3016. Minimum Number of Pushes to Type Word II LeetCode Solution
In this guide, you will get 3016. Minimum Number of Pushes to Type Word II LeetCode Solution with the best time and space complexity. The solution to Minimum Number of Pushes to Type Word II problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Number of Pushes to Type Word II
You are given a string word containing 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.
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 = “xyzxyzxyzxyz”
Output: 12
Explanation: The remapped keypad given in the image provides the minimum cost.
“x” -> one push on key 2
“y” -> one push on key 3
“z” -> one push on key 4
Total cost is 1 * 4 + 1 * 4 + 1 * 4 = 12
It can be shown that no other mapping can provide a lower cost.
Note that the key 9 is not mapped to any letter: it is not necessary to map letters to every key, but to map all the letters.
Example 3:
Input: word = “aabbccddeeffgghhiiiiii”
Output: 24
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
“f” -> one push on key 7
“g” -> one push on key 8
“h” -> two pushes on key 9
“i” -> one push on key 9
Total cost is 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 1 * 2 + 2 * 2 + 6 * 1 = 24.
It can be shown that no other mapping can provide a lower cost.
Constraints:
1 <= word.length <= 105
word consists of lowercase English letters.
3016. Minimum Number of Pushes to Type Word II LeetCode Solution in C++
class Solution {
public:
// Same as 3014. Minimum Number of Pushes to Type Word I
int minimumPushes(string word) {
int ans = 0;
vector<int> count(26);
for (const char c : word)
++count[c - 'a'];
ranges::sort(count, greater<>());
for (int i = 0; i < 26; ++i)
ans += count[i] * (i / 8 + 1);
return ans;
}
}; /* code provided by PROGIEZ */
3016. Minimum Number of Pushes to Type Word II LeetCode Solution in Java
class Solution {
// Same as 3014. Minimum Number of Pushes to Type Word I
public int minimumPushes(String word) {
int ans = 0;
int[] count = new int[26];
for (final char c : word.toCharArray())
++count[c - 'a'];
Arrays.sort(count);
for (int i = 0; i < 26; ++i)
ans += count[26 - i - 1] * (i / 8 + 1);
return ans;
}
} // code provided by PROGIEZ
3016. Minimum Number of Pushes to Type Word II LeetCode Solution in Python
class Solution:
# Same as 3014. Minimum Number of Pushes to Type Word I
def minimumPushes(self, word: str) -> int:
freqs = sorted(collections.Counter(word).values(), reverse=True)
return sum(freq * (i // 8 + 1) for i, freq in enumerate(freqs)) # code by PROGIEZ
