Array Programs in Java | Beginner to Expert Level
Array programs in Java traverse from basic single-dimensional arrays to complex multi-dimensional arrays and dynamic arrays using ArrayList. From initializing and accessing array elements to advanced operations like sorting and searching, arrays facilitate efficient data management and manipulation, forming a fundamental concept for every aspiring Java developer. Let's understand more!
An Array Program typically refers to a program that utilizes array data structures that can hold more than one value at a time. Arrays are used in programming to organize a series of items sequentially.
Read more: Arrays in Java.
This blog will help you learn Array Programs in Java from Beginner (Level 1) to Expert (Level 4).
Table of Content
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Level 1 (Beginners)
Here, we will see some very basic array programs!
Question 1: Creating and Initializing an Array
public class Main { public static void main(String[] args) { int[] myArray = {1, 2, 3, 4, 5}; System.out.println("Array created with elements: " + java.util.Arrays.toString(myArray)); }}
Output
Array created with elements: [1, 2, 3, 4, 5]
Question 2: Accessing Elements of an Array
public class Main { public static void main(String[] args) { int[] myArray = {1, 2, 3, 4, 5}; System.out.println("The first element is: " + myArray[1]); System.out.println("The fourth element is: " + myArray[4]); }}
Output
The first element is: 2 The fourth element is: 5
Question 3: Modifying Elements of an Array
public class Main { public static void main(String[] args) { int[] myArray = {1, 2, 3, 4, 5}; myArray[1] = 10; System.out.println("Array after modification: " + java.util.Arrays.toString(myArray)); }}
Output
Array after modification: [1, 10, 3, 4, 5]
Question 4: Finding the Length of an Array
public class Main { public static void main(String[] args) { int[] myArray = {1, 2, 3, 4, 6}; System.out.println("The length of the array is: " + myArray.length); }}
Output
The length of the array is: 5
Question 5: Looping through an Array
public class Main { public static void main(String[] args) { int[] myArray = {1, 2, 3, 4, 5}; for (int i = 0; i < myArray.length; i++) { System.out.println("Element at index " + i + ": " + myArray[i]); } }}
Output
Element at index 0: 1 Element at index 1: 2 Element at index 2: 3 Element at index 3: 4 Element at index 4: 5
Question 6: Multi-dimensional Arrays
public class Main { public static void main(String[] args) { int[][] multiArray = { {1, 2}, {3, 4}, {5, 6} }; System.out.println("Element at row 1 column 1: " + multiArray[0][0]); }}
Output
Element at row 1 column 1: 1
Question 7: Copying Arrays
public class Main { public static void main(String[] args) { int[] myArray = {1, 2, 3, 4, 5}; int[] newArray = java.util.Arrays.copyOf(myArray, myArray.length); System.out.println("Copied array: " + java.util.Arrays.toString(newArray)); }}
Output
Copied array: [1, 2, 3, 4, 5]
Question 8: Sorting Arrays
public class Main { public static void main(String[] args) { int[] myArray = {5, 3, 4, 1, 2}; java.util.Arrays.sort(myArray); System.out.println("Sorted array: " + java.util.Arrays.toString(myArray)); }}
Output
Sorted array: [1, 2, 3, 4, 5]
Level 2 (Intermediate Level)
Question 1: Find the Second Largest Number in an Array
public class Main { public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5, 6}; int largest = Integer.MIN_VALUE; int secondLargest = Integer.MIN_VALUE;
for (int num : arr) { if (num > largest) { secondLargest = largest; largest = num; } else if (num > secondLargest && num != largest) { secondLargest = num; } }
System.out.println("The second largest number is: " + secondLargest); }}
Output
The second largest number is: 5
Question 2: Reverse an Array
public class Main { public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5}; int n = arr.length;
