Can we pass a character as an index for integer array in java

Question

I am new to java and practicing some programs.Can someone please explain me the following program at //line 10

public static void main(String[] args) {
        String str1 = "xxyz";
        String str2 = "yxzx";
        System.out.println("Original strings: " + str1 + "  " + str2);
        System.out.println(stringPermutation(str1, str2));
    }

    public static void stringPermutation(String str1, String str2) {
        int[] arr = new int[500];
        for (int i = 0; i < str1.length(); i++) {
     System.out.println(arr[(int) str1.charAt(i)] += 1); //line 10

        } 
}

it's displaying the below output :

Original strings: xxyz  yxzx
1
2
1
1

Am trying to understand, how arr[(int) str1.charAt(i)]i.e, arr['x'] is possible. Please help me understanding this.

Characters are basicly int values, for example x==120. You can check some ASCII table to see values of all the characters. — Amongalen
– Amongalen, Commented Feb 27, 2020 at 12:23

Thecarisma · Accepted Answer · 2020-02-27 13:49:58Z

2

When you initialise an empty integer array the initial value for each of the item is zero, there for each index of arr will contain 0

int[] arr = new int[500];
arr[0] = 0;
arr[1] = 0;
//...
arr[500] = 0;

The ASCII value for x is 120, y is 121 and z is 122, since the arr field contains 500 item then 120, 121 and 122 is in range.

In your loop you are adding 1 to each of the element. therefore in your str1 = "xxyz" when the first x is encounter 1 is added to index arr[120] so arr[120] becomes 1 and when x is encounter again 1 is added to the value which makes arr[120] becomes 2.

arr[(int) 'x'] += 1 //=> arr[120] + 1 = (0 + 1) = 1
arr[(int) 'x'] += 1 //=> arr[120] + 1 = (1 + 1) = 2
arr[(int) 'y'] += 1 //=> arr[121] + 1 = (0 + 1) = 1
arr[(int) 'z'] += 1 //=> arr[122] + 1 = (0 + 1) = 1

arr['x'] is possible because in java char data type is a single 16-bit integer and int is 32-bit signed integer.

Update:

In the continuation for the program the second loop on str2 is deducting one at the char index, if str2 is a permutation of str1 the value all the item in arr should reset to 0.

After the loop on str1. The values in the array are

arr[0] = 0
//...
arr[120] = 2 //arr['x']
arr[121] = 1 //arr['y']
arr[122] = 1 //arr['z']
//...
arr[500] = 0

When str2 = "yxzx"

arr[(int) 'y'] -= 1 //=> arr[121] - 1 = (1 - 1) = 0
arr[(int) 'x'] -= 1 //=> arr[120] - 1 = (2 - 1) = 1
arr[(int) 'z'] -= 1 //=> arr[122] - 1 = (1 - 1) = 0
arr[(int) 'x'] -= 1 //=> arr[120] - 1 = (1 + 1) = 0

After the loop on str2 the values will be reset to 0

arr[0] = 0
//...
arr[120] = 0 //arr['x']
arr[121] = 0 //arr['y']
arr[122] = 0 //arr['z']
//...
arr[500] = 0

Hence looping through all the array if all the values are zeros then str2 is a permutation of str1.

edited Feb 27, 2020 at 13:49

answered Feb 27, 2020 at 12:48

Thecarisma

1,43318 silver badges19 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

Rahul Over a year ago

Thank you for the detailed explanation.

Rahul Over a year ago

In Continuation to this program to 'Check if a given string is a permutation of another specified string' , am unable to understand the below logic

for (int i = 0; i < str2.length(); i++) { 			System.out.println(arr[(int) str2.charAt(i)] -= 1); 		} 		for (int i = 0; i < arr.length; i++) { 			if (arr[i] != 0) 				return false; 		} 		return true; 	}

Rahul Over a year ago

can you please explain me how for (int i = 0; i < arr.length; i++) { if (arr[i] != 0) return false; } is used to check if string str2 is a perumtation of string str1.

Ashish Karn · Accepted Answer · 2020-02-27 12:25:00Z

1

Because of ASCII value of x is 120, y is 121 and z is 122. So, Now

arr[x] => arr[120] = 1
arr[x] => arr[120] = 2
arr[y] => arr[121] = 1
arr[z] => arr[122] = 1

answered Feb 27, 2020 at 12:25

Ashish Karn

1,1531 gold badge9 silver badges20 bronze badges

2 Comments

Rahul Over a year ago

Thank you.Though some people downvoted it, i understood the concept now.

Avv Over a year ago

Thank you. Never thought that char will be converted to ASCII integer.

Arvind Kumar Avinash · Accepted Answer · 2020-02-27 14:18:16Z

1

Yes, you can. A char variable can hold any integer from 0 to 65,535. The value of a char literal is equal to an integer from the ASCII table.

public class Main {
    public static void main(String[] args) {
        char[] charArr = { '0', 'A', 'z' };
        int[] anArr = new int[123];

        // Initialize
        for (int i = 1; i < 123; i++) {
            anArr[i] = i * 10;
        }

        // Display
        for (char c : charArr) {
            System.out.print(anArr[c] + "\t");// Will display 480   650 1220
        }
    }
}

Explanation:

anArr['0'] = anArr[48] which has 48*10 stored in it
anArr['A'] = anArr[65] which has 65*10 stored in it
anArr['z'] = anArr[122] which has 122*10 stored in it

edited Feb 27, 2020 at 14:18

answered Feb 27, 2020 at 14:09

Arvind Kumar Avinash

81k10 gold badges98 silver badges144 bronze badges

Collectives™ on Stack Overflow

Can we pass a character as an index for integer array in java

3 Answers 3

3 Comments

2 Comments

Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

3 Comments

2 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related