This problem simulates an old-fashioned mobile keypad where each of the 8 keys (2-9) can be assigned a set of letters. When you type a word, the number of "pushes" for a letter depends on its position on the key. For example, if 'a' is the first letter on key 2, it takes 1 push. If 'b' is second, it takes 2. In "Word I," all characters in the word are unique, and you want to map them to the 8 keys to minimize the total pushes.
Companies like Snapchat and Google use this to test basic greedy logic and greedy optimization. It's a fundamental problem that asks: how do you distribute items (letters) into bins (keys) to minimize a cost function? Since all letters appear once, the goal is to fill the "first position" of all 8 keys before moving to the "second position."
The "Minimum Number of Pushes to Type Word I coding problem" follows a Greedy approach. Since every letter in the input word has the same frequency (one), you simply need to distribute the letters as evenly as possible across the 8 available keys. The first 8 letters will cost 1 push each, the next 8 will cost 2 pushes each, and so on.
Suppose the word has 10 unique letters.
Candidates often over-engineer this by trying to track which specific letter goes to which key. Since all letters are unique and frequencies are the same, the actual mapping doesn't matter—only the distribution counts. Another mistake is forgetting that there are only 8 keys (2-9), not 10 or 26.
Always look at the constraints. In "Word I," the "all unique letters" constraint is the key. It simplifies the problem drastically because you don't have to sort by frequency. When you see "minimum cost" and "even distribution," greedy is usually the right path.
| Title | Difficulty | Topics | LeetCode |
|---|---|---|---|
| Maximum Odd Binary Number | Easy | Solve | |
| Largest Odd Number in String | Easy | Solve | |
| Minimum Swaps to Make Strings Equal | Medium | Solve | |
| Remove Colored Pieces if Both Neighbors are the Same Color | Medium | Solve | |
| Sum Game | Medium | Solve |
