Minimum Distance to Type a Word Using Two Fingers - Solution & Explanation

This approach involves using a dynamic programming technique to keep track of the minimum cost for typing the word with two fingers. We define a function that returns the minimum cost to type from a given position with a specific state defined by finger positions.

The position of each character on the keyboard can be represented in terms of X and Y coordinates. Movements from character to character can thus be calculated using Manhattan distance.

We use a memoization technique to save previously computed states to avoid redundant calculations, ensuring efficiency.

This code uses a recursive function `dp_solve` which takes the current index `i` and the positions of the two fingers `f1`, `f2` and returns the minimum cost to type the word from that point onwards. The base case is when `i` equals the length of the word, returning 0 as there's nothing left to type.

The function `dist` calculates the Manhattan distance between two positions. The memoization technique uses a dictionary `dp` to store already computed results for specific states (i, f1, f2), thus reducing redundant computations.

This approach is a variation of the dynamic programming solution, using a two-dimensional array to maintain a tabulation of results instead of recursion with memoization. We precompute distances and use a bottom-up DP approach to fill our dp table.

At each step, we decide the best possible move for each finger to type the character at the current index with minimum cost.

This C++ solution employs a bottom-up dynamic programming approach, using a matrix to track the minimum cost of reaching each letter with each finger. We iteratively build the solution by updating this matrix based on potential movements from one letter to the next.

We define f[i][j][k] to represent the minimum distance after typing word[i], with finger 1 at position j and finger 2 at position k. Here, positions j and k represent the numbers corresponding to the letters, ranging from [0,..25]. Initially, f[i][j][k] = infty.

We implement a function dist(a, b) to represent the distance between positions a and b, i.e., dist(a, b) = |\frac{a}{6} - \frac{b}{6}| + |a bmod 6 - b bmod 6|.

Next, we consider typing word[0], i.e., the case with only one letter. There are two choices:

• Finger 1 is at the position of word[0], and finger 2 is at any position. In this case, f[0][word[0]][k] = 0, where k \in [0,..25].
• Finger 2 is at the position of word[0], and finger 1 is at any position. In this case, f[0][k][word[0]] = 0, where k \in [0,..25].

Then we consider typing word[1,..n-1]. Let the positions of the previous letter and the current letter be a and b, respectively. Next, we discuss the following cases:

If the current finger 1 is at position b, we enumerate the position j of finger 2. If the previous position a was also the position of finger 1, then f[i][b][j] = min(f[i][b][j], f[i-1][a][j] + dist(a, b)). If the position of finger 2 is the same as the previous position a, i.e., j = a, then we enumerate the position k of finger 1 in the previous position. In this case, f[i][b][j] = min(f[i][b][j], f[i-1][k][a] + dist(k, b)).

Similarly, if the current finger 2 is at position b, we enumerate the position j of finger 1. If the previous position a was also the position of finger 2, then f[i][j][b] = min(f[i][j][b], f[i-1][j][a] + dist(a, b)). If the position of finger 1 is the same as the previous position a, i.e., j = a, then we enumerate the position k of finger 2 in the previous position. In this case, f[i][j][b] = min(f[i][j][b], f[i-1][a][k] + dist(k, b)).

Finally, we enumerate the positions of finger 1 and finger 2 at the last position and take the minimum value as the answer.

The time complexity is O(n times |\Sigma|^2), and the space complexity is O(n times |\Sigma|^2). Here, n is the length of the string word, and |\Sigma| is the size of the alphabet, which is 26 in this problem.