Find smallest integer value in array list in Java without Arrays.sort

Question

How can I find the smallest value in a int array without changing the array order?

code snippet:

    int[] tenIntArray = new int [10];
    int i, userIn;

    Scanner KyBdIn = new Scanner(System.in);
    System.out.println("Please enter 10 integer numbers ");

    for(i = 0; i < tenIntArray.length; i++){
        System.out.println("Please enter integer " + i);
        userIn = KyBdIn.nextInt();
        tenIntArray[i] = userIn;
    }

I am not sure how I can find the smallest array value in the tenIntArray and display the position

For example the array holds - [50, 8, 2, 3, 1, 9, 8, 7 ,54, 10]

The output should say "The smallest value is 1 at position 5 in array"

ouch. Index counting starts at 0, not 1. As a condition you should use i < tenIntArray.length and not !=, just to be safe when you start using doubles or floats. — extraneon
– extraneon, Commented Dec 14, 2013 at 16:51
@extraneon It's not "just to be safe", it's the more logical, generally accepted way of doing things. — Bernhard Barker
– Bernhard Barker, Commented Dec 14, 2013 at 17:08

Alexis C. · Accepted Answer · 2014-05-11 14:01:22Z

16

This figure should be helpful :

enter image description here

Then to answer your question, what would you do on paper ?

Create and initialize the min value at tenIntArray[0]
Create a variable to hold the index of the min value in the array and initialize it to 0 (because we said in 1. to initialize the min at tenIntArray[0])
Loop through the elements of your array
If you find an element inferior than the current min, update the minimum value with this element and update the index with the corresponding index of this element
You're done

Writing the algorithm should be straightforward now.

edited May 11, 2014 at 14:01

answered Dec 14, 2013 at 16:53

Alexis C.

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Comments

TruVortex_07 · Accepted Answer · 2018-08-01 00:30:10Z

4

Try this:

//Let arr be your array of integers
if (arr.length == 0)
    return;
int small = arr[0];
int index = 0;
for (int i = 0; i < arr.length; i++) {
    if (arr[i] < small) {
        small = arr[i];
        index = i;
    }
}

edited Aug 1, 2018 at 0:30

TruVortex_07

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answered Dec 14, 2013 at 18:57

user3102604

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J.R · Accepted Answer · 2017-04-11 07:36:37Z

2

Using Java 8 Streams you can create a Binary operator which compares two integers and returns smallest among them.

Let arr is your array

int[] arr = new int[]{54,234,1,45,14,54};
int small = Arrays.stream(arr).reduce((x, y) -> x < y ? x : y).getAsInt();

answered Apr 11, 2017 at 7:36

J.R

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2 Comments

Srikanth R Over a year ago

May I know the time complexity of this approach?

J.R Over a year ago

Will be O(n) I guess

akhil_mittal · Accepted Answer · 2015-05-11 09:37:11Z

The method I am proposing will find both min and max.

public static void main(String[] args) {
    findMinMax(new int[] {10,40,50,20,69,37});
}
public static void findMinMax(int[] array) {
    if (array == null || array.length < 1)
        return;
    int min = array[0];
    int max = array[0];

    for (int i = 1; i <= array.length - 1; i++) {
        if (max < array[i]) {
            max = array[i];
        }

        if (min > array[i]) {
            min = array[i];
        }
    }
    System.out.println("min: " + min + "\nmax: " + max);
}

Obviously this is not going to one of the most optimized solution but it will work for you. It uses simple comparison to track min and max values. Output is:

min: 10 max: 69

Dark Army · Accepted Answer · 2015-07-30 10:22:38Z

0

    int[] input = {12,9,33,14,5,4};

    int max = 0;
    int index = 0;
    int indexOne = 0; 
    int min = input[0];
    for(int i = 0;i<input.length;i++)
    {
        if(max<input[i])
        {
            max = input[i];
            indexOne = i;
        }

        if(min>input[i])
        {
            min = input[i];
            index = i;
        }

    }
    System.out.println(max);
    System.out.println(indexOne);
    System.out.println(min);
    System.out.println(index);

answered Jul 30, 2015 at 10:22

Dark Army

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Comments

SAM · Accepted Answer · 2016-08-31 16:15:39Z

0

Here is the function

public int getIndexOfMin(ArrayList<Integer> arr){
    int minVal = arr.get(0); // take first as minVal
    int indexOfMin = -1; //returns -1 if all elements are equal
    for (int i = 0; i < arr.size(); i++) {
        //if current is less then minVal
        if(arr.get(i) < minVal ){
            minVal = arr.get(i); // put it in minVal
            indexOfMin = i; // put index of current min
        }
    }
    return indexOfMin;  
}

answered Aug 31, 2016 at 16:15

SAM

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Comments

user3101937 · Accepted Answer · 2013-12-16 11:11:23Z

-2

the first index of a array is zero. not one.

for(i = 0; i < tenIntArray.length; i++)

so correct this. the code that you asked is :

    int small = Integer.MAX_VALUE;
    int i = 0;
    int index = 0;
    for(int j : tenIntArray){
        if(j < small){
            small = j;
            i++;
            index = i;
        }
    }
    System.out.print("The smallest value is"+small+"at position"+ index +"in array");

edited Dec 16, 2013 at 11:11

answered Dec 14, 2013 at 17:14

user3101937

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4 Comments

user3101937 Over a year ago

Yes I did. can you tell me about the fault?

Alexis C. Over a year ago

It's find the smallest but not the correct index. For example with the input int [] tenIntArray = {1,-8,2,2,0,-15};, you will increment the index variable 3 times. But that won't give you the correct index of the minimum value in the array. You have to keep track of the current index, and update another variable when you have found a smallest value (like you did for small).

user3101937 Over a year ago

ah I got that.thanks.if I start with int index = -1; , is that correct?

Jens Kooij Over a year ago

You should increase i outside of the if statement, because even if the current index is not the smallest, it should increase the i for keeping track of the current index.

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