Min Cost Climbing Stairs LeetCode Solution

Difficulty Level Easy
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Problem Statement

Min Cost Climbing Stairs LeetCode Solution – An integer array cost  is given, where cost[i] is the cost of ith step on a staircase. Once you pay the cost, you can either climb one or two steps.

You can either start from the step with index 0, or the step with index. 1. Return the minimum cost to reach the top of the floor.

Example 1:

Input:

 cost = [10,15,20]

Output:

 15

Explanation:

 You will start at index 1.
- Pay 15 and climb two steps to reach the top.
The total cost is 15.

Example 2:

Input:

 cost = [1,100,1,1,1,100,1,1,100,1]

Output:

 6

Explanation:

 You will start at index 0.
- Pay 1 and climb two steps to reach index 2.
- Pay 1 and climb two steps to reach index 4.
- Pay 1 and climb two steps to reach index 6.
- Pay 1 and climb one step to reach index 7.
- Pay 1 and climb two steps to reach index 9.
- Pay 1 and climb one step to reach the top.
The total cost is 6.

 

Constraints:

  • 2 <= cost.length <= 1000
  • 0 <= cost[i] <= 999

Approach

1. The Main idea of this problem is to use Dynamic Programming to find minimum cost climbing stairs.

2. We will discuss recursion with memoization as it is beginner-friendly.

3. Now, rec(i) is the minimum cost to the ith step. we can take one step or two-step from index i.

4. Now, recurrence will be rec(i)=cost[i]+min(rec(i+1),rec(i+2))

5. Base condition will be rec(i)=0 if i>=n where n is the length of the array or our estimated path because we reach the destination so no further movement is possible.

 

Code:

C++ Leetcode Solution:

class Solution {
public:
    #define ll int
    ll dp[1005];
    ll n;
    ll rec(ll i,vector<ll>&a)
    {
        if(i>=n)
            return 0;
        ll &ans=dp[i];
        if(ans!=-1)
            return ans;
        ans=a[i]+min(rec(i+2,a),rec(i+1,a));
       
        return ans;

    }
    int minCostClimbingStairs(vector<int>& cost) {
        memset(dp,-1,sizeof(dp));
         n=cost.size();
        
        return min(rec(1,cost),rec(0,cost));
    }
};

Java Leetcode Solution:

class Solution {
    int rec(int i,int n,int [] a,int[] dp)
    {
        
        if(i>=n)
            return 0;
        
        if(dp[i]!=-1)
            return dp[i];
        dp[i]=a[i]+Math.min(rec(i+2,n,a,dp),rec(i+1,n,a,dp));
       
        return (int)dp[i];
            
    }
    public int minCostClimbingStairs(int[] cost) {
        int n=cost.length;
        int[] dp = new int[n+11];
        Arrays.fill(dp, -1); 
        int a=rec(0,n,cost,dp);
        int b=rec(1,n,cost,dp);
        return (int)(Math.min(a,b));
    }
}

Complexity Analysis of Min Cost Climbing Stairs Leetcode Solution:

Time Complexity

Time complexity will be o(n). Because we are traversing the whole array.

 Space Complexity

Space complexity will be o(n). We are creating dp table of length n to store the value in every index.

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