How do you solve Minimum Cost to Reach Every Position on LeetCode?

Minimum Cost to Reach Every Position (LeetCode #3502) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Swapping with lower cost positions reduces future costs.

What is the time complexity of Minimum Cost to Reach Every Position?

The optimal solution for Minimum Cost to Reach Every Position runs in O(n) time and uses O(n) space. We make a single pass through the array, updating the minimum cost efficiently.

What is the brute force solution for Minimum Cost to Reach Every Position?

The brute force approach for Minimum Cost to Reach Every Position is Brute Force, with O(n²) time complexity and O(1) space complexity. We can calculate the cost to reach each position by checking all previous positions. This involves summing the costs for each possible path to each position. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Cost to Reach Every Position?

Minimum Cost to Reach Every Position uses the following concepts: Array. The recommended patterns to study are: Prefix Sum, Dynamic Programming.

What are the interview tips for Minimum Cost to Reach Every Position?

Explain your thought process clearly. Consider edge cases, like empty arrays. Discuss time and space complexities.

What are common mistakes when solving Minimum Cost to Reach Every Position?

Not updating the minimum cost correctly. Overlooking that swaps with people behind are free.

#3502

Minimum Cost to Reach Every Position

Easy

ArrayPrefix SumDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

By using a prefix minimum array, we can efficiently track the minimum cost to reach each position without redundant calculations.

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Algorithm

3 steps

1Step 1: Initialize an answer array and a min prefix array to store the minimum cost seen so far.
2Step 2: Iterate through the cost array, updating the prefix minimum and filling the answer array with the minimum cost for each position.
3Step 3: Return the answer array.

solution.py8 lines

1def minCost(cost):
2    n = len(cost)
3    answer = [0] * n
4    min_cost = float('inf')
5    for i in range(n):
6        min_cost = min(min_cost, cost[i])
7        answer[i] = min_cost
8    return answer

ℹ

Complexity note: We make a single pass through the array, updating the minimum cost efficiently.

1Swapping with lower cost positions reduces future costs.
2Using a prefix minimum helps avoid redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.