LeetCode 771. Jewels and Stones Java Solution

Problem

Jewels and Stones - LeetCode

Solution Approach

• This problem is about checking whether each character from one input string exists in another input string.

• Since each character needs to be checked at least once, the time complexity is O(n^2), where n is the length of the input string.

• It tests your basic string handling skills.

• To solve it, you need to understand the toCharArray() method and how to compare characters using the char data type in Java.

Java's Primitive Data Types and Reference Data Types

Github Link

https://github.com/eunhanlee/LeetCode_771_JewelsandStones_Solution.git

Time Complexity: O(n^2), Space Complexity: O(1)

class Solution {
    /**
     * Method to count the number of jewels in the stones.
     *
     * @param jewels The string representing types of jewels.
     * @param stones The string representing the stones you have.
     * @return The number of jewels found in the stones.
     */
    public int numJewelsInStones(String jewels, String stones) {
        int counter = 0;

        for (char stone : stones.toCharArray()) {
            for (char jewel : jewels.toCharArray()) {
                if (stone == jewel) {
                    counter++;
                }
            }
        }

        return counter;
    }
}