How to find largest and smallest number from integer array - Java Solution

Hello guys, if you have gone through any coding interview or have done some professional Software development then you know that a good understanding of array data structure is crucial for any software developer but it doesn't come for free, you need to spend time and effort. The best way to develop this understanding by solving coding problems and there are lots of programming exercises beginners can do. One of them is writing a program to find the smallest and largest number in an integer array. Java programmers are no different than others, so they can do this program in Java, not just to understand array but also different relational operators available in Java.  

In this program, you need to write a method, yes we call the function a method in Java, which will accept an integer array and then print the largest and smallest number from that array. Use of any third-party library or API method is not allowed, which means you need to do this exercise by using essential tools of Java programming language, which includes operators, control statements, keywords, and some classes from java.lang package.

This problem is also known as finding maximum and minimum numbers in an array, and the technique mentioned here can be used in any other programming language as well. As a bonus point, you can also write JUnit test cases to test your method, I have not done so and relied on a simple main method to test my code to show the output and keep it short, essential for any example or demo.

Btw, if you preparing for a programming job interview, then let me repeat that a good knowledge of essential data structures like an array, linked list, binary tree, the hash table goes a long way in doing well on the interview. You should spend some time learning those and filling gaps in your understanding.





Java Program to find the smallest and largest number in an integer array 

Here is a full code example of a Java program to find the smallest and largest number from an integer array. You can create a Java source file with the name MaximumMinimumArrayDemo.java and copy code there to compile and execute it in your favorite IDE. 

If you don't have an IDE setup, you can also compile and run this program by following the steps I have shown on HelloWorld in Java.

If you look at the code here, we have created a method called the largestAndSmallest(int[] numbers)  to print the largest and smallest number from the int array passed to the program.  We have used two variables largest and smallest, to store the maximum and minimum values from the array.

Initially, the largest is initialized with Integer.MIN_VALUE and smallest are initialized with Integer.MAX_VALUE.In each iteration of the loop, we compare the current number with the largest and smallest and update them accordingly.

Since if a number is larger than the largest, it can't be smaller than the smallest, which means you don't need to check if the first condition is true, that's why we have used if-else code block, where else part will only execute if the first condition is not true.

Here is another logical diagram or flow chart to find the largest element from an array in Java, here instead of assigning the variable with Integer.MAX_VALUE, we have assigned the first element from the array.

How to find maximum and minimum number from integer array Java



Since the array doesn't override the toString method in Java, we have used Arrays.toString() to print the contents of an array. Remember, this function is outside of core logic, so it's ok to use it. Since this is a static method, we can directly call this from the main method in Java, and so does our test code.

We pass the random array to this method and see if the largest and smallest number returned by the method is correct or not. For automated testing, a Unit test is better, but for a demonstration, you can use the main method.

Btw, if you preparing for a programming job interview, then don't forget to check the Cracking the Coding Interview book. It contains 189 Programming Questions and Solutions, which is more than enough for many coding interviews.


Java Program to find the largest and smallest element in an array:

Here is the Java program I am talking about. This shows you how to find the maximum and minimum number in a given array in Java, without using any library method. 
import java.util.Arrays;
/**
 * Java program to find largest and smallest number from an array in Java.
 * You cannot use any library method both from Java and third-party library.
 *
 * @author http://java67.com
 */
public class MaximumMinimumArrayDemo{

    public static void main(String args[]) {
        largestAndSmallest(new int[]{-20, 34, 21, -87, 92,
                             Integer.MAX_VALUE});
        largestAndSmallest(new int[]{10, Integer.MIN_VALUE, -2});
        largestAndSmallest(new int[]{Integer.MAX_VALUE, 40,
                             Integer.MAX_VALUE});
        largestAndSmallest(new int[]{1, -1, 0});
    }

    public static void largestAndSmallest(int[] numbers) {
        int largest = Integer.MIN_VALUE;
        int smallest = Integer.MAX_VALUE;
        for (int number : numbers) {
            if (number > largest) {
                largest = number;
            } else if (number < smallest) {
                smallest = number;
            }
        }

