Walmart Labs Data Structures & Algorithms Interview Questions

Sravanthi M. - "public static boolean isPalindrome(String str){ boolean flag = true; int len = str.length()-1; int j = len; for(int i=0;i<=len/2;i++){ if(str.charAt(i)!=str.charAt(j--)){ flag = false; break; } } return flag; }"See full answer

VContaineers - " Compare alternate houses i.e for each house starting from the third, calculate the maximum money that can be stolen up to that house by choosing between: Skipping the current house and taking the maximum money stolen up to the previous house. Robbing the current house and adding its value to the maximum money stolen up to the house two steps back. package main import ( "fmt" ) // rob function calculates the maximum money a robber can steal func maxRob(nums []int) int { ln"See full answer

Alfred O. - "We can use dictionary to store cache items so that our read / write operations will be O(1). Each time we read or update an existing record, we have to ensure the item is moved to the back of the cache. This will allow us to evict the first item in the cache whenever the cache is full and we need to add new records also making our eviction O(1) Instead of normal dictionary, we will use ordered dictionary to store cache items. This will allow us to efficiently move items to back of the cache a"See full answer

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

Kishor J. - "we can use two pointer + set like maintain i,j and also insert jth character to set like while set size is equal to our window j-i+1 then maximize our answer and increase jth pointer till last index"See full answer

Kofi N. - "const mergeIntervals = (intervals) => { const compare = (a, b) => { if(a[0] b[0]) return 1 else if(a[0] === b[0]) { return a[1] - b[1] } } let current = [] const result = [] const sorted = intervals.sort(compare) for(let i = 0; i = b[0]) current[1] = b[1] els"See full answer

Anonymous Roadrunner - "from typing import List def three_sum(nums: List[int]) -> List[List[int]]: nums.sort() triplets = set() for i in range(len(nums) - 2): firstNum = nums[i] l = i + 1 r = len(nums) - 1 while l 0: r -= 1 elif potentialSum < 0: l += 1 "See full answer

Prajwal M. - "from typing import List def maxprofitgreedy(stock_prices: List[int]) -> int: l=0 # buying r=1 # selling max_profit=0 while rstock_prices[l]: profit=stockprices[r]-stockprices[l] maxprofit=max(maxprofit,profit) else: l=r r+=1 return max_profit debug your code below print(maxprofitgreedy([7, 1, 5, 3, 6, 4])) `"See full answer

Tiago R. - "function generateParentheses(n) { if (n < 1) { return []; } if (n === 1) { return ["()"]; } const combinations = new Set(); let previousCombinations = generateParentheses(n-1); for (let prev of previousCombinations) { for (let i=0; i < prev.length; i++) { combinations.add(prev.slice(0, i+1) + "()" + prev.slice(i+1)); } } return [...combinations]; } `"See full answer

Ritaban M. - " Brute Force Two Pointer Solution: from typing import List def two_sum(nums, target): for i in range(len(nums)): for j in range(i+1, len(nums)): if nums[i]+nums[j]==target: return [i,j] return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"See full answer

Guilherme F. - "A much better solution than the one in the article, below: It looks like the ones writing articles here in Javascript do not understand the time/space complexity of javascript methods. shift, splice, sort, etc... In the solution article you have a shift and a sort being done inside a while, that is, the multiplication of Ns. My solution, below, iterates through the list once and then sorts it, separately. It´s O(N+Log(N)) class ListNode { constructor(val = 0, next = null) { th"See full answer

Bhaskar B. - "This could be done using two-pointer approach assuming array is sorted: left and right pointers. We need track two sums (left and right) as we move pointers. For moving pointers we will move left to right by 1 (increment) when right sum is greater. We will move right pointer to left by 1 (decrement) when left sum is greater. at some point we will either get the sum same and that's when we exit from the loop. 0-left will be one array and right-(n-1) will be another array. We are not going to mo"See full answer

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

Divyani .. - "Did the code in Python"See full answer