for (int i = 0; i < n / 2; i++) { int temp = arr[i]; arr[i] = arr[n - i - 1]; arr[n - i - 1] = temp; }
System.out.println("Reversed array: " + java.util.Arrays.toString(arr)); }}
Output
Reversed array: [5, 4, 3, 2, 1]
Question 3: Find the Duplicate Elements
import java.util.HashSet;import java.util.Set;
public class Main { public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 2, 5, 6, 3}; Set<Integer> set = new HashSet<>(); Set<Integer> duplicates = new HashSet<>();
for (int num : arr) { if (!set.add(num)) { duplicates.add(num); } }
System.out.println("Duplicate elements: " + duplicates); }}
Output
Duplicate elements: [2, 3]
Question 4: Find the Kth Largest and Smallest Element in an Array
public class Main { public static void main(String[] args) { int[] arr = {7, 5, 9, 3, 2, 8, 1, 6}; int k = 5; java.util.Arrays.sort(arr); System.out.println(k + "th largest element: " + arr[arr.length - k]); System.out.println(k + "th smallest element: " + arr[k - 1]); }}
Output
5th largest element: 5 5th smallest element: 6
Question 5: Move all Zeroes to the End of the Array
public class Main { public static void main(String[] args) { int[] arr = {1, 0, 2, 0, 3, 0, 4, 0}; int n = arr.length; int count = 0;
for (int i = 0; i < n; i++) { if (arr[i] != 0) { arr[count++] = arr[i]; } } while (count < n) { arr[count++] = 0; }
System.out.println("Array after moving zeroes: " + java.util.Arrays.toString(arr)); }}
Output
Array after moving zeroes: [1, 2, 3, 4, 0, 0, 0, 0]
Question 6: Find the “Leaders” in an Array
public class Main { public static void main(String[] args) { int[] arr = {16, 17, 4, 3, 5, 2}; int n = arr.length; int maxFromRight = arr[n - 1];
System.out.print("Leaders: " + maxFromRight + " ");
for (int i = n - 2; i >= 0; i--) { if (arr[i] > maxFromRight) { maxFromRight = arr[i]; System.out.print(maxFromRight + " "); } } }}
Output
Leaders: 2 5 17
A leader in an array is an element which is larger than all the elements to its right side, which is what you can see in the above output.
Question 7: Find the Majority Element
public class Main { public static void main(String[] args) { int[] arr = {3, 3, 4, 2, 4, 4, 2, 4, 4}; int candidate = findCandidate(arr); if (isMajority(arr, candidate)) { System.out.println("Majority element is: " + candidate); } else { System.out.println("No majority element found"); } }
static int findCandidate(int[] arr) { int majIndex = 0, count = 1; for (int i = 1; i < arr.length; i++) { if (arr[majIndex] == arr[i]) { count++; } else { count--; } if (count == 0) { majIndex = i; count = 1; } } return arr[majIndex]; }
static boolean isMajority(int[] arr, int candidate) { int count = 0; for (int num : arr) { if (num == candidate) { count++; } } return count > arr.length / 2; }}
Output
Majority element is: 4
A majority element in an array is an element that appears more than n/2 times where n is the size of the array, which is what you can see in the above output.
Question 8: Rotate Array by N Elements
public class Main { public static void main(String[] args) { int[] arr = {1, 2, 3, 4, 5, 6, 7}; int n = 2; // Number of positions to rotate reverseArray(arr, 0, n - 1); reverseArray(arr, n, arr.length - 1); reverseArray(arr, 0, arr.length - 1);
System.out.println("Rotated array: " + java.util.Arrays.toString(arr)); }
static void reverseArray(int[] arr, int start, int end) { while (start < end) { int temp = arr[start]; arr[start] = arr[end]; arr[end] = temp; start++; end--; } }}
Output
Rotated array: [3, 4, 5, 6, 7, 1, 2]
Level 3 (Advanced Level)
Question 1: Maximum Subarray Sum (Kadane’s Algorithm)
public class Main { public static void main(String[] args) { int[] nums = {-2, 1, -3, 4, -1, 2, 1, -5, 4}; System.out.println(maxSubArray(nums)); }
