Dynamic Programming Questions Medium
Dynamic Programming can be used to solve the Climbing Stairs problem by breaking it down into smaller subproblems and storing the solutions to these subproblems in a table or array.
To solve the Climbing Stairs problem, we can define a recursive function, let's call it climbStairs(n), which represents the number of distinct ways to climb to the top of a staircase with n steps.
The base cases for this recursive function are:
- If n is 0, there is only one way to climb the stairs (by not taking any steps).
- If n is 1, there is only one way to climb the stairs (by taking one step).
For any other value of n, we can calculate the number of distinct ways to climb the stairs by considering the two possible choices at each step:
- If we take one step at the current position, we need to calculate the number of distinct ways to climb the remaining n-1 steps.
- If we take two steps at the current position, we need to calculate the number of distinct ways to climb the remaining n-2 steps.
Therefore, the recursive formula for climbStairs(n) can be defined as:
climbStairs(n) = climbStairs(n-1) + climbStairs(n-2)
To avoid redundant calculations, we can use dynamic programming to store the solutions to the subproblems in an array. We start by initializing the array with the base cases (climbStairs(0) = 1 and climbStairs(1) = 1). Then, we iterate from 2 to n, calculating the number of distinct ways to climb the stairs at each step using the recursive formula and storing the result in the array.
Finally, we return the value of climbStairs(n) from the array, which represents the number of distinct ways to climb to the top of the staircase with n steps.