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Examples where the uniqueness of the solution makes it easier to find
The complexity class $\mathsf{UP}$ consists of those $\mathsf{NP}$problems that can be decided by a polynomial time nondeterministic Turing machine which has at most one accepting computational path. That is, the solution, if any, is unique in this sense.
It is thought highly unlikely that all $\mathsf{UP}$problems are in $\mathsf{P}$, because by the ValiantVazirani Theorem this would imply the collapse $\mathsf{NP}=\mathsf{RP}$.
On the other hand, no $\mathsf{UP}$problem is known to be $\mathsf{NP}$complete, which suggests that the unique solution requirement still somehow makes them easier.
I am looking for examples, where the uniqueness assumption leads to a faster algorithm.
For example, looking at graph problems, can a maximum clique in a graph be found faster (though possibly still in exponential time), if we know that the graph has a unique maximum clique? How about unique $k$colorability, unique Hamiltonian path, unique minimum dominating set etc.?

3SAT may be one such problem. Currently the best upper bound for Unique 3SAT is exponentially faster than for general 3SAT. (The speedup is exponential, although the reduction in the exponent is tiny.) The recordholder for the unique case is this paper by Timon Hertli.
Hertli's algorithm builds upon the important PPSZ algorithm of Paturi, Pudlák, Saks, and Zane for $k$SAT, which I believe is still the fastest for $k \geq 5$ (see also this encyclopedia article). The original analysis showed better bounds for Unique $k$SAT than for general $k$SAT when $k = 3, 4$; subsequently, however, Hertli showed in a different paper that you could get the same bounds for (a slightly tweaked) PPSZ algorithm, without assuming uniqueness. So, maybe uniqueness helps, and it can definitely simplify the analysis of some algorithms, but our understanding of the role of uniqueness for $k$SAT is still growing.
There is evidence that Unique $k$SAT is not too much easier than general $k$SAT.
20180620 19:20:54 
Shortest 2Vertex disjoint path problem in undirected graphs recently solved (ICALP14) by A. Bjorklund and T. Husfeldt. But the deterministic solution is for the case of existence of a unique solution. In the case that there are more than one solution, they showed that the problem belongs to RP. As authors of the paper mentioned, it is not known if the problem is in P in general scenario.
20180620 19:56:33 
Mentioning another result by Björklund, if you are guaranteed that there is at most one Hamiltonian cycle in a graph, you can decide if a graph $G$ is Hamiltonian faster than you can in general.
The uniqeness assumption means that the parity of the number of Ham. paths is the same as deciding if the graph is Hamiltonian.
Björklund's method deterministically computes the parity of the number of Hamiltonian cycles in $O(1.619^n)$ while the best known randomized algorithm for Undirected Hamiltonicity runs in $O^*(1.657^n)$ , and the best deterministic algorithm for Directed Hamiltonicity (to the best of my knowledge) is still the 50 years old $O(n^22^n)$ dynamic programming algorithm by Bellman, Held and Karp.
20180620 20:06:03 
Outside of complexity theory and the analysis of algorithms, the assumption that there can be only one solution forms the basis for some of the standard rules used to deduce the solution in Sudoku puzzles. These rules generally involve looking for ways in which parts of the puzzle might be able to have two or more solutions that don't interact with the rest of the puzzle. That can't happen in the actual solution, so if a pattern that threatens to cause this is found, then it must be broken, allowing the solver to deduce constraints on what the actual solution can look like. See http://www.brainbashers.com/sudokuuniquerectangles.asp for some examples of deduction rules based on uniqueness.
20180620 20:24:31