Algorithm with time complexity o(n^2)

In summary, the conversation discusses finding an algorithm with time complexity $o(n^2)$ to return two elements from a set $M$ such that their absolute difference is the minimum among all possible pairs. One method suggested is to use heapsort to sort the values in $M$ and then find the desired elements in one pass, resulting in a total time complexity of $n^2$. The correctness of the algorithm could potentially be proven using induction on $i$.
  • #1
evinda
Gold Member
MHB
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Hi! (Smile)

I am looking at the following exercise:

Let $M=\{ y_1, y_2, \dots, y_n \}$ a set of real numbers, where $n \geq 2$. Describe an algorithm, that has time complexity $o(n^2)$ and that finds and returns two elements $y_k$ and $y_l$ of $M$, such that:

$$|y_k-y_l|=\min_{1 \leq i,j \leq n} |y_i-y_j|$$I think that we cannot do this with the use of two while loops, because the time complexity will be $\leq cn^2$, but we want it to be $<cn^2$, or am I wrong? (Thinking)
How else could we do this? (Worried)
 
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  • #2
First sort the values (with say heap sort) of time complexity $n\ln\,n$. Then in the sorted list find the desired elements in one pass of the n elements. Total complexity "little oh" $n^2$.
 
  • #3
johng said:
First sort the values (with say heap sort) of time complexity $n\ln\,n$. Then in the sorted list find the desired elements in one pass of the n elements. Total complexity "little oh" $n^2$.

So, is it right like that? (Thinking)

Code:
Heapsort(A){
  Buildheap(A);
  for (i=size(A); i>1; --i){
       swap(A[i],A[1]);
       heap_size(A)=heap_size(A)-1;
       Heapify(A,1);
   }
}
 
Elements(S,n){
  int i,min,k;
  Heapsort(S)
  min=abs(S[1]-S[2]);
  for (i=2; i<n; i++){ 
      if (abs(S[i]-S[i+1])<min)
          min=abs(S[i]-S[i+1]);
          k=i;
  }
  return S[k],S[k+1];
}
 
  • #4
Looks right to me.
 
  • #5
How could we show the correctness of the algorithm?
Could we prove that it is correct, using induction on [m] i [/m] ?
 

Related to Algorithm with time complexity o(n^2)

1. What does an algorithm with time complexity o(n^2) mean?

An algorithm with time complexity o(n^2) means that the time it takes to run the algorithm increases exponentially as the input size increases. This can be visualized as a parabola on a graph, where the time taken to run the algorithm increases rapidly as the input size increases.

2. How can I determine the time complexity of an algorithm?

To determine the time complexity of an algorithm, you can analyze the number of operations or steps it takes to complete for different input sizes. The general rule is that the time complexity is o(n^2) if there are two nested loops, o(n) if there is one loop, and o(1) if there are no loops.

3. What is the significance of having a time complexity of o(n^2)?

An algorithm with a time complexity of o(n^2) can be considered inefficient because it takes a significantly longer time to run as the input size increases. This means that for larger input sizes, the algorithm will take a lot longer to complete compared to algorithms with a lower time complexity.

4. How can I improve the time complexity of an algorithm with o(n^2)?

The time complexity of an algorithm with o(n^2) can be improved by using more efficient algorithms or data structures. For example, using a hashing algorithm or a binary search tree can significantly reduce the time complexity compared to using a simple linear search algorithm.

5. Can an algorithm with a time complexity of o(n^2) ever be faster than an algorithm with a lower time complexity?

No, an algorithm with a time complexity of o(n^2) will always be slower than an algorithm with a lower time complexity for larger input sizes. However, for very small input sizes, the difference in performance may not be significant.

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