LeetCode #3019 — EASY

Number of Changing Keys

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

The Problem

Problem Statement

You are given a 0-indexed string s typed by a user. Changing a key is defined as using a key different from the last used key. For example, s = "ab" has a change of a key while s = "bBBb" does not have any.

Return the number of times the user had to change the key.

Note: Modifiers like shift or caps lock won't be counted in changing the key that is if a user typed the letter 'a' and then the letter 'A' then it will not be considered as a changing of key.

Example 1:

Input: s = "aAbBcC"
Output: 2
Explanation: 
From s[0] = 'a' to s[1] = 'A', there is no change of key as caps lock or shift is not counted.
From s[1] = 'A' to s[2] = 'b', there is a change of key.
From s[2] = 'b' to s[3] = 'B', there is no change of key as caps lock or shift is not counted.
From s[3] = 'B' to s[4] = 'c', there is a change of key.
From s[4] = 'c' to s[5] = 'C', there is no change of key as caps lock or shift is not counted.

Example 2:

Input: s = "AaAaAaaA"
Output: 0
Explanation: There is no change of key since only the letters 'a' and 'A' are pressed which does not require change of key.

Constraints:

1 <= s.length <= 100
s consists of only upper case and lower case English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string s typed by a user. Changing a key is defined as using a key different from the last used key. For example, s = "ab" has a change of a key while s = "bBBb" does not have any. Return the number of times the user had to change the key. Note: Modifiers like shift or caps lock won't be counted in changing the key that is if a user typed the letter 'a' and then the letter 'A' then it will not be considered as a changing of key.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: General problem-solving

Example 1

"aAbBcC"

Example 2

"AaAaAaaA"

Step 02

Core Insight

What unlocks the optimal approach

Change all the characters to lowercase and then return the number of indices where the character does not match with the last index character.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3019: Number of Changing Keys
class Solution {
    public int countKeyChanges(String s) {
        int ans = 0;
        for (int i = 1; i < s.length(); ++i) {
            if (Character.toLowerCase(s.charAt(i)) != Character.toLowerCase(s.charAt(i - 1))) {
                ++ans;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3019: Number of Changing Keys
func countKeyChanges(s string) (ans int) {
	s = strings.ToLower(s)
	for i, c := range s[1:] {
		if byte(c) != s[i] {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #3019: Number of Changing Keys
class Solution:
    def countKeyChanges(self, s: str) -> int:
        return sum(a != b for a, b in pairwise(s.lower()))

// Accepted solution for LeetCode #3019: Number of Changing Keys
impl Solution {
    pub fn count_key_changes(s: String) -> i32 {
        let s = s.to_lowercase();
        let bytes = s.as_bytes();
        let mut ans = 0;
        for i in 1..bytes.len() {
            if bytes[i] != bytes[i - 1] {
                ans += 1;
            }
        }
        ans
    }
}

// Accepted solution for LeetCode #3019: Number of Changing Keys
function countKeyChanges(s: string): number {
    s = s.toLowerCase();
    let ans = 0;
    for (let i = 1; i < s.length; ++i) {
        if (s[i] !== s[i - 1]) {
            ++ans;
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.