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Can we sort using a single loop?

 
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Hi All,

I want to know if this is possible:" Writing a program that sorts an array by using only a single loop."; I dont want the solution but I wanted to know if this is really feasible.

Thanks,
Pavan.
 
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where 'x' is the integer array name, 'n' is the number of elements in 'x'.
This for loop sorts the elements of the array in ascending order. To sort in descending order, change "if ( x[ i ] < x[ i + 1 ] )" to "if ( x[ i ] > x[ i + 1 ] )"

I have no way of testing all of these, but I'm pretty sure they'll work.
 
Pavan Kumar Dittakavi
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I think I have got it.


@anirudh jagithyala: The code you provided fails for input 5 1 7 3 9. Thanks anyway.

Thanks,
-Pavan.
 
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Pavan Kumar Dittakavi wrote:Hi All,

I want to know if this is possible:" Writing a program that sorts an array by using only a single loop."; I dont want the solution but I wanted to know if this is really feasible.

Thanks,
Pavan.



Short answer: Any algorithm can be reworked as a single loop.

For this problems, you can implement what amounts to a bubble sort by using two indices (say i and j) and just have all the initialization, incrementing and termination expressions in a handful of if/then/else statements at the bottom of the single loop. It would still execute in O(n^2) time, but it would technically be a single loop.
 
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Hi All,

I think we can sort the following way. Not sure about the complexity.

public class SingleLoopSorting {
public static void main(String[] args) {
int arr[] = { 5, 1, 7, 3, 9 };
for (int i = 1; i < arr.length; i++) {
if (arr[i] < arr[i - 1]) {
arr[i] = arr[i] + arr[i - 1];
arr[i - 1] = arr[i] - arr[i - 1];
arr[i] = arr[i] - arr[i - 1];
i = 0;
}
}
System.out.print("Sorted Array : ");
for (int i = 0; i < arr.length; i++) {
System.out.print(arr[i] + " ");
}
}
}
 
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Hi Abhishek,

Welcome to the Ranch!

I imagine the OP from 3 years ago has moved on, but thank you for contributing.
In the future, please wrap your code using Code Tags (← click here if you are not sure how).

Code Tags will make your code much easier for everyone to read, as demonstrated here:

Cheers!
Chris
 
Abhishek K Sharma
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Hey Chris,
Sure I will do that. I posted the reply thinking that it might be helpful to some other person.
 
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