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So ... I have : int array[] = {-8,2,0,5,-3,6,0,9};

I want to find the smallest positive number ( which in the list above is 2 )

This is what i am doing :

int array[] = {-8,2,0,5,-3,6,0,9}; int smallest=0; for(int i=0;i<array.length;i++) // Find the first number in array>0 (as initial // value for int smallest) { if(array[i]>0) { smallest=array[i]; break;// Break out of loop, when you find the first number >0 } } for(int i=0;i<array.length;i++) // Loop to find the smallest number in array[] { if(smallest>array[i]&&array[i]>0) { smallest=array[i]; } } System.out.println(smallest); }

My question is :

Can we reduce the number of steps ? Is there any smarter/shorter way of doing it, with no other data structure.

Thanks.

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3
  • You can sort the array using Arrays.sort and then take the first or last item as you need. Commented Aug 26, 2014 at 7:27
  • That will give -8, not 2, in the example. Commented Aug 26, 2014 at 7:28
  • stackoverflow.com/questions/1484347/… Commented Aug 26, 2014 at 7:31

11 Answers 11

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7

is there any smarter/shorter way of doing it?

If you want shorter, with Java 8, you can use a stream of ints:

int min = Arrays.stream(array).filter(i -> i >= 0).min().orElse(0);

(assuming you are happy with a min of 0 when the array is empty).

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7 Comments

so much overhead for a so simple task
How do you handle the only positive numebrs limiation without adding a filter, which practically means another iteration?
@assylias Is it optimized to overcome the slow-down of multiple iterations?
@amit there will only be one iteration.
@amit it is by specification: "Laziness-seeking. Many stream operations, such as filtering [...] can be implemented lazily, exposing opportunities for optimization. For example, "find the first String with three consecutive vowels" need not examine all the input strings. Stream operations are divided into intermediate (Stream-producing) operations and terminal (value- or side-effect-producing) operations. Intermediate operations are always lazy.". See also stackoverflow.com/a/21223645/829571.
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You do not need to have smallest=array[i], just initialize a variable with INTEGER.MAX_VALUE or array[0] and iterate over the array comparing the value with this variable.

This is achieved in O(n) time and O(1) space and thats the best you can get! :)

a simpler way would be

 int[] array ={-1, 2, 1}; boolean max_val_present = false; int min = Integer.MAX_VALUE; for(int i=0;i<array.length;i++) // Loop to find the smallest number in array[] { if(min > array[i] && array[i] > 0) min=array[i]; //Edge Case, if all numbers are negative and a MAX value is present if(array[i] == Integer.MAX_VALUE) max_val_present = true; } if(min == Integer.MAX_VALUE && !max_val_present) //no positive value found and Integer.MAX_VALUE //is also not present, return -1 as indicator return -1; return min; //return min positive if -1 is not returned
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12 Comments

You still return an INTEGER.MAX_VALUE when all items are non-positive.
@SeyfülislamÖzdemir : good point, updated code..thanks fella! :)
Now you return -1 when the input is an array which contains only non-positive numbers and at least one INTEGER.MAX_VALUE. You should return an INTEGER.MAX_VALUE for such a case. :)
@SeyfülislamÖzdemir : Edge cases.. :) updated code, validate now! :)
Why min <= array[i]? If min == array[i] there is no point of assign min=array[i]. I would change it with min < array[i].
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Without any prioe knowledge there is no way to avoid iterating the entire array.

You can however make sure you iterate it only once by removing the first loop, and instead just assign smallest = Integer.MAX_VALUE. You can also add a boolean that indicates the array was changed to distinguish between cases where there is no positive integer, and cases where the only positive integer is Integer.MAX_VALUE

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4 Comments

50sec behind...i am getting old!! :)
Wondering who downvoted this answer. Please provide a comment why this answer is wrong.
not really sure who downvoted you..but may be lack of code might be reason (i too hate those kind of visitors) :)
@NoobEditor I really prefer words over code in many situations, as it lets the OP understand the answer, and implement it by his own. I hope this is not the issue :|
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You don't need two loops for this purpose.

int smallest = Integer.MAX_VALUE; for(int i=0;i<array.length;i++) { if(array[i] > 0 && smallest > array[i]) { smallest = array[i]; } }

The only problem with this code is that after the loop you can't know that whether all elements are non-positive or at least one of them is Integer.MAX_INT and remaining is non-positive. You should add controls if you think that such a case is possible.

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A sample for you:

 int array[] = {-8,2,0,5,-3,6,0,9}; int minPosNum = Integer.MAX_VALUE; for (int i = 0; i < array.length; i++) { if(array[i] > 0 && array[i] < minPosNum){ minPosNum = array[i]; } } System.out.println(minPosNum);
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int array[] = {-8,2,0,5,-3,6,0,9}; int minPos = Integer.MAX_VALUE; for (int number : array) { if (number > 0) minPos = Math.min(number, minPos); }
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0

You can try this to do it with 1 loop:

int array[] = {8,2,0,5,-3,6,0,9}; int smallestPositive = array[0]; for(int i = 1; i < array.length; i++){ if(smallestPositive > 0){ if(array[i] < smallestPositive && array[i] > 0) smallestPositive = array[i]; }else smallestPositive = array[i]; }

System.out.println(smallestPositive); will print 2

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0

You can try the following code

import java.util.*; public class SmallItem { public static void main(String args[]) { int array[] = {-8,2,0,5,-3,6,0,9}; int number = Integer.MAX_VALUE; ArrayList al = new ArrayList(); //Here we add all positive items into a ArrayList al for(int i=1;i<array.length;i++){ if(array[i]>0){ al.add(new Integer(array[i])); } } Iterator it = al.iterator(); while(it.hasNext()){ int n = ((Integer)it.next()).intValue(); if(n<number){ number = n; } } System.out.println("Smallest number : " + number); } }
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For C++

int smallest (int array[]){ int a; for(int i=0;i<n;i++){ if(array[i]>0){ a=array[i]; break; } } for (int i=0; i<n; i++){ if(array[i]<=a && array[i]>0){ a=array[i]; } } return a; }
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Please add a line of explanation for the code you have provided
0 
namespace ConsoleApplication3 { class Program { static void Main(string[] args) { int[] array = { -1, 2, 1, -2, 11, 23, 44 }; int min = MinArray(array); Console.WriteLine(min.ToString() + " is the minimum number!"); Console.ReadKey(); } public static int MinArray(int[] array) { int minimumNumber = Int32.MaxValue; foreach (int i in array) { if (i < minimumNumber && i > 0) { minimumNumber = i; } } return minimumNumber; } } }
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In JavaScript

var inputArray = [-1,2,1,5,-20]; var arrayWithOnlyPositiveValue = []; for(var i=0;i<inputArray.length;i++){ if(inputArray[i]>=0){ arrayWithOnlyPositiveValue.push(inputArray[i]) } } var min_of_array = Math.min.apply(Math, arrayWithOnlyPositiveValue); console.log('---smallestPositiveValue----',min_of_array);
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1 Comment

you are effectively looping twice

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