2268. Minimum Number of Keypresses

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2268. Minimum Number of Keypresses

Description

You have a keypad with 9 buttons, numbered from 1 to 9, each mapped to lowercase English letters. You can choose which characters each button is matched to as long as:

  • All 26 lowercase English letters are mapped to.
  • Each character is mapped to by exactly 1 button.
  • Each button maps to at most 3 characters.

To type the first character matched to a button, you press the button once. To type the second character, you press the button twice, and so on.

Given a string s, return the minimum number of keypresses needed to type s using your keypad.

Note that the characters mapped to by each button, and the order they are mapped in cannot be changed.

Example 1:

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Input: s = "apple"
Output: 5
Explanation: One optimal way to setup your keypad is shown above.
Type 'a' by pressing button 1 once.
Type 'p' by pressing button 6 once.
Type 'p' by pressing button 6 once.
Type 'l' by pressing button 5 once.
Type 'e' by pressing button 3 once.
A total of 5 button presses are needed, so return 5.

Example 2:

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Input: s = "abcdefghijkl"
Output: 15
Explanation: One optimal way to setup your keypad is shown above.
The letters 'a' to 'i' can each be typed by pressing a button once.
Type 'j' by pressing button 1 twice.
Type 'k' by pressing button 2 twice.
Type 'l' by pressing button 3 twice.
A total of 15 button presses are needed, so return 15.

Constraints:

  • 1 <= s.length <= 10^5
  • s consists of lowercase English letters.

Hints/Notes

  • 2025/02/08 Q2
  • sorting
  • No solution from 0x3F or Leetcode

Solution

Language: C++

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class Solution {
public:
    int minimumKeypresses(string s) {
        int count[26]{};
        for (char& c : s) {
            count[c - 'a']++;
        }
        priority_queue<int> pq;
        for (int i = 0; i < 26; i++) {
            if (count[i] > 0) {
                pq.push(count[i]);
            }
        }
        int sum = 0, k = 0;
        while (!pq.empty()) {
            int freq = pq.top();
            pq.pop();
            sum += freq * (k / 9 + 1);
            k++;
        }
        return sum;
    }
};