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Friday, October 13, 2023

LeetCode Daily Challenges: 746. Min Cost Climbing Stairs

CP , LeedCode
LeetCode Daily Challenges: 746. Min Cost Climbing Stairs

746. Min Cost Climbing Stairs

You are given an integer array cost 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.


Intuition:

The problem is about finding the minimum cost to reach the top of the staircase by either climbing one or two steps at a time. To solve this, we can use dynamic programming to keep track of the minimum cost to reach each step.


Approach:

  1. Initialize an array dp of the same length as the cost array to store the minimum cost for each step.
  2. Initialize dp[0] and dp[1] with the costs of the first two steps, as you can start from either step 0 or step 1.
  3. Iterate through the array starting from step 2 and for each step i, update dp[i] as the minimum of the cost to reach step i-1 plus the cost of step i and the cost to reach step i-2 plus the cost of step i. This ensures that you choose the minimum cost path to reach the current step.
  4. After iterating through all the steps, the minimum cost to reach the top of the staircase is the minimum of the last two elements in the dp array.

Complexity:

Time complexity: The algorithm iterates through the cost array once, so the time complexity is O(n), where n is the number of steps in the staircase.

Space complexity: We use an additional array dp of the same length as the cost array, so the space complexity is O(n) to store the dynamic programming table. 


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Thursday, October 12, 2023

LeetCode Daily Challenges: 557. Reverse Words in a String III

CP , LeedCode
LeetCode Daily Challenges: 557. Reverse Words in a String III

 557. Reverse Words in a String III



Problem Description:

Given a string s, reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

Input: s = "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"
Example 2:

Input: s = "God Ding"
Output: "doG gniD"


Intuition

The problem requires us to reverse the order of characters in each word within a sentence while preserving whitespace and the initial word order. To achieve this, we can split the sentence into words, reverse each word individually, and then join them back together with whitespace to form the reversed sentence.

Approach

  1. Split the input sentence into words: We can split the sentence into words using whitespace as the delimiter, which will give us an array of words.
  2. Reverse each word: For each word in the array, we reverse its characters. We can use a StringBuilder to efficiently reverse the characters of a word.
  3. Join the reversed words: Finally, we join the reversed words back together with whitespace to form the reversed sentence.


Complexity

  1. Time complexity: The time complexity is determined by the length of the input sentence and the number of words it contains. Let's denote the length of the input sentence as n and the average word length as k. Splitting the sentence takes O(n) time, and reversing each word takes O(k) time, so the overall time complexity is O(n * k).
  2. Space complexity: The space complexity depends on the storage required for the array of words and the reversed sentence. Since we are not using any additional data structures apart from the input and output, the space complexity is O(n), where n is the length of the input sentence.

Code Snapshots:













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