IBM Software Engineer Interview Questions

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

Jean-pierre C. - "function twoSum(nums, target) { let complements = new Map(); for (let i = 0; i < nums.length; i++) { let diff = target - nums[i]; if (complements.has(diff)) { return [complements.get(diff), i]; } complements.set(nums[i], i); } return []; } console.log(twoSum([2, 7, 11, 15], 9)); `"See full answer