Is Minimum Number of Pushes to Type Word I easy or hard?

Minimum Number of Pushes to Type Word I is classified as an Easy problem. The main challenge is recognizing that assigning lower push costs to higher-frequency letters minimizes the total operations. Once that greedy insight is clear, the implementation is straightforward.

Minimum Number of Pushes to Type Word I Python/Java solution

Python and Java implementations follow the same idea: compute a frequency array for 26 letters, sort it in descending order, and assign push costs based on index groups of size eight. Each frequency is multiplied by its push cost tier and summed. The logic is short and runs in linear time relative to the input size.

How to solve Minimum Number of Pushes to Type Word I in O(n)?

Scan the word and count the frequency of each character. Sort the 26 frequencies in descending order and assign push costs based on position: the first 8 letters cost 1 push, the next 8 cost 2 pushes, the next 8 cost 3 pushes, and the remaining letters cost 4 pushes. Multiply each frequency by its assigned push cost and sum the results to get the minimum total pushes.

What is the best approach for Minimum Number of Pushes to Type Word I?

The greedy frequency assignment approach is the most efficient. Count how often each letter appears, sort the frequencies in descending order, and assign the smallest push counts to the most frequent letters. Because only 8 letters can have the same push cost tier, the first 8 get cost 1, the next 8 cost 2, and so on. This minimizes the total number of pushes required to type the word.

Is Minimum Number of Pushes to Type Word I asked at Google/Amazon/Meta?

Problems involving greedy frequency assignment and keypad optimization commonly appear in interviews at companies like Amazon, Google, and Meta. Variations of this question test your ability to minimize weighted costs using greedy ordering and frequency counting.

What data structure is used in Minimum Number of Pushes to Type Word I?

The main data structure is a frequency array or hash map to count occurrences of each letter. After counting, the frequencies are sorted to apply the greedy assignment of push costs. The algorithm also relies on simple array iteration and arithmetic calculations.

What is the time complexity of Minimum Number of Pushes to Type Word I?

The optimal solution runs in O(n + 26 log 26) time, where n is the length of the word. The algorithm scans the string once to compute character frequencies and then sorts at most 26 values. Space complexity is O(1) because the alphabet size is fixed.

Description Hints 2 Solution Video 31:28

3014. Minimum Number of Pushes to Type Word I

Easy77.1% acceptance4 companies

You are given a string word containing distinct lowercase English letters.

Telephone keypads have keys mapped with distinct collections of lowercase English letters, which can be used to form words by pushing them. For example, the key 2 is mapped with ["a","b","c"], we need to push the key one time to type "a", two times to type "b", and three times to type "c" .

It is allowed to remap the keys numbered 2 to 9 to distinct collections of letters. The keys can be remapped to any amount of letters, but each letter must be mapped to exactly one key. You need to find the minimum number of times the keys will be pushed to type the string word.

Return the minimum number of pushes needed to type word after remapping the keys.

An example mapping of letters to keys on a telephone keypad is given below. Note that 1, *, #, and 0 do not map to any letters.

Example 1:

Input: word = "abcde"
Output: 5
Explanation: The remapped keypad given in the image provides the minimum cost.
"a" -> one push on key 2
"b" -> one push on key 3
"c" -> one push on key 4
"d" -> one push on key 5
"e" -> one push on key 6
Total cost is 1 + 1 + 1 + 1 + 1 = 5.
It can be shown that no other mapping can provide a lower cost.

Example 2:

Input: word = "xycdefghij"
Output: 12
Explanation: The remapped keypad given in the image provides the minimum cost.
"x" -> one push on key 2
"y" -> two pushes on key 2
"c" -> one push on key 3
"d" -> two pushes on key 3
"e" -> one push on key 4
"f" -> one push on key 5
"g" -> one push on key 6
"h" -> one push on key 7
"i" -> one push on key 8
"j" -> one push on key 9
Total cost is 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 12.
It can be shown that no other mapping can provide a lower cost.

Constraints: