Description: Solution: row = [1] * n => The bottom row: [1, 1, 1, 1, 1, 1, 1] We start from (n-2) with reverse order. When we complete the second row[7, 6, 5, 4, 3, 2, 1], we update this newRow as row. Time Complexity: O(n*m) Space Complexity: O(n)
Description: Solution: We should consider if the number is two digits case: 10, 20 Since A-Z is 26, 20-26 is the right range for decode. ex) s = "112" dp[3] = 1 dp[2] = 1 dp[1] = 1 + 1 = 2 dp[0] = 2 + 1 = 3 Space Complexity: O(1)
Description: Solution: temp is for rob2. rob1 & rob2 is adjacent pointer. We are gonna compare n to (n - 1) + (n + 1)!
Description: Solution: For this problem, we are gonna use hashmap. ex) nums = [1, 2, 3], target = 4 dp[0] = 1 dp[1] = dp[1] + dp[1-1] + dp[1-2] + dp[1-3] = 0 + 1 + 0 + 0 = 1 dp[2] = dp[2] + dp[2-1] + dp[2-2] + dp[2-3] = 0 + 1 + 1 + 0 = 2 dp[3] = dp[3] + dp[3-1] + dp[3-2] + dp[3-3] = 0 + 2 + 1 + 1 = 4 dp[4] = dp[4] + dp[4-1] + dp[4-2] + dp[4-3] = 0 + 4 + 2 + 1 = 7 โต dp[target] = 7 Time Complexity..
Description: Solution: ex) s = "leetcode" wordDict = ["leet", "code"] In s, we start from the end of the string. dp[8] = True do[7] = False dp[6] = dp[5] = False dp[4] = True = dp[4 + 4] dp[3] = dp[2] = dp[1] = False dp[0] = True = dp[0 + 4] If dp[0] is True, it means we can break the string with wordDict!
Descrtiption: Solution: We are gonna start from the last value of array. ex) nums = [1 ,2 ,4, 3] LIS[3] = 1 LIS[2] = max(1, 1 + LIS[3]) = 1 (nums[2] > nums[3]) LIS[1] = max(1, 1 + LIS[2], 1 + LIS[3]) = 2 LIS[0] = max(1, 1+1, 1+1, 1+2) = 3 So, the answer is 3. Time Complexity: O(n^2)
Description: Solution: We have two list: start & end. ex) start: [0, 5, 15] end: [10, 20, 30] We shift the start pointer when start[s] is less than end[e] => count += 1 We shift the end pointer when start[s] is equal or more thn end[e] => count -= 1 Time Complexity: O(nlogn) Space Complexity: O(n)
Description: Solution: Note that the range should begin from index 1. Time Complexity: O(nlogn)
Description: Solution: In this problem, we are gonna compare the end of the previous interval to the start of the current interval. If the new start point is bigger than prevEnd, we don't count. Else, it is overlapping, so that we count them as 1. And update prevEnd. Since, new intervals are still overlapping, we should use the min(end, prevEnd). Time Complexity: O(nlogn)
