LeetCode 3502: Minimum Cost to Reach Every Position Solution

Master LeetCode problem 3502 (Minimum Cost to Reach Every Position), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3502. Minimum Cost to Reach Every Position

Easy

Array

Problem Explanation

Explanation:

To solve this problem, we can iterate through the cost array from the end to the beginning. At each position, we calculate the minimum cost to reach that position by considering two possibilities:

Swapping with the person at the current position.
Not swapping with the person at the current position.

We maintain an array answer to store the minimum total cost to reach each position in the line.

Algorithm:

Initialize an array answer of size n, where n is the length of the cost array.
Set answer[n-1] = cost[n-1] as the cost to reach the last position is equal to the cost at that position.
Iterate from n-2 to 0:
- Calculate answer[i] as the minimum of cost[i] and answer[i+1] (cost at current position and cost at next position).
Return the answer array as the result.

Time Complexity:

The time complexity of this solution is O(n), where n is the length of the cost array.

Space Complexity:

The space complexity of this solution is O(n) for the answer array.

: :

Solution Code

class Solution {
    public int[] minCostToReachEveryPosition(int[] cost) {
        int n = cost.length;
        int[] answer = new int[n];
        answer[n-1] = cost[n-1];
        
        for (int i = n-2; i >= 0; i--) {
            answer[i] = Math.min(cost[i], answer[i+1]);
        }
        
        return answer;
    }
}

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