Reverse Array

Reverse Array _____

            I'm using increment for loop to solve this problem

Another Approach to solve this problem: You can solve this problem by traiverse the reverse loop also.

in both condition Time Complexity will be 0(N);

        

 public class ReverseArrayExtraArray {

public static void reverseArrayExtraArray(int[] arr) {

int[] reversedArr = new int[arr.length];

for (int i = 0; i < arr.length; i++) {

reversedArr[i] = arr[arr.length - i - 1];

}

// Print reversed array

System.out.print("Reversed Array: ");

for (int i : reversedArr) {

System.out.print(i + " ");

}

}

public static void main(String[] args) {

int[] originalArr = {1, 2, 3, 4, 5};

reverseArrayExtraArray(originalArr);

}

}

Dry run for this -----

Let's break it down:

• int i = 0;: Hum loop ko index 0 se start kar rahe hain, jo reversedArr ka pehla element hoga.

• i < arr.length;: Hum loop ko chalate hain jab tak humare original array arr ke saare elements iterate nahi ho jaate.

• reversedArr[i] = arr[arr.length - i - 1];: Yeh line kya kar rahi hai? Dekhiye:

  • arr.length - i - 1: Yeh expression humein reverse order mein arr ke elements ko access karne mein madad karta hai. Jaise agar arr ka length 5 hai, aur i ki value 0 hai to, (5 - 0 - 1) = 4, matlab ki hum arr ke last element pe pahunch gaye hain. Agar i ki value 1 hai, to (5 - 1 - 1) = 3, to hum ab second last element pe hain.

  • reversedArr[i] = arr[arr.length - i - 1];: Ab, jo element arr[arr.length - i - 1] hai, matlab vo original array arr ka reverse element hai. Aur hum is reverse element ko reversedArr ke ith index pe store kar rahe hain.

Yeh process hum arr ke saare elements ke liye repeat karte hain, jisse reversedArr mein arr ke elements reverse order mein store ho jaate hain.