Step 1: Use the Gell-Mann-Nishijima relation instead of adding quark charges one by one.
$Q = I_3 + \frac{Y}{2}$, where $Y = B + S$ is the hypercharge, $B$ is baryon number, and $I_3$ is the third component of isospin. This gives a second, independent way to pin down $Q$ and $S$ for each particle from its known isospin and baryon number, as a check on the quark-counting method.
Step 2: Apply it to $\Sigma^{*-}$.
The $\Sigma^*$ triplet has isospin 1, with $\Sigma^{*-}$ sitting at $I_3 = -1$; it is a baryon so $B=1$, and its strangeness is $S=-1$ (one strange quark). So $Y = 1 + (-1) = 0$, giving $Q = -1 + 0 = -1$. Then $Q - S = -1 - (-1) = 0$, TRUE.
Step 3: Apply it to $K^+$.
$K^+$ belongs to an isospin doublet with $I_3 = +\frac{1}{2}$, it is a meson so $B=0$, and its strangeness is $S=+1$. So $Y = 0 + 1 = 1$, giving $Q = \frac{1}{2} + \frac{1}{2} = 1$. Then $Q - S = 1 - 1 = 0$, TRUE.
Step 4: Apply it to $\Omega^-$.
$\Omega^-$ is an isospin singlet, $I_3 = 0$, baryon so $B=1$, strangeness $S=-3$. So $Y = 1 - 3 = -2$, giving $Q = 0 + (-1) = -1$. Then $Q - S = -1 - (-3) = 2 \ne 0$, FALSE.
Step 5: Apply it to $\Delta^{++}$.
$\Delta^{++}$ sits at $I_3 = +\frac{3}{2}$ in the isospin-3/2 quartet, baryon so $B=1$, and it carries no strangeness, $S=0$. So $Y = 1 + 0 = 1$, giving $Q = \frac{3}{2} + \frac{1}{2} = 2$. Then $Q - S = 2 - 0 = 2 \ne 0$, FALSE.
Final Answer:
The isospin-based route gives exactly the same two survivors as direct quark counting.\[ \boxed{\Sigma^{*-}\text{ and } K^+} \]