Given a list of costs, find the minimum total cost needed to cross the array or reach the end of the array.

Explanation:

• the movement rules in this problem use 1-based indexing
• index 1 means the first element of the input list
• from any visited index, you may move exactly two indices forward or exactly one index backward
• if you land on an index, you must add that index cost to your total
• each index may be visited at most once
• if costs = List.of(2, 5, 8), then:
  • index 1 has cost 2
  • index 2 has cost 5
  • index 3 has cost 8

You begin already standing on index 1, so the cost at index 1 is included in the total.

You succeed if:

• you land on index n, where n is the number of elements, or
• you move to any index greater than n, which means you have crossed the array

Moving to an index less than 1 is not allowed.

Return the minimum possible total cost.

Method Signature

int minimumCost(List<Integer> costs)

Parameters

• costs is a list of positive integers
• costs.get(0) represents the cost at index 1
• costs.get(i) represents the cost at index i + 1

Return Value

• return the minimum total cost needed to reach index n or cross beyond it

Rules

• you start at index 1
• starting on index 1 adds its cost immediately
• from index i, you may move only to:
  • i + 2
  • i - 1
• you may visit any valid array index at most one time
• if you move to an index greater than n, no additional cost is added for that move
• all costs are positive, so revisiting indices would never help and is explicitly forbidden

Suggested Constraints

• 1 ≤ costs.size() ≤ 100,000
• 1 ≤ costs.get(i) ≤ 10,000

Notes

• because the problem starts at index 1, the input list must contain at least one value
• comparison is based on the actual minimum total cost
• you must never use a path that visits the same valid index more than once

Examples

Example 1

Example 2

Example 3

Example 4

Example 5