        System.out.println("Given integer array : " + Arrays.toString(numbers));
        System.out.println("Largest number in array is : " + largest);
        System.out.println("Smallest number in array is : " + smallest);
    }
}
Output:
Given integer array : [-20, 34, 21, -87, 92, 2147483647]
Largest number in array is : 2147483647
Smallest number in array is : -87
Given integer array : [10, -2147483648, -2]
Largest number in array is : 10
Smallest number in array is : -2147483648
Given integer array : [2147483647, 40, 2147483647]
Largest number in array is : 2147483647
Smallest number in array is : 40
Given integer array : [1, -1, 0]
Largest number in array is : 1
Smallest number in array is : -1


That's all about How to find the largest and smallest number from integer array in Java. As I said, this question can also be asked to find the maximum and minimum numbers in an Array in Java, so don't get confused there. By the way, there are more ways to do the same task, and you can practice it to code solutions differently. Can you write a solution that is different than this? Go ahead and give it a try.

Some more programs for coding interviews:

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  • Top 10 Programming problems from Java Interviews? (article)

P. S. - If you are preparing for a programming job interview, then you must prepare for an all-important topic like data structure, String, array, etc. One book which can help you with this task is the Grokking the Coding Interview: Patterns for Coding Questions course on . It contains popular coding interview patterns which will help you to solve most of the problems in your coding interviews.

48 comments:

  1. Replies
    1. package Sankey;

      public class smalestno
      {public static void main(String[] args) {


      int a[]={-12,-1,-13,22,54,65,4,7,9,5,765,765567};
      int n=0;
      n=a.length-1;
      //System.out.println(n);
      for(int i=0;ia[j])
      {
      int temp;
      temp=a[i];
      a[i]=a[j];
      a[j]=temp;

      }

      }

      }
      for(int i=0;i<a.length;i++)
      {
      System.out.println(a[i]);
      }
      System.out.println("smallest " +a[0]);
      System.out.println("largest "+a[n]);

      }
      }

      Delete
    2. this can easily done with python.

      import array as arr
      x=arr.array("i",[2,3,4,5,543,556])
      print(max(x))
      print(min(x))

      output:
      556
      2

      Delete
  2. This code doesn't look right, it should be changed to (remove "else"):
    if (number > largest) {
    largest = number;
    }

    if (number < smallest) {
    smallest = number;
    }

    If you try {1, 2, 3}, you will see the difference.

    ReplyDelete
    Replies
    1. Agreed. I ran the code example and encountered the incorrect result. It should be updated to have two separate if statements just as shown above.

      Delete
    2. Why not just set largest/smallest to array[0] instead of INT_MAX, INT_MIN? Then it'll work with the `else`. Setting to INT_MAX/INT_MIN is a rather rookie mistake.

      Delete
    3. when smallest is the first element if does not give the right answer so there should be two if statements in place of if else if.

      Delete
  3. Remove "else" because it fails at both places.
    1. for largest try ascending order digit (1,2,3)
    2. for smallest try descending order digit in negative(-3,-2,-1)

    ReplyDelete
    Replies
    1. Yes, the else looks like a typo, it should be removed otherwise solution will not produce correct result for all outputs.

      Delete
  4. //programm to find largest no in an given array.
    public class Larg {

    public static void main(String[] args) {
    int max=0;
    int arr[]={900,2,54,15,40,100,20,011,299,30,499,699,66,77};
    max=arr[0];
    for(int i=0;i<arr.length-1;i++)
    {
    if(max<arr[i+1])
    {
    max=arr[i+1];
    }
    }
    System.out.println(max);
    }
    }

    ReplyDelete
  5. Can we convert into array into string without API or without Arrays.to String(arr)

    ReplyDelete
  6. Can we convert int array to string without using Arrays.to string()

    ReplyDelete
  7. private static void largestAndSmallest(int[] numbers) {
    int largest = numbers[0];
    int smallest = numbers[0];

    for (int i = 1; i < numbers.length; i++) {
    if(numbers[i]>largest){
    largest=numbers[i];
    }else if(numbers[i]<smallest){
    smallest=numbers[i];
    }
    }

    System.out.println("\nGiven integer array : " + Arrays.toString(numbers));
    System.out.println("Largest number in array is : " + largest);
    System.out.println("Smallest number in array is : " + smallest);
    }

    ReplyDelete

  8. import java.util.Scanner;

    class exam
    {
    public static void main(String[] args)
    {
    Scanner input = new Scanner(System.in);

