IBM Interview Questions

Salome L. - "SELECT MIN(id) AS id, TRIM(LOWER(email)) AS cleaned_email FROM users GROUP BY cleaned_email ORDER BY id `"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

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

Avon T. - "Initialize left pointer: Set a left pointer left to 0. Iterate through the array: Iterate through the array from left to right. If the current element is not 0, swap it with the element at the left pointer and increment left. Time complexity: O(n). The loop iterates through the entire array once, making it linear time. Space complexity: O(1). The algorithm operates in-place, modifying the input array directly without using additional data structures. "See full answer

Anonymous Roadrunner - "from typing import List def two_sum(nums: List[int], target: int) -> List[int]: prevMap = {} for i, n in enumerate(nums): diff = target - n if diff in prevMap: return [prevMap[diff], i] else: prevMap[n] = i return [] debug your code below print(two_sum([2, 7, 11, 15], 9)) `"See full answer

Alessandra L. - "My answer followed the framework: Users -> Goals -> Main Tasks -> Requirements -> Design -> Implementation -> Launch"See full answer