Question: I'm trying to find an efficient way to compute maximum of a signal using its DFT. More formally:

$$\max\left\{ \mathcal F^{-1}\left(X\_k\right)\right\}, X\_k\text{ is the DFT of the signal and } \mathcal F^{-1} \text{ is the IDFT}$$

The original signal $x(n)$ is real, and it has some noise when sometimes there are wide peaks.

I'm looking for a solution that is quicker than the actual IFFT since the signal is very long (but we have its DFT).

Answer: 
This sounds like a good candidate for the Sparse Fast Fourier Transform (sFFT). Have a look at H. Hassanieh, P. Indyk, D. Katabi, and E. Price, **[sFFT: Sparse Fast Fourier Transform](http://groups.csail.mit.edu/netmit/sFFT/)**, 2012. I don't know details of the algorithm. You have the domains swapped, which should be fine as the discrete Fourier transform (DFT) and its inverse (IDFT) are nearly identical.

[sFFT has been discussed](https://dsp.stackexchange.com/questions/2317/what-is-the-sparse-fourier-transform) also here on dsp.stackexchange.com.
Generally:

There can be no shortcut here – every single element of the (inverse or forward, doesn't matter) DFT needs every element of the input. So, the FFT is already pretty much as fast as you can go.

However, you say:

> 
> The original signal $x(n)$ is real, and it has some noise when sometimes there are wide peaks.
> 
> 
> 

Aha! That means that your $X$ is (hermitian) symmetric. Which means you might be able to pick an FFT algorithm that benefits from that knowledge and can omit quite a few calculation – but I don't think that'll change the general $\mathcal O(N \log N)$ complexity of the IFFT, nor the $\mathcal O(N)$ of finding the maximum.