Number Of Ways To Form A Target String Given A Dictionary

Solution For Number Of Ways To Form A Target String Given A Dictionary

Problem Statement:
Given a list of strings “words” and a string “target”, find the number of ways that a target string can be formed by concatenating words from the list words, where each word can be used any number of times.

Example:
Input: words = [“abc”, “ab”, “ac”], target = “abc”
Output: 4
Explanation:
The possible ways to form “abc” are:
– “abc” (using “abc”)
– “ab” + “c” (using “ab” and “c”)
– “ac” + “b” (using “ac” and “b”)
– “a” + “b” + “c” (using “ab” and “ac”)

Solution:
This problem can be solved using dynamic programming. Let dp[i] be the number of ways to form a prefix of target of length i, and dp[0] is initialized as 1. We can then iterate through all possible prefixes of target and check if any word in the dictionary can be used to form that prefix. If a word can be used for a given prefix, we can add the number of ways to form the remaining prefix (which is target[0:i-len(word)]) to dp[i].

For example, for the input example above, the dp array would be:
dp[0] = 1 (empty prefix)
dp[1] = 1 (“a” can form “a”)
dp[2] = 2 (“ab” can be formed using “ab” or “a” and “b”)
dp[3] = 4 (“abc” can be formed using “abc”, “ab” and “c”, “ac” and “b”, or “a”, “b”, and “c”)

The final answer is dp[len(target)].

Here’s the Python code for the solution:

def numWays(words, target):
dp = [1] + [0] * len(target)
for i in range(len(target)):
for word in words:
if len(word) <= i+1 and target[i+1-len(word):i+1] == word:
dp[i+1] += dp[i-len(word)+1] return dp[len(target)]

The time complexity of this solution is O(len(target) * len(words) * max_word_length), where max_word_length is the maximum length of a word in words.

Step by Step Implementation For Number Of Ways To Form A Target String Given A Dictionary

There are several ways to solve this problem:

1) Use a HashMap to keep track of the number of ways to form each target string. For each string in the dictionary, check if it can be formed by concatenating other strings in the dictionary. If so, update the count for the target string in the HashMap.

2) Use a Trie to keep track of all the strings in the dictionary. For each string in the dictionary, check if it can be formed by concatenating other strings in the dictionary. If so, update the count for the target string in the Trie.

3) Use a Set to keep track of all the strings in the dictionary. For each string in the dictionary, check if it can be formed by concatenating other strings in the dictionary. If so, update the count for the target string in the Set.
Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

The same word in the dictionary may be reused multiple times in the segmentation.
You may assume the dictionary does not contain duplicate words.
Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".
Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.
Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false
There are a few ways to solve this problem. One way would be to use a brute force approach and try every possible combination of characters from the dictionary until you find a match with the target string. However, this could be very time-consuming and is not very efficient.

A better way to solve this problem would be to use a dynamic programming approach. We can create a 2D array where each cell represents the number of ways to form the target string using the characters up to that index. Then, we can fill in the array by iterating through the characters in the target string and the dictionary. For each character in the target string, we will check to see if it is in the dictionary. If it is, we will update the number of ways to form the target string using that character.

For example, if the target string is "abc" and the dictionary contains the words "a", "b", and "c", then we would have the following 2D array:

[

[1, 0, 0], // ways to form "a"

[1, 1, 0], // ways to form "ab"

[1, 1, 1]  // ways to form "abc"

]

If the target string is "abcd" and the dictionary contains the words "a", "b", "c", and "d", then we would have the following 2D array:

[

[1, 0, 0, 0], // ways to form "a"

[1, 1, 0, 0], // ways to form "ab"

[1, 1, 1, 0], // ways to form "abc"

[1, 1, 1, 1]  // ways to form "abcd"

]

The number of ways to form the target string "abcd" would be the value in the last cell of the array, which is 1.
There are several ways to solve this problem in C++. One approach would be to use a hashtable to keep track of all the words in the dictionary, and then use a backtracking algorithm to try to construct the target string using the words in the hashtable. Another approach would be to use a Trie data structure to store the dictionary words, and then use a recursive algorithm to try to construct the target string using the words in the Trie.
There are many ways to solve this problem, but one way to do it would be to use a hashtable to keep track of all the words in the dictionary, and then use a recursive function to try to find all the ways to form the target string.

First, we would create a hashtable that maps each letter to the set of words that start with that letter. Then, we would create a recursive function that takes in the current index of the target string, the current word, and the current set of words that we have used.

If the current index is equal to the length of the target string, then we know we have found a valid way to form the string, so we would add the current word to our list of ways to form the string.

Otherwise, we would loop through all the letters in the current word, and for each letter, we would get the set of words that start with that letter from our hashtable. Then, for each word in that set, we would call our recursive function with the next index of the target string, the current word concatenated with the new word, and the current set of words concatenated with the new word.

This should give us all the ways to form the target string.


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