList[0]);
		while(cInd < input.length() && cInd != lastIndex){
			cInd = input.indexOf(keywordsList[0], cInd+1);			
			firstWordIndices.add(cInd);
		}
		cInd = -1;
		lastIndex = input.lastIndexOf(keywordsList[1]);
		while(cInd < input.length() && cInd != lastIndex){
			cInd = input.indexOf(keywordsList[1], cInd+1);
			lastWordIndices.add(cInd);
		}
		Map<Integer, Integer> lastFirstMap = new HashMap<>();
		for(Integer index: lastWordIndices){
			for(Integer findex: firstWordIndices){
				if(findex<index){
					if(lastFirstMap.get(index) != null){
						if(findex > lastFirstMap.get(index)){
							lastFirstMap.put(index, findex);
						}
					}
					else{
						lastFirstMap.put(index, findex);
					}
				}
			}
		}
		Entry<Integer, Integer> currentSmallestEntry = null;
		int currentDiff = 0;
		for(Entry<Integer, Integer> entry: lastFirstMap.entrySet()){
			if(currentSmallestEntry == null){
				currentSmallestEntry = entry;
				currentDiff = currentSmallestEntry.getKey() - currentSmallestEntry.getValue();
				continue;
			}
			if((entry.getKey() - entry.getValue()) < currentDiff){
				currentSmallestEntry = entry;
			}
		}
		System.out.println("currentSmallestEntry: "+input.substring(currentSmallestEntry.getValue(), currentSmallestEntry.getKey()+keywordsList[1].length()));
	}

}

- xxx October 06, 2018 | Flag Reply
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0
of 0 vote

public class ShortestString {
	/*Give a  string [] words,
Find the shortest string [] containing the keyword inside.
example:
words: sky cloud google search sky work blue
keywords: sky blue
return: sky work blue*/
	
	public static void main(String args[]){
		String input = "sky cloud google search sky work blue";
		String keywords = "sky blue";
		String[] keywordsList = keywords.split(" ");
		List<Integer> firstWordIndices = new ArrayList<>();
		List<Integer> lastWordIndices = new ArrayList<>();
		int cInd = -1;
		int lastIndex = input.lastIndexOf(keywordsList[0]);
		while(cInd < input.length() && cInd != lastIndex){
			cInd = input.indexOf(keywordsList[0], cInd+1);			
			firstWordIndices.add(cInd);
		}
		cInd = -1;
		lastIndex = input.lastIndexOf(keywordsList[1]);
		while(cInd < input.length() && cInd != lastIndex){
			cInd = input.indexOf(keywordsList[1], cInd+1);
			lastWordIndices.add(cInd);
		}
		Map<Integer, Integer> lastFirstMap = new HashMap<>();
		for(Integer index: lastWordIndices){
			for(Integer findex: firstWordIndices){
				if(findex<index){
					if(lastFirstMap.get(index) != null){
						if(findex > lastFirstMap.get(index)){
							lastFirstMap.put(index, findex);
						}
					}
					else{
						lastFirstMap.put(index, findex);
					}
				}
			}
		}
		Entry<Integer, Integer> currentSmallestEntry = null;
		int currentDiff = 0;
		for(Entry<Integer, Integer> entry: lastFirstMap.entrySet()){
			if(currentSmallestEntry == null){
				currentSmallestEntry = entry;
				currentDiff = currentSmallestEntry.getKey() - currentSmallestEntry.getValue();
				continue;
			}
			if((entry.getKey() - entry.getValue()) < currentDiff){
				currentSmallestEntry = entry;
			}
		}
		System.out.println("currentSmallestEntry: "+input.substring(currentSmallestEntry.getValue(), currentSmallestEntry.getKey()+keywordsList[1].length()));
	}

}

- xxx October 06, 2018 | Flag Reply
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0
of 0 vote

Here is my answer in python.

Running time: O(N Log N)

import sys

def find_closest_bigger_number(array, target, next_closest):
    if(len(array) == 1):
        return array[0] if (array[0] > target) and (next_closest - target) > (array[0] - target) else next_closest
    
    half = len(array)/2

    if array[half] > target:
        next_closest = array[half] if (next_closest - target) > (array[half] - target) else next_closest
        return find_closest_bigger_number(array[:half], target, next_closest)
    else:
        return find_closest_bigger_number(array[half+1:], target, next_closest)

def find_shortest_sentence(keywords, words):
    keywordHash ={}
    answer =[]
    shortest = sys.maxint
    keyword_occurrence_map = []

    for pos, keyword in enumerate(keywords):
        keywordHash[keyword] = pos
        keyword_occurrence_map.append([])

    #go through the words and count the occurrence, O(N)
    for p, word in enumerate(words):
        matched_key_word_pos = keywordHash.get(word, -1)
        if matched_key_word_pos == -1:
            continue
        keyword_occurrence_map[matched_key_word_pos].append(p)

    for first_pos in keyword_occurrence_map[0]:
        prev_pos = first_pos
        last_pos = sys.maxint
        for next_word_occurrence_map in keyword_occurrence_map[1:] :
            next_word_pos = find_closest_bigger_number(next_word_occurrence_map, prev_pos, sys.maxint)
            if (next_word_pos == sys.maxint):
                last_pos = sys.maxint
                break
            last_pos = next_word_pos

        if last_pos == sys.maxint:
            continue

        if (last_pos - first_pos < shortest):
            shortest = last_pos - first_pos
            answer= [first_pos, last_pos]
    return answer

- hankm2004 January 22, 2019 | Flag Reply


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