Minimum-Cost Color Sequence

For each day i, costs[i][0], costs[i][1], and costs[i][2] are the costs of choosing blue, green, or red.
Choose exactly one color per day, and never choose the same color on consecutive days. Return the unique minimum-cost color sequence as a string using b, g, and r.

Function
minimumCostColorSequence(costs: int[][]) → String

Examples
Example 1
costs = [[1,5,9],[4,2,8],[7,6,1]]
return = "bgr"
Choosing blue, green, then red costs 1 + 2 + 1 = 4, which is the unique minimum.
Example 2
costs = [[7,2,5]]
return = "g"
With one day, the least expensive choice is green.

Constraints
1 <= costs.length <= 100000.
costs[i].length == 3.
1 <= costs[i][j] <= 1000000.
The minimum-cost valid sequence is unique.

The trick is dynamic programming with three states per day. Let dp[i][c] be the minimum cost to finish day i ending on color c. Then dp[i][c] = costs[i][c] + min(dp[i-1][other two colors]). You only need the previous row, so space is O(1) for cost. But you must return the sequence, so store a parent pointer per day and color, which is O(n) memory with n up to 100000. Pick the best final color, then walk the parents backward and reverse. The common pitfall is greedy, and the second is forgetting to map indices correctly: 0 is b, 1 is g, 2 is r. Skip recursion here, since depth 100000 will blow the stack. The uniqueness guarantee means no tie-break logic is needed. If you freeze on the reconstruction, StealthCoder is the hedge during the live OA.

StealthCoder is the hedge for the one pattern you didn't drill. It runs invisibly during the screen share.