Leetcode #1320: Minimum Distance to Type a Word Using Two Fingers
In this guide, we solve Leetcode #1320 Minimum Distance to Type a Word Using Two Fingers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You have a keyboard layout as shown above in the X-Y plane, where each English uppercase letter is located at some coordinate. For example, the letter 'A' is located at coordinate (0, 0), the letter 'B' is located at coordinate (0, 1), the letter 'P' is located at coordinate (2, 3) and the letter 'Z' is located at coordinate (4, 1).
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: String, Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: word = "CAKE"
Output: 3
Explanation: Using two fingers, one optimal way to type "CAKE" is:
Finger 1 on letter 'C' -> cost = 0
Finger 1 on letter 'A' -> cost = Distance from letter 'C' to letter 'A' = 2
Finger 2 on letter 'K' -> cost = 0
Finger 2 on letter 'E' -> cost = Distance from letter 'K' to letter 'E' = 1
Total distance = 3
Python Solution
class Solution:
def minimumDistance(self, word: str) -> int:
def dist(a: int, b: int) -> int:
x1, y1 = divmod(a, 6)
x2, y2 = divmod(b, 6)
return abs(x1 - x2) + abs(y1 - y2)
n = len(word)
f = [[[inf] * 26 for _ in range(26)] for _ in range(n)]
for j in range(26):
f[0][ord(word[0]) - ord('A')][j] = 0
f[0][j][ord(word[0]) - ord('A')] = 0
for i in range(1, n):
a, b = ord(word[i - 1]) - ord('A'), ord(word[i]) - ord('A')
d = dist(a, b)
for j in range(26):
f[i][b][j] = min(f[i][b][j], f[i - 1][a][j] + d)
f[i][j][b] = min(f[i][j][b], f[i - 1][j][a] + d)
if j == a:
for k in range(26):
t = dist(k, b)
f[i][b][j] = min(f[i][b][j], f[i - 1][k][a] + t)
f[i][j][b] = min(f[i][j][b], f[i - 1][a][k] + t)
a = min(f[n - 1][ord(word[-1]) - ord('A')])
b = min(f[n - 1][j][ord(word[-1]) - ord('A')] for j in range(26))
return int(min(a, b))
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.