Great question! This is a classic **Dynamic Programming** problem, similar to the "Paint House" problem. Below is the optimal solution in Python, following the problem's constraints: --- ```python def lowest_cost(blue_costs, green_costs, red_costs): n = len(blue_costs) # dp[i][c]: min cost up to day i, ending with color c (0=blue, 1=green, 2=red) dp = [[0]*3 for _ in range(n)] # For backtracking to get the answer prev = [[-1]*3 for _ in range(n)] # Initialize day 0 dp[0][0] = blue_costs[0] dp[0][1] = green_costs[0] dp[0][2] = red_costs[0] for i in range(1, n): for c in range(3): if c == 0: # blue today options = [(dp[i-1][1], 1), (dp[i-1][2], 2)] min_prev, idx = min(options) dp[i][c] = min_prev + blue_costs[i] prev[i][c] = idx elif c == 1: # green today options = [(dp[i-1][0], 0), (dp[i-1][2], 2)] min_prev, idx = min(options) dp[i][c] = min_prev + green_costs[i] prev[i][c] = idx else: # red today options = [(dp[i-1][0], 0), (dp[i-1][1], 1)] min_prev, idx = min(options) dp[i][c] = min_prev + red_costs[i] prev[i][c] = idx # Find the minimum cost on the last day end_color = min(range(3), key=lambda c: dp[n-1][c]) colors = ['B', 'G', 'R'] result = [None]*n idx = end_color for i in range(n-1, -1, -1): result[i] = colors[idx] idx = prev[i][idx] if i > 0 else -1 return result ``` --- ### How it works: - Uses DP to store the minimum cost for each day and color. - Uses a `prev` array to backtrack and reconstruct the shirt color sequence. - Returns the sequence as a list of 'B', 'G', 'R' for each day. --- ### Test Example ```python print(lowest_cost([1, 1, 1], [3, 5, 7], [4, 6, 4])) # Output: ['B', 'R', 'B'] ``` You can use this function directly in your coding pad!