Number of Changing Keys

Number of Changing Keys - Problem

String Easy

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.

Input & Output

Example 1 — Basic Case with Changes

$ Input: s = "aAbBcC"

› Output: 2

💡 Note: Changes occur at positions 2 (A→b) and 4 (B→c). Characters 'a'/'A', 'b'/'B', and 'c'/'C' are the same keys respectively.

Example 2 — No Changes

$ Input: s = "AaAaAaaA"

› Output: 0

💡 Note: All characters represent the same key 'a', so no key changes are needed regardless of case.

Example 3 — All Different Keys

$ Input: s = "abcd"

› Output: 3

💡 Note: Every adjacent pair represents different keys: a→b, b→c, c→d, resulting in 3 changes.

Constraints

1 ≤ s.length ≤ 100
s consists of only English letters

Visualization

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Asked in

M Microsoft 15 🍎 Apple 12 G Google 8

The key insight is to treat uppercase and lowercase letters as the same key by converting to lowercase before comparison. Best approach is a single pass through the string comparing adjacent characters. Time: O(n), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force - Character by Character Comparison	O(n)	O(1)	Compare each character with the previous one, converting to lowercase
Optimized Single Pass	O(n)	O(1)	Single pass with early termination for edge cases

Brute Force - Character by Character Comparison — Algorithm Steps

Start from index 1
Compare current character with previous
Convert both to lowercase
Count when they differ

Visualization

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Step-by-Step Walkthrough

Start

Begin from second character

Compare

Check if current differs from previous (case-insensitive)

Count

Increment counter when keys change

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <ctype.h>

int solution(char* s) {
    int len = strlen(s);
    if (len <= 1) {
        return 0;
    }
    
    int changes = 0;
    for (int i = 1; i < len; i++) {
        if (tolower(s[i]) != tolower(s[i-1])) {
            changes++;
        }
    }
    
    return changes;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    
    int result = solution(s);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Single pass through the string of length n

✓ Linear Growth

Space Complexity

O(1)

Only using a counter variable and character comparisons

✓ Linear Space

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Character by Character Comparison — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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