Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. For example, given the following triangle [ [2], [3,4], [6,5,7], [4,1,8,3] ] The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). Note: Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. ...
Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n. Example 1: Input: n = 12 Output: 3 Explanation: 12 = 4 + 4 + 4. Example 2: Input: n = 13 Output: 2 Explanation: 13 = 4 + 9. Solution class Solution { public int numSquares(int n) { int[] dp = new int[n + 1]; dp[0] = 0; dp[1] = 1; for (int i = 2; i < dp.length; i++) { dp[i] = Integer.MAX_VALUE; for (int j = 1; j * j <= i; j++) { dp[i] = Math.min(dp[i], dp[i - j * j] + 1); } } return dp[n]; } } Solution 2021-11-30 class Solution { public int numSquares(int n) { int[] dp = new int[n + 1]; Arrays.fill(dp, Integer.MAX_VALUE); dp[0] = 0; int max_square_index = (int) Math.sqrt(n) + 1; int[] squares = new int[max_square_index]; for (int i = 1; i < squares.length; i++) { squares[i] = i * i; } for (int i = 1; i < dp.length; i++) { for (int j = max_square_index - 1; j > 0; j--) { if (i - squares[j] >= 0) { int at = i - squares[j]; dp[i] = Math.min(dp[i], dp[at] + 1); } } } return dp[n]; } }
You are climbing a stair case. It takes n steps to reach to the top. Each time you can either climb 1 or 2 steps. In how many distinct ways can you climb to the top? Note: Given n will be a positive integer. Example 1: Input: 2 Output: 2 Explanation: There are two ways to climb to the top. 1. 1 step + 1 step 2. 2 steps Example 2: Input: 3 Output: 3 Explanation: There are three ways to climb to the top. 1. 1 step + 1 step + 1 step 2. 1 step + 2 steps 3. 2 steps + 1 step Recursive Solution: class Solution { public int climbStairs(int n) { return climbStairs(0, n); } int climbStairs(int i, int n) { if (i > n) return 0; if (i == n) return 1; return climbStairs(i + 1, n) + climbStairs(i + 2, n); } } Recursive Solution with Memo class Solution { public int climbStairs(int n) { int[] memo = new int[n + 1]; return climbStairs(0, n, memo); } int climbStairs(int i, int n, int[] memo) { if (i > n) return 0; if (i == n) return 1; if (memo[i] > 0) { return memo[i]; } return memo[i] = climbStairs(i + 1, n, memo) + climbStairs(i + 2, n, memo); } }
A robot is located at the top-left corner of a m x n grid (marked ‘Start’ in the diagram below). The robot can only move either down or right at any point in time. The robot is trying to reach the bottom-right corner of the grid (marked ‘Finish’ in the diagram below). How many possible unique paths are there? Above is a 7 x 3 grid. How many possible unique paths are there? ...
You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin. Example 1: Input: amount = 5, coins = [1, 2, 5] Output: 4 Explanation: there are four ways to make up the amount: 5=5 5=2+2+1 5=2+1+1+1 5=1+1+1+1+1 Example 2: Input: amount = 3, coins = [2] Output: 0 Explanation: the amount of 3 cannot be made up just with coins of 2. Example 3: Input: amount = 10, coins = [10] Output: 1 Note: ...