Solutions consultant Interview Questions | Glassdoor.co.uk

# Solutions consultant Interview Questions

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## Top Interview Questions

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22 Dec 2015

22 Dec 2015
 Boggle game - given a board of letters (2d array) and a word (string), return whether the word exists in the board. From each letter you can move in all directions (including diagonals), but you cannot use the same letter twice.4 AnswersFind all occurrences of the first letter in the word, then use a recursive function to move to all valid directions to the next letter until you get past the last letter. Use a hashtable to remember the used positions, or mark each letter as some unique string and revert it back later.//Boggle game - given a board of letters (2d array) and a word (string), return whether the word exists in the board. //From each letter you can move in all directions (including diagonals), but you cannot use the same letter twice. // For each index that has first character in the matrix call the de function. Then for each de function call de again recursively for its neighbors with some logic as written below. Use a hashtable to make sure you do not use same letter twice. Clear has for each master call form main program here. unordered_map seen; bool de(vector> x, int row, int col, string s, int r, int c) { if(s.size() == 0) return true; if(seen.find(s[0]) != seen.end()) return false; if(r = row || c >= col ) return false; if(x[c][r] == s[0]){ seen[s[0]] = 1; string temp = s.assign(s.begin()+1, s.end()) ; s = temp; } else return false; return ( de(x, row, col, s, r-1, c-1) || de(x, row, col, s, r-1, c) || de(x, row, col, s, r-1, c+1) || de(x, row, col, s, r, c-1) || de(x, row, col, s, r+1, c-1 ) || de(x, row, col, s, r+1, c) || de(x, row, col, s, r+1, c+1) || de(x, row, col, s, r, c+1)) ; } int main (int argc, char *argv[]) { map locs; vector> x = {{'A', 'B', 'O', 'P', 'L'}, {'X', 'Y', 'P', 'X', 'R'}, {'Q', 'S', 'T', 'A', 'L'}, {'R', 'T', '5', 'M', 'L'}, {'S', 'M', 'Y', 'P', 'Q'}}; int row = (int)x[0].size(); int col = (int)x.size(); string s = "SPAM"; for(int i = 0 ; i first, it->second); seen.clear(); //refresh seen vector to restart cache of seen characters if(retval) break; } if(retval) cout << "String " << s << " Found in the matrix" << "\n" ; else cout << "Too bad not found" << "\n" ; return 0; }Doesn't the above approach have a time complexity of pow(8,(m,n))?Show more responsesimport java.util.*; public class Game { public boolean check(char[][]game, String s) { boolean result = false; // Compute the position of initial character HashMap> hmap = new HashMap(); char[] characters = s.toCharArray(); for(int i = 0; i < game.length; i++) { for (int j = 0; j < game[0].length; j++) { if(characters[0]==game[i][j]) { // Call Horizontal result = HorizontalLR(characters, game, i, j); if(result) { return true; } // Call Vertical result = VerticalTB (characters, game, i, j ); if(result) { return true; } result = DiagonalLR (characters, game, i, j ); if(result) { return true; } } } } return false; } // Horizontal R-L public boolean HorizontalLR(char [] characters, char[][]game, int i, int j) { // check if the remaining length of the board is greater than or equal to length of string if(game[0].length - j < characters.length) { return false; } for (int p = j, q = 0; p < game[0].length && q < characters.length ; p++, q++) { if(characters[q] != game[i][p]) { return false; } } return true; } // Vertical Top to Bottom public boolean VerticalTB(char [] characters, char[][]game, int i, int j) { if(game.length - i< characters.length) { return false; } for(int p = i, q = 0; p < game.length && q < characters.length ; p++, q ++) { if(characters[q] != game[p][j]) { return false; } } return true; } // Diagonal L-R public boolean DiagonalLR(char [] characters, char[][]game, int i, int j) { if(game[0].length - j < characters.length || game.length-i< characters.length) { return false; } for(int p = i , q = j , r = 0 ; p< game[0].length && q < game.length && r < characters.length; p++, q++, r++) { if(characters[r] != game[p][q]) { return false; } } return true; } public static void main(String [] args) { Game g = new Game(); char [][] game = {{'B','A','T','A'}, {'A','N','M','N'}, {'G','G','H','T'}, {'M','E','T','E'}}; System.out.println(g.check(game, "AMTE")); } }

