How do you solve Minimum Distance to Type a Word Using Two Fingers on LeetCode?

Minimum Distance to Type a Word Using Two Fingers (LeetCode #1320) can be solved using Brute Force. The optimal approach is Brute Force with undefined time complexity and undefined space complexity. Key insight: The brute force approach considers all possible placements of the two fingers for each character in the word. It calculates the distance for each possible combination, which can be quite inefficient.

What is the time complexity of Minimum Distance to Type a Word Using Two Fingers?

The optimal solution for Minimum Distance to Type a Word Using Two Fingers runs in undefined time and uses undefined space. undefined

What algorithm or data structure is used to solve Minimum Distance to Type a Word Using Two Fingers?

Minimum Distance to Type a Word Using Two Fingers uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: .

#1320

Minimum Distance to Type a Word Using Two Fingers

Hard

StringDynamic Programming

LeetCode ↗

Approaches

💡

Intuition

Time Space

The brute force approach considers all possible placements of the two fingers for each character in the word. It calculates the distance for each possible combination, which can be quite inefficient.

⚙️

Algorithm

3 steps

1Step 1: Initialize a variable to store the minimum distance.
2Step 2: For each character in the word, iterate through all possible positions of the two fingers.
3Step 3: Calculate the distance for each finger movement and update the minimum distance accordingly.

solution.py26 lines

1# Full working Python code
2from itertools import product
3
4def minDistance(word):
5    # Keyboard layout coordinates
6    coords = {'A': (0, 0), 'B': (0, 1), 'C': (0, 2), 'D': (0, 3), 'E': (0, 4),
7              'F': (1, 0), 'G': (1, 1), 'H': (1, 2), 'I': (1, 3), 'J': (1, 4),
8              'K': (2, 0), 'L': (2, 1), 'M': (2, 2), 'N': (2, 3), 'O': (2, 4),
9              'P': (3, 0), 'Q': (3, 1), 'R': (3, 2), 'S': (3, 3), 'T': (3, 4),
10              'U': (4, 0), 'V': (4, 1), 'W': (4, 2), 'X': (4, 3), 'Y': (4, 4), 'Z': (4, 5)}
11    n = len(word)
12    min_dist = float('inf')
13
14    for i in range(n):
15        for j in range(i + 1, n):
16            finger1_pos = coords[word[i]]
17            finger2_pos = coords[word[j]]
18            total_distance = 0
19            for k in range(n):
20                if k == i:
21                    total_distance += abs(finger1_pos[0] - coords[word[k]][0]) + abs(finger1_pos[1] - coords[word[k]][1])
22                elif k == j:
23                    total_distance += abs(finger2_pos[0] - coords[word[k]][0]) + abs(finger2_pos[1] - coords[word[k]][1])
24            min_dist = min(min_dist, total_distance)
25    return min_dist
26minDistance('CAKE')

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