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Service Now Interview Questions

Review this list of 13 Service Now interview questions and answers verified by hiring managers and candidates.
  • Service Now logoAsked at Service Now 
    5 answers
    +2

    "This is a fantastic question and one that comes up very often in behavioral rounds. When interviewers ask you to talk about something you built end to end without relying on others, they are not looking for a lone wolf who cannot collaborate. Instead, they are deeply testing your Ownership and trying to determine your engineering level. Here is how different levels typically approach this exact same question: Junior to Mid Level: Focuses on independently completing a well defined UI component"

    Ingrid Z. - "This is a fantastic question and one that comes up very often in behavioral rounds. When interviewers ask you to talk about something you built end to end without relying on others, they are not looking for a lone wolf who cannot collaborate. Instead, they are deeply testing your Ownership and trying to determine your engineering level. Here is how different levels typically approach this exact same question: Junior to Mid Level: Focuses on independently completing a well defined UI component"See full answer

    Product Manager
    Behavioral
    +1 more
  • Service Now logoAsked at Service Now 
    126 answers
    Video answer for 'Tell me about yourself.'
    +118

    "As you know, this is the most important question for any interview. Here is a structure I like to follow, Start with 'I'm currently a SDE/PM/TPM etc with XYZ company.... ' Mention how you got into PM/TPM/SDE field (explaining your journey) Mention 1 or 2 accomplishments Mention what you do outside work (blogging, volunteer etc) Share why are you looking for a new role Ask the interviewer if they have any questions or will like to dive deep into any of your experience"

    Bipin R. - "As you know, this is the most important question for any interview. Here is a structure I like to follow, Start with 'I'm currently a SDE/PM/TPM etc with XYZ company.... ' Mention how you got into PM/TPM/SDE field (explaining your journey) Mention 1 or 2 accomplishments Mention what you do outside work (blogging, volunteer etc) Share why are you looking for a new role Ask the interviewer if they have any questions or will like to dive deep into any of your experience"See full answer

    Product Manager
    Behavioral
    +19 more
  • Service Now logoAsked at Service Now 
    16 answers
    Video answer for 'Given an integer array nums and an integer k, return true if nums has a subarray of at least two elements whose sum is a multiple of k.'
    +12

    "Would be better to adjust resolution in the video player directly."

    Anonymous Prawn - "Would be better to adjust resolution in the video player directly."See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    1 answer

    "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."

    Gaston B. - "Use a representative of each, e.g. sort the string and add it to the value of a hashmap> where we put all the words that belong to the same anagram together."See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    24 answers
    +21

    "def is_valid(s: str) -> bool: stack = [] closeToOpen = { ")" : "(", "]" : "[", "}" : "{" } for c in s: if c in closeToOpen: if stack and stack[-1] == closeToOpen[c]: stack.pop() else: return False else: stack.append(c) return True if not stack else False debug your code below print(is_valid("()[]")) `"

    Anonymous Roadrunner - "def is_valid(s: str) -> bool: stack = [] closeToOpen = { ")" : "(", "]" : "[", "}" : "{" } for c in s: if c in closeToOpen: if stack and stack[-1] == closeToOpen[c]: stack.pop() else: return False else: stack.append(c) return True if not stack else False debug your code below print(is_valid("()[]")) `"See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
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  • Service Now logoAsked at Service Now 
    19 answers
    Video answer for 'Given stock prices for the next n days, how can you maximize your profit by buying or selling one share per day?'
    +14

    "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "

    Laksitha R. - "public static int maxProfitGreedy(int[] stockPrices) { int maxProfit = 0; for(int i = 1; i todayPrice) { maxProfit += tomorrowPrice - todayPrice; } } return maxProfit; } "See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    Add answer
    Video answer for 'Find the median of two sorted arrays.'
    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    18 answers
    Video answer for 'Given an nxn grid of 1s and 0s, return the number of islands in the input.'
    +15

    " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "

    Rick E. - " from typing import List def getnumberof_islands(binaryMatrix: List[List[int]]) -> int: if not binaryMatrix: return 0 rows = len(binaryMatrix) cols = len(binaryMatrix[0]) islands = 0 for r in range(rows): for c in range(cols): if binaryMatrixr == 1: islands += 1 dfs(binaryMatrix, r, c) return islands def dfs(grid, r, c): if ( r = len(grid) "See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    30 answers
    +22

    " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"

    Rick E. - " O(n) time, O(1) space from typing import List def maxsubarraysum(nums: List[int]) -> int: if len(nums) == 0: return 0 maxsum = currsum = nums[0] for i in range(1, len(nums)): currsum = max(currsum + nums[i], nums[i]) maxsum = max(currsum, max_sum) return max_sum debug your code below print(maxsubarraysum([-1, 2, -3, 4])) `"See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    Add answer
    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    13 answers
    +10

    "from typing import List def traprainwater(height: List[int]) -> int: if not height: return 0 l, r = 0, len(height) - 1 leftMax, rightMax = height[l], height[r] res = 0 while l < r: if leftMax < rightMax: l += 1 leftMax = max(leftMax, height[l]) res += leftMax - height[l] else: r -= 1 rightMax = max(rightMax, height[r]) "

    Anonymous Roadrunner - "from typing import List def traprainwater(height: List[int]) -> int: if not height: return 0 l, r = 0, len(height) - 1 leftMax, rightMax = height[l], height[r] res = 0 while l < r: if leftMax < rightMax: l += 1 leftMax = max(leftMax, height[l]) res += leftMax - height[l] else: r -= 1 rightMax = max(rightMax, height[r]) "See full answer

    Software Engineer
    Data Structures & Algorithms
    +4 more
  • Service Now logoAsked at Service Now 
    Add answer
    Product Manager
    Behavioral
    +1 more
  • Service Now logoAsked at Service Now 
    3 answers

    "boolean isMatch(String s, String p) { int i=0; int j=0; int sID=-1; int prevM=-1; while(i<s.length()){ if(j<p.length() && (s.charAt(i)==p.charAt(j) || p.charAt(j)=='?')){ i++; j++; }else if(j<p.length() && p.charAt(j)=='*'){ sID=j; prevM=i; j++; }else if(sID!=-1){ j=sID+1; prevM++; i=prevM; }else{ return false; } } while(j<p.length() && p.charAt(j)=='*') j++; if(i!=s.length() || j!=p.leng"

    Ravi C. - "boolean isMatch(String s, String p) { int i=0; int j=0; int sID=-1; int prevM=-1; while(i<s.length()){ if(j<p.length() && (s.charAt(i)==p.charAt(j) || p.charAt(j)=='?')){ i++; j++; }else if(j<p.length() && p.charAt(j)=='*'){ sID=j; prevM=i; j++; }else if(sID!=-1){ j=sID+1; prevM++; i=prevM; }else{ return false; } } while(j<p.length() && p.charAt(j)=='*') j++; if(i!=s.length() || j!=p.leng"See full answer

    Software Engineer
    Data Structures & Algorithms
    +1 more
Showing 1-13 of 13
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