Problem Description

In this problem, you are given a list of strings called words, where each string is of the same length, and a target string target. Your goal is to construct the target string character by character, from left to right, using characters from the strings in words. The key rules to follow while forming the target string are:

• You may select the character at position k from the jth string in words to form the ith character of target, if and only if target[i] matches words[j][k].
• After using the kth character from jth string of words, all characters at or to the left of position k in any string of words can no longer be used. This means you have to choose subsequent characters from positions greater than k.
• You can repeat this process of selecting characters from words until you completely form the target string.
• It is allowed to use multiple characters from the same string in words as long as they meet the above conditions.

The problem asks for the number of distinct ways to form the target from words. As the total number could be very large, you are required to return this number modulo 10^9 + 7.

Intuition

To solve this problem, we make use of dynamic programming or a depth-first search approach to count all possible ways to construct the target.

Firstly, we process the words array to calculate the frequency of each character at each position (count array cnt). This preprocessing step simplifies the dynamic programming or DFS implementation by giving quick access to the number of times a particular character appears in each position across all strings.

The next step is defining a recursive function, let's call it dfs(i, j), representing the number of ways to construct the substring of target starting at index i with the starting character position in words being j. Our base cases are when we've successfully formed the target string (all characters are chosen) or when we can no longer choose characters from words (out of bounds).

The transitions depend on our choices at each step:

• We can decide not to pick a character at position j in any string of words, in which case we simply move to the next character position in words (dfs(i, j + 1)).
• Alternatively, we can pick the character if it matches the current target character we want to form, and then move to the next character in both target and words, while also multiplying by the count of the selected character (dfs(i + 1, j + 1) * count).

Finally, we initiate our search/recursion from the beginning of the target string and the words array (dfs(0, 0)), and continuously add the number of ways at each new choice we can make.

In the case of dynamic programming, we employ a bottom-up approach to store and reuse the results of subproblems, using a 2D table f[i][j] to represent the number of ways to build up to the ith character of target considering characters upto jth position in the strings of words.

Overall, the problem is approached via breaking down the complex task of string construction into smaller subproblems, calculating the number of ways for each subproblem, and then using these to build up to the final solution.

Learn more about Dynamic Programming patterns.

Solution Approach

The problem lends itself nicely to a depth-first search (DFS) with memoization or dynamic programming to avoid redundant computations. In both the recursive (DFS) and iterative (dynamic programming) approach, a crucial step is preprocessing, and an essential data structure used is the count array cnt.

Preprocessing

Before we dive into building the target string, we calculate the frequency of each character at every position across all strings in words. We create a 2D array cnt of size n × 26, where n is the length of the strings in words. cnt[j][c] keeps track of how many times the character c appears in the jth position in all strings of words.

Recursive DFS with Memoization

In the DFS approach, we define a function dfs(i, j) that computes the number of ways to construct substring target[i...] given that we are currently looking to match characters starting from position j in the strings of words.

The recursion has two base cases:

• When i >= m, which means the entire target is formed, the number of ways is 1.
• When j >= n, which implies we can't select more characters from words and haven't formed the target, the number of ways is 0.

For other cases, we have two choices:

• We can opt not to choose a character from position j in words, so we explore further with dfs(i, j + 1).
• If we pick the character, we use the count of that character at position j and proceed to the next characters in both target and words (dfs(i + 1, j + 1) * cnt[j][target[i] - 'a']).

The results of these calls are added together to get the final answer for dfs(i, j) and memoized to optimize overlapping subproblems. The modulo operation ensures the answer does not overflow the limits.

Dynamic Programming

In dynamic programming, we also take advantage of the count array cnt. The dp array f[i][j] represents the number of ways to construct the first i characters of target while restricted to using up to the first j characters from each string in words.

We initialize f[0][j] = 1 for all 0 ≤ j ≤ n because there's always one way to construct an empty target regardless of words.

The dp transitions are straightforward: for each cell f[i][j], we consider two cases:

• Not selecting the j-th character from words: we just carry over the previous count f[i][j - 1].
• Selecting the j-th character: the number of ways increases by the number of ways to make the prefix target[...i - 1] while considering characters only up to j - 1th position multiplied by the count of the target[i - 1] at this position cnt[j - 1][target[i - 1] - 'a'].