    System.out.println("Enter 5 numbers:");

    int[] numbers = new int[5];
    int sum = 0;
    int max = numbers[0];
    int min = numbers[0];

    for(int i=0; i<5; i++){
    numbers[i] = input.nextInt();
    sum = sum + numbers[i];

    if (numbers[i] > max){
    max = numbers[i];
    }
    else if (numbers[i] < min){
    min = numbers[i];
    }

    }

    int average = sum / 5;

    System.out.println("Sum: " + sum);
    System.out.println("Average: " + average);
    System.out.println("Max: " + max);
    System.out.println("Min: " + min );
    System.out.println("Display sorted data : " + numbers[0] );
    }
    }

    How come the min is always displaying 0...Please can someone help me...Thanks in advance

    ReplyDelete
    Replies
    1. Else in this snippet "else if (numbers[i] < min){" is the culprit

      Delete
  9. public void doAlgorithm(int a[]){
    int big = 0, temp = 0;
    for (int i = 0; i < a.length; i++) {
    for (int j = i+1; j < a.length; j++) {
    big = (a[i] > a[j]) ? a[i] : a[j];
    if(temp < big)
    temp = big;
    }
    }
    System.out.println(temp);
    }

    ReplyDelete
  10. laughable

    why all this hassle with Integer.MAX_VALUE and Integer.MIN_VALUE?
    Simply make your largest and smallest values equal to the numbers[0]. Then you can save one iteration.

    What if the numbers is empty? You will still get Integer.MIN_VALUE and Integer.MAX_VALUE which obviously would be incorrect. You need to check for empty and null array.
    A programmer should learn to check his inputs way earlier, than he would learn algorithms. It's a matter of simple hygiene - learn to brush your teeth before learning how to assemble fusion reactor.

    ReplyDelete
  11. int[] arr = {1,3,2,6,3};
    Arrays.sort(arr);
    int min = arr[0];
    int max = arr[arr.length-1];
    System.out.println(min + ":" + max);

    ReplyDelete
  12. Why not you correct code in this post..please replace else if with if condition..

    ReplyDelete
  13. int a[]= {10,3,0,1,45,22,667,8,99,0,3};
    int min = a[0];
    int max = a[1];
    int minPos = 0;
    int maxPos = 1;
    if(maxa[i]) {
    min = a[i];
    minPos = i;
    }
    if(max<a[i]) {
    max = a[i];
    maxPos = i;
    }

    }
    System.out.println("min : "+ min +" Pos : "+(minPos+1));
    System.out.println("max : "+ max +" Pos : "+(maxPos+1));

    ReplyDelete
  14. import java.util.Arrays;
    import java.util.Collections;
    import java.util.List;

    import org.apache.commons.lang3.ArrayUtils;

    public class Test{

    int []arr = {-1 , 3,2, -9, 5 ,6 ,-4};
    System.out.println("Largest is "+findLargest(arr));
    System.out.println("Smallest is "+findSmallest(arr));
    }
    private static int findSmallest(int[] arr) {
    List b = Arrays.asList(ArrayUtils.toObject(arr));


    return (int) Collections.min(b);
    }
    public static int findLargest(int [] arr){
    List b = Arrays.asList(ArrayUtils.toObject(arr));

    return (int) Collections.max(b);
    }
    }

    ReplyDelete
  15. public static void main (String []args){

    largestAndSmallest(new Integer[]{-20, 34, 21, -87, 92,
    10,100,-120});

    }
    public static void largestAndSmallest(Integer[] numbers)
    {
    int largest = 0;
    int smallest = 0;
    int temp = 0;

    /*
    //// simple using builtin function for sorting array in ascending order ////
    Arrays.sort(numbers);
    System.out.println("In asscending order : " + Arrays.toString(numbers));

    //// simple using builtin function for sorting array in descending order ////
    Arrays.sort(numbers, Collections.reverseOrder());
    System.out.println("In descending order : " + Arrays.toString(numbers));

    */

    for(int i=0; i numbers[j]){
    // temp = numbers[i];
    // numbers[i]=numbers[j];
    // numbers[j]=temp;
    // }
    // largest = numbers[0];
    // smallest = numbers[j];
    }
    }

    System.out.println("Given integer array : " + Arrays.toString(numbers));
    System.out.println("Largest number in array is : " + largest);
    System.out.println("Smallest number in array is : " + smallest);

