It's possible to do this using just a one-dimensional array. Since we only need the last column of the matrix we calculated to calculate the next column, the ar

It's possible to do this using just a one-dimensional array. Since we only need the last column of the matrix we calculated to calculate the next column, the array can be updated in-place. coins = [1, 2, 5, 10, 20, 50, 100, 200] total = 200 matrix = [0] * (total + 1) matrix[0] = 1 for coin in coins: for j in range(coin, len(matrix)): matrix[j] += matrix[j - coin]