{"post":"

Java Program to Find Maximum and Minimum Numbers in an Array

Problem Statement

Write a Java program that accepts N integer elements from the user, stores them in an array, and finds both the maximum and minimum values in the array.

Algorithm

1. Read the size of the array N from the user

2. Create an integer array of size N

3. Read N integers from the user and store them in the array

4. Initialize two variables:

- max_element with the first array element

- min_element with the first array element

5. Iterate through the array starting from the second element:

- Update max_element if current element is larger

- Update min_element if current element is smaller

6. Display both maximum and minimum values

Java Implementation

import java.util.Scanner;\n\npublic class ArrayMinMax {\n    public static void main(String[] args) {\n        Scanner scanner = new Scanner(System.in);\n        \n        // Read array size\n        System.out.print(\"Enter the size of the array: \");\n        int n = scanner.nextInt();\n        \n        // Input validation\n        if (n <= 0) {\n            System.out.println(\"Array size must be a positive integer.\");\n            return;\n        }\n        \n        int[] numbers = new int[n];\n        \n        // Read array elements\n        System.out.println(\"Enter \" + n + \" integers:\");\n        for (int i = 0; i < n; i++) {\n            System.out.print(\"Enter element \" + (i + 1) + \": \");\n            numbers[i] = scanner.nextInt();\n        }\n        \n        // Initialize min and max with first element\n        int maxElement = numbers[0];\n        int minElement = numbers[0];\n        \n        // Find min and max\n        for (int i = 1; i < n; i++) {\n            if (numbers[i] > maxElement) {\n                maxElement = numbers[i];\n            }\n            if (numbers[i] < minElement) {\n                minElement = numbers[i];\n            }\n        }\n        \n        // Display results\n        System.out.println(\"\\nResults:\");\n        System.out.println(\"Maximum number: \" + maxElement);\n        System.out.println(\"Minimum number: \" + minElement);\n        \n        scanner.close();\n    }\n}\n\n

Sample Output

Enter the size of the array: 6\nEnter 6 integers:\nEnter element 1: 12\nEnter element 2: 36\nEnter element 3: 5\nEnter element 4: 41\nEnter element 5: 20\nEnter element 6: 36\n\nResults:\nMaximum number: 41\nMinimum number: 5\n

Time Complexity

\n
Time: O(n) - Single pass through the array
\n
\n
Space: O(n) - Storage for the array elements
\n

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