    }

    ReplyDelete
  16. import java.util.BitSet;

    public class SmallestLargest {
    public static void main(String[] args) {
    int array[] = new int[] {4,64,3,67,12};
    BitSet bitSet = new BitSet();
    for(int i = 0; i < array.length; i++) {
    bitSet.set(array[i] - 1);
    }
    System.out.println("Smallest Number is " + (bitSet.nextSetBit(0) + 1));
    System.out.println("Largest Number is: " + bitSet.length());
    }
    }

    ReplyDelete
  17. you can just replace the max and the min with the first element and it works.
    just use arr[0] in both places

    ReplyDelete
  18. public class FindMaxSmall {


    public static void main(String[] args) {
    // TODO Auto-generated method stub

    int[] arrayValue = { 2, 3, 1, 4, 7, 8, 5, -1, -2, 99, 2, 2, 99 };

    int maximum = arrayValue[0];
    int minimum = arrayValue[0];

    for (int i=1;i maximum) {
    maximum = arrayValue[i];
    }

    if (arrayValue[i] < minimum) {

    minimum = arrayValue[i];
    }
    }

    System.out.println("Maximum Value" + maximum);

    System.out.println("Minimum Value" + minimum);

    }

    }

    ReplyDelete
  19. public class MyClass {
    public static void main(String args[]) {
    int[] array = {2,1,3,4,5,6,7,8,10,9};

    int min=array[0];
    int max=0;
    for(int i=0;iarray[i])
    {
    min=array[i];
    }else if(max<array[i]){
    max=array[i];
    }

    }
    System.out.println("Min Number = " + min);
    System.out.println("Max Number = " + max);

    }
    }

    ReplyDelete
  20. lol
    int [] array = {3,2,5,1,6};
    Arrays.sort(array);
    int min = array[0];
    int max = array[array.length - 1];
    System.out.println("min = " + min + " max = " + max);

    ReplyDelete
  21. static void maxMin(int[] arr){

    int min = arr[0];
    int max = arr[0];

    for(int i = 1; i < arr.length;i++){
    if(max < arr[i]){
    max = arr[i];
    }
    if(min > arr[i]){
    min = arr[i];
    }
    }
    System.out.println(String.format("Max = %s, Min = %s", max, min));
    }

    ReplyDelete
  22. Using Binary method:

    private void minAndMax(int[] intArray) {
    int middle = intArray.length / 2;
    int k = intArray.length - 1;
    int minVal = Integer.MIN_VALUE;
    int maxVal = Integer.MAX_VALUE;

    for (int i = 0; i < middle; i++) {

    if(intArray[i] >= minVal){
    minVal = intArray[i];
    }else if(intArray[i] < maxVal){
    maxVal = intArray[i];
    }


    if(intArray[k] >= minVal){
    minVal = intArray[k];
    }else if(intArray[k] < maxVal){
    maxVal = intArray[k];
    }
    k--;
    }

    System.out.println("minVal -->"+minVal);
    System.out.println("maxVal -->"+maxVal);
    }

    ReplyDelete
  23. private void minAndMax(int[] intArray) {

    int middle = intArray.length / 2;
    int k = intArray.length - 1;
    int minVal = Integer.MIN_VALUE;
    int maxVal = Integer.MAX_VALUE;

    for (int i = 0; i < middle; i++) {

    if(intArray[i] >= minVal){
    minVal = intArray[i];
    }else if(intArray[i] < maxVal){
    maxVal = intArray[i];
    }


    if(intArray[k] >= minVal){
    minVal = intArray[k];
    }else if(intArray[k] < maxVal){
    maxVal = intArray[k];
    }
    k--;
    }

    System.out.println("minVal -->"+minVal);
    System.out.println("maxVal -->"+maxVal);
    }

    ReplyDelete
  24. public static void main(String args[]) {
    int[] arr = { 5, 2, 3, 41, -95, 530, 6, 42, -361, 81, 8, 19, 90 };
    int smallest = arr[0];
    int largest = arr[0];

    for (int i = 0; i < arr.length; i++) {
    if (arr[i] > largest) {
    largest = arr[i];
    }
    }
    for (int i = 1; i < arr.length; i++) {

    if (smallest >= arr[i]) {
    smallest = arr[i];