The total number of ways for f[i][j] is the sum of the counts from the two cases, and we must again apply the modulo here. Once all cells are calculated, f[m][n] will contain the answer.

Both recursive and dynamic programming solutions ensure that we do not perform the same calculations repeatedly, and both require O(m * n) time and space, where m is the length of the target and n is the length of the strings in words.

Example Walkthrough

Let's go through an example to illustrate the solution approach described.

Suppose we have the following inputs:

• words = ["ax", "by", "cz"]
• target = "abc"

Our target is "abc", and our words list contains strings of the same length, i.e., 2.

Preprocessing Step

Firstly, we create the count array cnt from words. Since all strings in words are of length 2, we will have a 2 by 26 (number of letters in the alphabet) count matrix.

cnt would be as follows:

• For the first position (index 0) in our strings:
  • a appears once (in "ax"),
  • b appears once (in "by"),
  • c appears once (in "cz").
• For the second position (index 1) in our strings:
  • x, y, and z each appear once.

Our cnt array after processing:

cnt[0][0] = 1  // 'a' at position 0
cnt[0][1] = 1  // 'b' at position 0
cnt[0][2] = 1  // 'c' at position 0
...            // other letters at position 0 would be 0
cnt[1][23] = 1 // 'x' at position 1
cnt[1][24] = 1 // 'y' at position 1
cnt[1][25] = 1 // 'z' at position 1
...            // other letters at position 1 would be 0

Recursive DFS with Memoization

We define our dfs function to begin constructing the target.

For target[0] = 'a', the DFS explores:

• dfs(1, 1) since 'a' can be chosen from words[0] = "ax" at position 0. The count is cnt[0][0] = 1.

Next, for target[1] = 'b', from the new starting point j = 1, the DFS finds:

• dfs(2, 2) this time choosing 'b' from words[1] = "by" at position 1 (since 'b' doesn't appear at position 1 in "ax"). The count is cnt[1][1] = 1.

Finally, for target[2] = 'c':

• We would choose 'c' from words[2] = "cz" at position 2, which again has a count of cnt[2][2] = 1.

Since each step gives us a count of 1, and we can form the target with one way at each step, the DFS would end with a total of 1 way to create the target "abc".

Dynamic Programming

Using dynamic programming, we construct a DP table f that has m rows (target length + 1) and n columns (words string length + 1).

The base case is f[0][j] = 1 for all j, indicating that there is one way to construct an empty target.

The table is filled according to the discussed transitions. By examining choices at each step, the DP table for our example would look like:

f[0][0] = 1
f[0][1] = 1
f[0][2] = 1

From the transitions, we would find:
f[1][1] = f[0][0] * cnt[0]['a' - 'a'] = 1
f[1][2] = f[1][1]                  = 1
f[2][2] = f[1][1] * cnt[1]['b' - 'a'] = 1

Since f[2][2] gives the number of ways to form "ab" using up to the second position in strings of words and cnt[2]['c' - 'a'] = 1, the final count f[3][3] for forming the entire "abc" target will be f[2][2] * cnt[2]['c' - 'a'] = 1 * 1 = 1.

Hence, there is 1 distinct way to form target = "abc" using the strings in words. This result matches our DFS calculation.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(m * n * k), where m is the length of the target string, n is the length of the strings in the words array, and k is the number of recursive calls made, which does not exceed the length of the target string. Each function call iterates through the length of words[0], and there are m possible values for i and n possible values for j. The memoization through @cache ensures that each (i, j) pair's computation is done only once.

The space complexity is O(m * n) for the memoization cache, which stores a result for each pair (i, j). Since there are m possible values for i and n possible values for j, we have m * n pairs. Additionally, the cnt array uses O(n * 26) space, which simplifies to O(n) since 26 is a constant. As a result, the space complexity does not change due to the cnt array. The overall space complexity includes the stack space used by recursion calls, which could be O(m) in the worst case. Hence, the total space complexity is O(m * n) when considering the memoization and recursion stack space, where n is the number of indices in each word and m is the number of characters in the target.

Learn more about how to find time and space complexity quickly using problem constraints.