< Back 746. Min Cost Climbing Stairs Quentessential dynamic programming problem. We're asked to find the minimum number of steps to reach the top, where cost[i] is the cost to take a certain step. We can start from step 0 or step 1. We're allowed to make two choices during our ascent, picking to go up one stair or 2 stairs for each step. We intuit that we can start from index 2 since the 0th and 1st index are already completed for us. Starting from index 2, we look backwards one step and two steps, selecting the minimum cost. Making this optimal choice at each step in the cost[] array, and reusing already computing information, is central to our dynamic programming solution. Eventually, we reach the top, with cost 0, and cost[n - 1], with n being the size of the array, contains the minimum cost to reach the top of the stairs. The solution is as follows: class Solution: def minCostClimbingStairs(self, cost: List[int]) -> int: cost.append(0) n = len(cost) for i in range(2, n): cost[i] += min(cost[i - 1], cost[i - 2]) return cost[n - 1] _ Time Complexity: O(n) - Where n is the length of cost[], we iterate over the entire array. _ Space Complexity: O(1) - We append a single integer, 0, to the input array and use constant space to store the length of the input array. All operations are conducted in O(1) space on the input array.