22 Dec 2015
 Write a basic Regex engine implementing the "." (any character) and "*" (previous rule, 0 to many). The function receives a string (letters only, no need for escaping) and a string pattern. It returns a bool whether the string matches the pattern. For example, the pattern "AB.*E" should match both "ABCDE" and "ABEEE".3 AnswersUse recursion. Set 2 pointers at the beginning of both strings, advance each one by one. The trick is to look forward for a "*" and in this case fork to 2 recursive calls - one for a zero occurrences and one for multiple occurrences.Would this pattern(ab.*e) match str "abe"?not to be cheeky but if they only want dot and star couldn't you just write a function whose body consisted solely of the line "return true"? dot star matches every possible string

### Solution Architect at JLL was asked...

31 Mar 2015
 No, specific interview questions as such (tells you how inefficient the process is) but mostly talking about your CV and past experience.3 AnswersJust talk about your past projects andThere were a few red flags throughout the recruitment process but I chose to accept the job and only focus on the positives. You live and learnI had a similar recent experience with "JLL London" There were red flags throughout the recruitment process but I chose to accept the job and only focus on the positives. You live and learn

22 Dec 2015
 Given an array of integers, write an in-place function to bring all the non-zero elements to the left of the array keeping the original order.3 AnswersSet a pointer to the first index, then iterate over each element. If the element is a non-zero swap with the pointer index and advance it by 1.Keep running count of zeros at each index. If a value at an index is non zero move it to the left by current running count of zero at that index. If value is 0 just increase running count. In the end pad from right with 0 with number of running counts. void movetoleft(int arr[], int size) { int runnigcountzero = 0; for(int i = 0; i 0) { arr[i - runnigcountzero] = arr[i]; } } int x =0 ; while (x < runnigcountzero) { arr[size-1 - x] = 0; x++; } }Or just move the zeroes to the right... ```` function moveToLeft(\$array) { \$zeroes = 0; \$original_size = count(\$array); for (\$i = 0; \$i < \$original_size; \$i++) { if (0 === \$array[\$i]) { unset(\$array[\$i]); \$array[\$original_size+\$zeroes] = 0; \$zeroes++; } } return array_values(\$array); } ````

### Cloud Solution Architect at Microsoft was asked...

27 Oct 2020
 How do my ethics and values fit into Microsoft’s3 AnswersHI, I have recevied a call from Microsoft Recruiter last week. They mentioned there would be some technicall assessment and if I pass, will be called for some back to back interview panel, including presentations. Can you please advise further on the possible areas that the technical assesment would cover. types of questions there, and also some more details how the internal pannel was, like what type of presentation was asked, if any. Thanks,Hi I have gone through the technical screening now i have the assessment round, please can u let me know the examples of what to prepare for the Deep technical level, presentation( what topics would come) and commercial round. Please kindly provide your valuable suggestions. thanksGuys, can you shed some light on what kind of questions and preparation you had done in both rounds especially the initial technical assessment and after?

19 May 2012
 multiply two two-digit numbers together2 AnswersLet S= the set of nonnegative integers less than 10. Consider arbitrary elements (not necessarily distinct) a,b,c,d∈S but with nonzero elements a and c. Then (b×〖10〗^0+a×〖10〗^1 )=(b+10a) and (d×〖10〗^0+c×〖10〗^1 )=(d+10c) are two digits numbers; whose product is (b+10a)×(d+10c)=bd+10bc+10ad+100ac.This is ridiculous and out of the left field, but I like it!

### Multi Asset Stratergies (Model Portfolio Solutions) at BlackRock was asked...

13 Jan 2018
 What is CAPM?2 AnswersI have never heard of thatCapital Asset Pricing Model

### Solution Architect at SCC was asked...

8 Apr 2014
 SCC has a staff retention problem and will want to work hard to understand your reasons for having more than one job in your entire career...2 Answersbe prepared to justify your reasons... which they will probably ignore anywayThe HR team are muppets and their package are below market. Not forgetting they are based in a terrible location for commuting, but hey it’s cheap which means the family are milking it in.

### Client Solutions Summer Analyst at BlackRock was asked...

29 Nov 2017
 Tell me about a time you showed leadership?2 AnswersHow long have you waited after the 1st round before being called for the 2nd? Many thanks !what the competency based interview was about please
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