public static int maxSubArray(int[] nums) { int maxSoFar = nums[0], maxEndingHere = nums[0]; for (int i = 1; i < nums.length; i++) { maxEndingHere = Math.max(maxEndingHere + nums[i], nums[i]); maxSoFar = Math.max(maxSoFar, maxEndingHere); } return maxSoFar; }}
Output
6
Question 2: Longest Increasing Subsequence
public class Main { public static void main(String[] args) { int[] nums = {10, 22, 9, 33, 21, 50, 41, 60}; System.out.println(lengthOfLIS(nums)); }
public static int lengthOfLIS(int[] nums) { int[] dp = new int[nums.length]; int length = 0;
for (int num : nums) { int i = java.util.Arrays.binarySearch(dp, 0, length, num); if (i < 0) i = -(i + 1); dp[i] = num; if (i == length) length++; } return length; }}
Output
5
Question 3: Matrix Spiral Traversal
public class Main { public static void main(String[] args) { int[][] matrix = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}}; System.out.println(spiralOrder(matrix)); }
public static java.util.List<Integer> spiralOrder(int[][] matrix) { java.util.List<Integer> res = new java.util.ArrayList<>(); if (matrix.length == 0) return res; int rowBegin = 0, rowEnd = matrix.length - 1; int colBegin = 0, colEnd = matrix[0].length - 1;
while (rowBegin <= rowEnd && colBegin <= colEnd) { for (int j = colBegin; j <= colEnd; j++) { res.add(matrix[rowBegin][j]); } rowBegin++; for (int j = rowBegin; j <= rowEnd; j++) { res.add(matrix[j][colEnd]); } colEnd--; if (rowBegin <= rowEnd) { for (int j = colEnd; j >= colBegin; j--) { res.add(matrix[rowEnd][j]); } } rowEnd--; if (colBegin <= colEnd) { for (int j = rowEnd; j >= rowBegin; j--) { res.add(matrix[j][colBegin]); } } colBegin++; } return res; }}
Output
[1, 2, 3, 6, 9, 8, 7, 4, 5]
Question 4: Median of Two Sorted Arrays
public class Main { public static void main(String[] args) { int[] nums1 = {1, 3}; int[] nums2 = {2}; System.out.println(findMedianSortedArrays(nums1, nums2)); }
public static double findMedianSortedArrays(int[] nums1, int[] nums2) { if (nums1.length > nums2.length) { int[] temp = nums1; nums1 = nums2; nums2 = temp; } int m = nums1.length, n = nums2.length; int imin = 0, imax = m, halfLen = (m + n + 1) / 2; double median = 0.0;
while (imin <= imax) { int i = (imin + imax) / 2; int j = halfLen - i;
if (i < m && nums2[j - 1] > nums1[i]) { imin = i + 1; } else if (i > 0 && nums1[i - 1] > nums2[j]) { imax = i - 1; } else { int maxOfLeft = 0; if (i == 0) maxOfLeft = nums2[j - 1]; else if (j == 0) maxOfLeft = nums1[i - 1]; else maxOfLeft = Math.max(nums1[i - 1], nums2[j - 1]);
if ((m + n) % 2 == 1) return maxOfLeft;
int minOfRight = 0; if (i == m) minOfRight = nums2[j]; else if (j == n) minOfRight = nums1[i]; else minOfRight = Math.min(nums1[i], nums2[j]);
return (maxOfLeft + minOfRight) / 2.0; } } return median; }}
Output
2.0
Question 5: Maximum Product Subarray
public class Main { public static void main(String[] args) { int[] nums = {2, 3, -2, 4}; System.out.println(maxProduct(nums)); }
public static int maxProduct(int[] nums) { if (nums.length == 0) return 0; int maxSoFar = nums[0], minSoFar = nums[0], result = nums[0]; for (int i = 1; i < nums.length; i++) { if (nums[i] < 0) { int temp = maxSoFar; maxSoFar = minSoFar; minSoFar = temp; } maxSoFar = Math.max(nums[i], maxSoFar * nums[i]); minSoFar = Math.min(nums[i], minSoFar * nums[i]); result = Math.max(result, maxSoFar); } return result; }}
Output
6
Level 4 (Expert)
Problem Statement: Question 1
Given a list of buildings where each building is represented by triplets [Li,Ri,Hi] :
- Li: An integer, representing the left position of the building.
- Ri: An integer, representing the right position of the building.
- Hi: An integer, representing the height of the building.
Calculate the skyline formed by these buildings.
Input Format:
- A list of n building triplets. Each triplet contains three integers.
- 1≤n≤105
- 0≤Li,Ri,Hi≤104
Output Format:
- A list of key-height pairs representing the skyline.
