About "(...) the minima equation for double slit and single slit are opposite of each other, despite both are built on a similar concept." no one has answered yet.
The answer to this conundrum lies in the fact that two-slit interference we consider only the very edges of a slit to be wavelet sources separated by $D$ and in one-slit interference we consider infinite number of sources in between the edges of the slit (including those extreme points). Their average distance to a point on the screen is equal to the distance between the slit centre and that point on the screen.
If you want the reasoning behind this, read on. What you'll find below is heavily based on Crawford's "Waves".
So in the far field limit, the pattern for two sources is defined by their relative phase difference (optical path difference)
$$\Delta\varphi=k(r_2-r_1)=kD\sin(\theta)=\frac{2\pi}{\lambda}D\sin(\theta)$$ Two electric fields superpose giving their sum:
$$E(r,\theta,t)=A(r)\cos(\omega t+\varphi_1)+A(r)\cos(\omega t+\varphi_2)\\=A(r,\theta)\cos(\omega t+\varphi_{avg}-kr)$$
where $A(r,\theta)=2A(r)\cos(\omega t+\frac{1}{2}(\varphi_2-\varphi_1)+\frac{1}{2}\Delta\varphi)$.
Having that irradiance is $I(t)=n\varepsilon_0c|E(t)|^2$, and that the electric field amplitude averaged over time
$$\left<|E(t)|^2\right>=\frac{1}{2}A^2(r,\theta)$$
this gives averaged intensity
$$I(\theta)=I_{max}\cos^2\left(\frac{1}{2}(\varphi_2-\varphi_1)+\frac{1}{2}\Delta\varphi\right)$$
When the two sources are initially in phase this simplifies to
$$I(\theta)=I_{max}\cos^2\left(\frac{1}{2}\Delta\varphi\right)=I_{max}\cos^2\left(\frac{\pi}{\lambda}D\sin(\theta)\right)$$
From periodicity of $\cos^2$ function we get bright fringes at $\frac{m\pi}{\lambda}D\sin(\theta)=m\pi$ or $D\sin(\theta)=m\lambda$ for $m\in(0,\pm1,\pm2,...)$.
Now for diffraction on single slit. A slit of size $D$ can be divided into $N$ point sources (separated by $d$) of wavelets every one superposing with another to give average electric field (as in interference between two point sources) - so this will become generalized case of the previous one.
$$E=\sum_{j=1}^{N} A(r)\cos(kr_j-\omega t)$$
or in terms of complex amplitude:
$$E_c=A(r)e^{-i\omega t}(\sum_{j=1}^{N} e^{ikr_j})$$
Having all $N$ elements emitting rays at an angle $\theta$ (which is approximately valid in the far field limit), the $N$-th element travels distance $r_N=r_1+(N-1)d\sin(\theta)$. And so the complex amplitude becomes
$$E_c=A(r)e^{-i\omega t}e^{ikr_1}S$$
where $S=\sum_{j=1}^{N}e^{i\Delta\varphi(j-1)}$. Because $$S(e^{i\Delta\varphi}-1)=e^{i\Delta\varphi N}-1$$
dividing each side by $e^{i\Delta\varphi}-1$ leads to
$$S=\frac{e^{i\Delta\varphi N}-1}{e^{i\Delta\varphi}-1}$$
Multiplication and division by $e^{-i\Delta\varphi/2}$ gives
$$S=\frac{e^{i\Delta\varphi\left(N-1/2\right)}-e^{-i\Delta\varphi/2}}{e^{-i\Delta\varphi/2}\cdot\left(e^{i\Delta\varphi}-1\right)}=e^{i\Delta\varphi\left(N-1\right)/2}\cdot\frac{e^{i\Delta\varphi N/2}-e^{-i\Delta\varphi N/2}}{e^{i\Delta\varphi/2}-e^{-i\Delta\varphi/2}}$$
Since
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$S$ becomes equal to:
$$e^{i\frac{1}{2}(N-1)\Delta\varphi}\frac{\sin(\frac{1}{2}N\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}$$
and complex amplitude equals
$$E_c=A(r)e^{-i\omega t}e^{ikr}\frac{\sin(\frac{1}{2}N\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}$$
where $r = r_1+\frac{1}{2}(N-1)d\sin(\theta) = r_1+\frac{1}{2}D\sin(\theta)$ is the distance between the centre of the slit and a point on the screen at an angle $\theta$ created between the ray coming from the slit's centre and the line perpendicular to the screen.
Taking now only the real part of the complex amplitude we obtain
$$E(r,\theta,t)=\left[A(r)\frac{\sin(\frac{1}{2}N\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}\right]\cos(kr-\omega t)=A(r,\theta)\cos(kr-\omega t)$$
For $N=2$ as in the two-source problem of Young's experiment, this becomes
$$E(r,\theta,t)=\left[A(r)\frac{\sin(\Delta\varphi)}{\sin(\frac{1}{2}\Delta\varphi)}\right]\cos(kr-\omega t)=\left[2A(r)\cos\left(\frac{1}{2}\Delta\varphi\right)\right]\cos(kr-\omega t)$$
This is on par with the previous result where we assumed initial in-phase relation of the source wavelets.
Furthermore, from consideration of $N\rightarrow\infty$ we then have that the total phase difference across the slit is $\Phi=kD\sin(\theta)=(N-1)\Delta\varphi\approx N\Delta\varphi$.
Taking only the first term of $\frac{\sin(\Delta\varphi N/2)}{\sin(\Delta\varphi/2)}$ expanded at $N=\infty$ and having that $\Phi\rightarrow 0$ when $\theta\rightarrow 0$, electric field becomes:
$$E(r,\theta,t)=\left[A(r,0)\frac{\sin(\frac{1}{2}\Phi)}{\frac{1}{2}\Phi}\right]\cos(kr-\omega t)$$
And so the intensity profile on a screen:
$$I(\theta)=I_{max}\frac{\sin^2(\frac{1}{2}\Phi)}{(\frac{1}{2}\Phi)^2}$$
The dark fringes appear every time $\frac{1}{2}\Phi=m\pi=\frac{1}{2}\frac{2\pi}{\lambda}D\sin(\theta)$ or $m\lambda=D\sin(\theta)$, where $m\in(\pm1,\pm2,\pm3...)$.
So, the result is expected. The only difference between these two experiments is in fact spatial distribution. In reality these two phenomena coexist and there is no pure diffraction-free interference pattern.
