3014. Minimum Number of Pushes to Type Word I

Intuition

There are 8 usable keys (2 to 9). Each key's first assigned letter costs 1 push, its second costs 2, its third costs 3, etc. To minimize total pushes, fill all 8 keys with one letter each (cost 1) before giving any key a second letter (cost 2), and so on. Since all letters in word are distinct, only the count of letters matters, not which letters they are.

So the letter placed at index i (0-based) lands in "tier" i // 8, costing i // 8 + 1 pushes.

Approach

1. Let n = len(word).
2. Sum i // 8 + 1 for i from 0 to n - 1.

A closed form also works: full tiers of 8 plus the remainder, but the direct sum is clearest for n <= 26.

Complexity

• Time: O(n)
• Space: O(1)

Edge Cases

• n <= 8 → every letter costs 1 → answer n (Example 1)
• n = 26 → 8 letters at cost 1, 8 at cost 2, 8 at cost 3, 2 at cost 4
• Distinctness is guaranteed, so no letter frequency to weigh (that's part II)

class Solution:
    def minimumPushes(self, word: str) -> int:
        """
        3014. Minimum Number of Pushes to Type Word I
        Time: O(n)
        Space: O(1)
        """
        # Distinct letters: the i-th letter (0-based) costs i // 8 + 1 pushes.
        return sum(i // 8 + 1 for i in range(len(word)))

/**
 * 3014. Minimum Number of Pushes to Type Word I
 * Time: O(n)
 * Space: O(1)
 */
class Solution {
    public int minimumPushes(String word) {
        int total = 0;
        // Distinct letters: the i-th letter (0-based) costs i / 8 + 1 pushes.
        for (int i = 0; i < word.length(); i++) {
            total += i / 8 + 1;
        }
        return total;
    }
}