Solution:
import java.util.*;import java.util.stream.*;
public class Main {
public static void main(String[] args) { int[][] buildings = {{2, 9, 10}, {3, 7, 15}, {5, 12, 12}, {15, 20, 10}, {19, 24, 8}}; List<List<Integer>> result = getSkyline(buildings); for (List<Integer> point : result) { System.out.println(point); } }
public static List<List<Integer>> getSkyline(int[][] buildings) { List<int[]> heights = new ArrayList<>(); for (int[] b : buildings) { heights.add(new int[]{b[0], -b[2]}); // Start of a building heights.add(new int[]{b[1], b[2]}); // End of a building }
Collections.sort(heights, (a, b) -> { if (a[0] != b[0]) return a[0] - b[0]; return a[1] - b[1]; });
PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> (b - a)); pq.offer(0); int prev = 0; List<List<Integer>> result = new ArrayList<>();
for (int[] h : heights) { if (h[1] < 0) { pq.offer(-h[1]); // Height of a building start } else { pq.remove(h[1]); // Height of a building end } int cur = pq.peek(); if (prev != cur) { result.add(Arrays.asList(h[0], cur)); // A point in the skyline prev = cur; } }
return result; }}
Output
[2, 10] [3, 15] [7, 12] [12, 0] [15, 10] [20, 8] [24, 0]
Problem Statement: Question 2
Given a 2D matrix of dimensions m x n containing integers, find the rectangle (sub-matrix) with the maximum sum of its elements.
Input Format:
- A 2D matrix of size m x n where 1≤m,n≤100
- Matrix values: −104≤matrix[i][j]≤104
Output Format:
- An integer representing the maximum sum.
Solution:
import java.util.Arrays;import java.util.TreeSet;
public class Main { public static void main(String[] args) { int[][] matrix = { {1, 2, -1, -4, -20}, {-8, -3, 4, 2, 1}, {3, 8, 10, 1, 3}, {-4, -1, 1, 7, -6} }; System.out.println("Maximum Sum of Sub-matrix: " + maxSumRectangle(matrix)); }
public static int maxSumRectangle(int[][] M) { int m = M.length, n = M[0].length; int[] sums = new int[m]; int maxSum = Integer.MIN_VALUE;
for (int left = 0; left < n; left++) { Arrays.fill(sums, 0);
for (int right = left; right < n; right++) { for (int i = 0; i < m; i++) { sums[i] += M[i][right]; } TreeSet<Integer> set = new TreeSet<>(); set.add(0); int curSum = 0;
for (int sum : sums) { curSum += sum; Integer num = set.ceiling(curSum - maxSum); if (num != null) { maxSum = Math.max(maxSum, curSum - num); } set.add(curSum); } } } return maxSum; }}
Output
Maximum Sum of Sub-matrix: 1
Thus, I hope all the questions help you understand the concept better. Keep learning, Keep exploring!
FAQs
What is an array in Java and how do you declare one?
An array in Java is a container object that holds a fixed number of values of a single type. The length of an array is established when the array is created. Arrays are declared with a specific type followed by square brackets. For example, to declare an array of integers, you would write: int[] myArray;
How do you initialize an array in Java?
There are several ways to initialize an array in Java. You can do it at the time of declaration, like this: int[] myArray = {1, 2, 3, 4, 5}; or you can allocate memory and then initialize it, like this: int[] myArray = new int[5]; myArray[0] = 1; ... myArray[4] = 5;
What is a multidimensional array and how is it used in Java?
A multidimensional array in Java is an array of arrays. Each element of a multidimensional array is itself an array. The most common type is a 2D array, like int[][] my2DArray = new int[10][20]; which can be thought of as a table with 10 rows and 20 columns.
How do you iterate over an array in Java?
You can iterate over an array in Java using a traditional for loop, a for-each loop, or a while loop.
How do you handle exceptions when working with arrays in Java?
The most common exception when working with arrays is ArrayIndexOutOfBoundsException, which occurs when trying to access an element at an index that is outside the bounds of the array. This can be handled using a try-catch block.
Hello, world! I'm Esha Gupta, your go-to Technical Content Developer focusing on Java, Data Structures and Algorithms, and Front End Development. Alongside these specialities, I have a zest for immersing myself in v... Read Full Bio
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