Climbing Stairs: A Dynamic Programming Approach

The "Climbing Stairs" problem is a classic example used to illustrate dynamic programming in algorithmic problem-solving. It involves finding the number of distinct ways to climb a staircase with a given number of steps.

Problem Statement

Given a staircase with n steps, where each time you can either climb 1 or 2 steps, the task is to determine the total number of distinct ways to reach the top of the staircase.

Examples

• Input: n = 2 Output: 2 Explanation: There are two ways to climb to the top:

  1. 1 step + 1 step
  2. 2 steps

• Input: n = 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

Dynamic Programming Solution

Here's how you can solve this problem using dynamic programming in JavaScript:

function climbStairs(n) {
  if (n === 1) return 1;
  let dp = [1, 2];
  for (let i = 2; i < n; i++) {
    dp[i] = dp[i - 1] + dp[i - 2];
  }
  return dp[n - 1];
}

Breaking Down the Solution

• Base Cases: If n is 1, the answer is straightforwardly 1. The dynamic programming array dp is initialized with its base cases: dp[0] = 1 and dp[1] = 2. These represent the number of ways to climb a staircase with one step and two steps, respectively.

• Filling the dp Array: For each step from 2 to n - 1, the function calculates the number of ways to reach that step. The value of dp[i] is determined as the sum of dp[i - 1] and dp[i - 2]. This represents the total ways to climb to the current step, either by taking one step from the previous step or two steps from the step before that.

• Returning the Result: The final result is stored in dp[n - 1], which gives the total number of distinct ways to reach the top of the staircase. This value is returned as the solution to the problem.

Conclusion

The "Climbing Stairs" problem is a classic example that showcases the efficacy of dynamic programming in solving computational problems related to combinations and counting. It emphasizes the importance of breaking down the problem into smaller, overlapping subproblems and building up the solution by storing and reusing intermediate results. This approach not only makes the solution more efficient but also simplifies the process of solving complex problems.