LeetCode 3019: Number of Changing Keys Solution

Master LeetCode problem 3019 (Number of Changing Keys), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

3019. Number of Changing Keys

Easy

Problem Explanation

Explanation:

To solve this problem, we can iterate through the input string and count the number of times the user had to change the key based on the conditions provided. We need to compare each character with the previous character and check if a key change is required. We can ignore cases where caps lock or shift is used as they do not count as a key change.

  • Initialize a variable changes to 0 to keep track of the number of key changes.
  • Iterate through the string starting from index 1.
  • For each character, check if it is different from the previous character and both characters are not the same letter with different cases.
  • If the above condition is met, increment the changes count.
  • Finally, return the changes count as the result.

Time Complexity: O(n) where n is the length of the input string. Space Complexity: O(1)

Solution Code

class Solution {
    public int numberOfChangingKeys(String s) {
        int changes = 0;
        for (int i = 1; i < s.length(); i++) {
            if (s.charAt(i) != s.charAt(i - 1) && Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
                changes++;
            }
        }
        return changes;
    }
}

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