Day 2 of 100 Days of DSA Challenge: Arrays Basics

In today's journey of the 100 Days of DSA challenge, I explored fundamental array operations. Arrays are a versatile data structure, and mastering their operations is crucial for solving problems effectively. Here are the problems I tackled, along with their Java implementations.

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1. Find the Largest and Smallest Elements in an Array

The task is to traverse the array once, maintaining variables to track the largest and smallest values.

import java.util.*;

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

        Scanner sc = new Scanner(System.in);

        System.out.println("Enter number of elements: ");
        int count = sc.nextInt();
        int[] arr = new int[count];

        int min = 0;
        int max = 0;

        System.out.println("Enter elements of the array:");
        for(int i=0; i<count; i++){
            arr[i] = sc.nextInt();
            if(i==0){
                min = arr[i];
                max = arr[i];
            }
            if(arr[i]>max)
                max = arr[i];
            else if(arr[i]<min)
                min = arr[i];
        }

        System.out.println("Largest element is: ".concat(Integer.toString(max)));
        System.out.println("Smallest element is: ".concat(Integer.toString(min)));

    }
}

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2. Find the Second Largest Element in an Array

This involves maintaining two variables: one for the largest element and another for the second largest.

import java.util.*;

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

        Scanner sc = new Scanner(System.in);

        System.out.println("Enter number of elements: ");
        int count = sc.nextInt();
        int[] arr = new int[count];

        int ans = 0;
        int max = 0;

        System.out.println("Enter elements of the array:");
        for(int i=0; i<count; i++){
            arr[i] = sc.nextInt();
            if(i==0){
                ans = arr[i];
                max = arr[i];
            }
            if(arr[i]>max){
                ans = max;
                max = arr[i];
            }
            else if(arr[i]>ans)
                ans = arr[i];
        }

        sc.close();

        System.out.println("Secound Largest element is: ".concat(Integer.toString(ans)));
    }
}

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3. Rotate an Array to the Left by k Positions

To rotate an array to the left, shift elements k positions to the left, and wrap around the first k elements.

import java.util.*;

public class RotateArrayLeft {

    public static void printArray(int[] arr){
        for(int i=0; i<arr.length; i++)
            System.out.print(arr[i]+" ");
        System.out.println();
    }

    public static void rotateArray(int[] arr){
        int num = arr[0];
        for(int i=0; i<arr.length-1; i++)
            arr[i] = arr[i+1];
        arr[arr.length-1] = num;
    }

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);

        System.out.println("Enter number of elements in the array: ");
        int count = sc.nextInt();
        int[] arr = new int[count];

        System.out.println("Enter elements of the array: ");
        for(int i=0; i<count; i++)
            arr[i] = sc.nextInt();

        System.out.println("Enter the value of k(no. of positions): ");
        int k = sc.nextInt();

        if(count==1)
            printArray(arr);
        else{
            while(k>0){
                rotateArray(arr);
                k--;
            }
            printArray(arr);
        }

    }    
}

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4. Count the Occurrences of a Given Element in an Array

By iterating through the array, count how many times the target element appears.

import java.util.*;

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

        System.out.println("Enter number of elements: ");
        int count = sc.nextInt();
        int[] arr = new int[count];

        System.out.println("Enter elements of the array: ");
        for(int i=0; i<count; i++)
            arr[i] = sc.nextInt();

        System.out.println("Which element's occurrence do you want to find out?");
        int num = sc.nextInt();

        int ans = 0;

        for(int i=0; i<count; i++)
            ans = (arr[i]==num)?ans+1:ans;

        System.out.println("Element "+num+" has "+ans+" occurrences.");

    }   
}

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5. Find All Elements That Appear More Than Once in an Array

This problem can be solved by a brute force approach in the following way.

import java.util.*;

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

        Scanner sc = new Scanner(System.in);

        System.out.println("Enter number of elements: ");
        int count = sc.nextInt();
        int[] arr = new int[count];
        ArrayList<Integer> ans = new ArrayList<>();

        System.out.println("Enter elements of the array: ");
        for(int i=0; i<count; i++)
            arr[i] = sc.nextInt();

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

            int num = 0;

            for(int j=0; j<count; j++){
                if(i==j)
                    continue;
                else if(arr[i] == arr[j])
                    num++;                
            } 

            if(num>0 && !ans.contains(arr[i]))
                ans.add(arr[i]);

        }

        System.out.println(ans);

    }
}

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Reflection

Day 2 was a rewarding experience, as I got to practice various operations on arrays. These operations are essential in solving many real-world programming challenges and interview problems.

Stay tuned for Day 3, where I'll explore more advanced array operations! 🚀