    }
    }

    System.out.println(largest);
    System.out.println(smallest);
    }

    ReplyDelete

  25. var a = [100, 500, 1000, 5000, 350000, 100000, 200000, 15, 20, 30, 25, 2];
    var b = [];
    function largestNumber(a){
    for(let i=0; i<= a.length-2 ;i++){
    // console.log(a[i])

    if(i==0){
    b.push(a[i])
    }

    if (i > 0){
    if(b[0] <= a[i+1]){

    b.pop()
    b.push(a[i+1])
    }
    }
    else if(b[0] <= a[i+1]){
    b.pop()
    b.push(a[i+1])
    }
    console.log(b +" is bigger number than " + a[i+1])
    }
    }
    largestNumber(a)
    console.log(b)

    ReplyDelete
  26. Why not to use mergesort?.. Also,Merge sort time complexity is O(nlogn).. After merge sort, access first and last elements as smallest and largest elements

    ReplyDelete
    Replies
    1. IF sorting is allowed then yes you can use either quicksort or mergesort, but if sorting is not allowed then you need to write a different logic. Also, sorting is overkill because you don't need to sort whole array, you just need to find largest and smallest.

      Delete
  27. int min = 0;
    int max = 0;

    int arr[] = {3,2,6,9,1};
    for (int i = 0; i < arr.length;i++){
    min = arr[0];
    if (arr[i] <= min){
    min = arr[i];
    }else if(arr[i] >= max){
    max = arr[i];
    }
    }
    System.out.println(min + " " + max);

    ReplyDelete
  28. int arr[] = {90000000,25145,6221,90000,3213211};
    int min = arr[0];
    int max = arr[0];
    for (int i = 0; i < arr.length;i++){
    if (min >= arr[i]){
    min = arr[i];
    }
    if(arr[i] >= max){
    max = arr[i];
    }
    }
    System.out.println(min + " " + max);

    ReplyDelete
  29. int[] arrays = { 100, 1, 3, 4, 5, 6, 7, 8, 9, 2, 1 };
    Arrays.sort(arrays);
    System.out.println("Minimum value in Arrays : " + arrays[0]);
    System.out.println("Miximum value in Arrays : " + arrays[arrays.length - 1]);

    ReplyDelete
  30. I've found the minimum, but how can I square it?

    ReplyDelete
  31. import java.util.HashMap;
    import java.util.LinkedList;
    import java.util.List;
    import java.util.Map;

    public class maxMinimumArray {

    public static void main(String[] args) {

    int[] values = {-20, 34, 21, -87, 92};

    int[] sortedArr = sortValues(values);

    Map results = maxMinArr(sortedArr);

    for(Map.Entry entry : results.entrySet()) {
    System.out.println(entry.getKey() + " => " + entry.getValue());
    }

    }

    public static int[] sortValues(int[] arr) {
    // sort in asc first (any sort algo will do depending on the complexity you want
    // going with bubble sort
    for (int i = 0; i < arr.length; i++) {
    for (int j = 1; j < arr.length; j++) {
    if (arr[j - 1] > arr[j]) {
    int temp = arr[j - 1];
    arr[j - 1] = arr[j];
    arr[j] = temp;
    }
    }
    }
    return arr;
    }

    public static Map maxMinArr(int[] arr){
    Map result = new HashMap<>();

    result.put("MinimumValue", arr[0]);
    result.put("MaximumValue", arr[arr.length - 1]);

    return result;
    }

    }

    ReplyDelete
  32. public static void findLargestAndSmallestNumberInUnsortedIntArray (int [] unsortedInputArray) {
    int largest = unsortedInputArray[0];
    int smallest = unsortedInputArray[0];
    for(int number : unsortedInputArray) {
    if(largestnumber) {
    smallest=number;
    }
    }
    System.out.println("smallest : "+smallest);
    System.out.println("largest : "+largest);
    }

    ReplyDelete
  33. int low = Integer.MAX_VALUE; // Initialize with a large value
    int max = 0;

    for (int i = 0; i < arr.length; i++) {
    if (arr[i] > max) {
    max = arr[i];
    }

    if (arr[i] < low) {
    low = arr[i];
    }
    }

    System.out.println(max + " " + low);

    ReplyDelete
  34. int low = Integer.MAX_VALUE; // Initialize with a large value
    int max = 0;

    for (int i = 0; i < arr.length; i++) {
    if (arr[i] > max) {
    max = arr[i];
    }

    if (arr[i] < low) {
    low = arr[i];
    }
    }

    System.out.println(max + " " + low);

    ReplyDelete

Feel free to comment, ask questions if you have any doubt.