How do you solve Minimum Number of Pushes to Type Word I on LeetCode?

Minimum Number of Pushes to Type Word I (LeetCode #3014) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Even distribution of letters minimizes pushes.

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

The optimal solution for Minimum Number of Pushes to Type Word I runs in O(n) time and uses O(1) space. This complexity is linear because we only need to calculate the number of letters and perform a constant amount of arithmetic operations.

What is the brute force solution for Minimum Number of Pushes to Type Word I?

The brute force approach for Minimum Number of Pushes to Type Word I is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we try to assign letters to keys in every possible way to find the combination that results in the minimum number of pushes. This is straightforward but inefficient as it explores all mappings. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Minimum Number of Pushes to Type Word I?

Minimum Number of Pushes to Type Word I uses the following concepts: Math, String, Greedy. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Minimum Number of Pushes to Type Word I?

Explain your thought process clearly. Discuss potential edge cases. Be ready to optimize your solution.

What are common mistakes when solving Minimum Number of Pushes to Type Word I?

Not considering the distribution of letters across keys. Overcomplicating the mapping process.

#3014

Minimum Number of Pushes to Type Word I

Easy

MathStringGreedyHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach involves evenly distributing the letters across the keys to minimize the number of pushes. Since we have 8 keys, we can assign letters in groups to achieve the least number of total pushes.

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Algorithm

3 steps

1Step 1: Count the number of letters in the word.
2Step 2: Calculate how many full groups of 8 can be formed and how many letters will remain.
3Step 3: Calculate the total pushes based on the number of groups and remaining letters.

solution.py9 lines

1# Full working Python code
2
3def min_pushes_optimal(word):
4    n = len(word)
5    full_groups = n // 8
6    remaining = n % 8
7    pushes = (full_groups * 8 * (8 + 1)) // 2 + remaining * (full_groups + 1)
8    return pushes
9

ℹ

Complexity note: This complexity is linear because we only need to calculate the number of letters and perform a constant amount of arithmetic operations.

1Even distribution of letters minimizes pushes.
2Each key can handle 8 letters optimally.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.