What is the time complexity of Number of Ways to Form a Target String Given a Dictionary?

The optimal solution has a time complexity of O(m*n).

What data structures are used to solve Number of Ways to Form a Target String Given a Dictionary?

This problem uses Array, String, Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Number of Ways to Form a Target String Given a Dictionary?

Insight 1: Dynamic programming is crucial for efficiently exploring all possible combinations of character selections without redundant calculations. Insight 2: The restriction that once we use the k-th character of any word, no character at or to the left of k can be used again, allows us to process each column sequentially.

Which companies ask LeetCode 1639. Number of Ways to Form a Target String Given a Dictionary in interviews?

This problem is frequently asked at Dunzo, Snowflake.

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Number of Ways to Form a Target String Given a Dictionary - LeetCode 1639 Solution

Q: How to solve Number of Ways to Form a Target String Given a Dictionary?

The solution uses dynamic programming. We build a 2D array `dp` where `dp[i][j]` represents the number of ways to form the first `i` characters of the `target` string using the first `j` columns of the `words`. The key is to consider two options for each character in the `target` string: either we don't use the current column to form the character, or we use it. If we don't use the current column, the number of ways remains the same as using the previous columns. If we do use it, we multiply the number of ways to form the previous characters of the `target` with the count of the required character in the current column.

Number of Ways to Form a Target String Given a Dictionary - Complete Solution Guide

Number of Ways to Form a Target String Given a Dictionary is LeetCode problem 1639, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given a list of strings of the same length words and a string target . Your task is to form target using the given words under the following rules: target should be formed from left to right. To form the i th character ( 0-indexed ) of target , you can choose the k th character of the j th string in words if target[i] = words[j][k] . Once you use the k th character of the j th string of words , you can no longer use the x th character of any string in words where x <= k . In other words,

Detailed Explanation

The problem asks us to find the number of ways to form a target string by selecting characters from a given list of words. All words in the list have the same length. The target string is formed from left to right. When selecting a character at index 'k' from a word, all characters at indices less than or equal to 'k' in all words become unusable. The goal is to return the number of distinct ways to form the target string modulo 10^9 + 7.

Solution Approach

The solution uses dynamic programming. We build a 2D array `dp` where `dp[i][j]` represents the number of ways to form the first `i` characters of the `target` string using the first `j` columns of the `words`. The key is to consider two options for each character in the `target` string: either we don't use the current column to form the character, or we use it. If we don't use the current column, the number of ways remains the same as using the previous columns. If we do use it, we multiply the number of ways to form the previous characters of the `target` with the count of the required character in the current column.

Step-by-Step Algorithm

Step 1: Initialize a 2D array `counts` to store the frequency of each character (a-z) at each column index in the list of words.
Step 2: Initialize a 1D DP array `dp` of size `n+1` with all values set to 1. `dp[j]` represents the number of ways to form the target string up to the current character using the first `j` columns.
Step 3: Iterate through each character of the target string. For each character, create a new DP array `new_dp`.
Step 4: For each column index `j` from 0 to `n-1`, calculate `new_dp[j+1]` based on two options: not using column `j`, and using column `j` to match the current character of the target. If not using column `j`, `new_dp[j+1] = new_dp[j]`. If using column `j`, `new_dp[j+1] = (new_dp[j+1] + dp[j] * counts[j][char_code]) % MOD`, where `char_code` is the index of the target character in the alphabet.
Step 5: Update the `dp` array with the `new_dp` array after each iteration through the target string characters.
Step 6: Return `dp[n]`, which represents the total number of ways to form the entire target string using all columns of the words.

Key Insights

Insight 1: Dynamic programming is crucial for efficiently exploring all possible combinations of character selections without redundant calculations.
Insight 2: The restriction that once we use the k-th character of any word, no character at or to the left of k can be used again, allows us to process each column sequentially.
Insight 3: Counting the frequencies of characters at each column index of the words is necessary for efficiently determining the number of ways to match a particular character in the target string.

Complexity Analysis

Time Complexity: O(m*n)

Space Complexity: O(n)

Topics

This problem involves: Array, String, Dynamic Programming.

Companies

Asked at: Dunzo, Snowflake.

Number of Ways to Form a Target String Given a Dictionary | LeetCode 1639

Hard

Array String Dynamic Programming

Time: O(m*n)Space: O(n)

Solution Code

Available in:

class Solution {
    public int numWays(String[] words, String target) {
        int MOD = 1_000_000_007;
        int m = target.length();
        int n = words[0].length();

        int[][] counts = new int[n][26];
        for (String word : words) {
            for (int j = 0; j < word.length(); j++) {
                char charAtJ = word.charAt(j);
                counts[j][charAtJ - 'a']++;
            }
        }

        long[] dp = new long[n + 1];
        for (int i = 0; i <= n; i++) {
            dp[i] = 1;
        }

        for (int i = 0; i < m; i++) {
            long[] new_dp = new long[n + 1];
            int char_code = target.charAt(i) - 'a';
            for (int j = 0; j < n; j++) {
                // Option 1: Don't use column j
                // Ways to form target[:i+1] with j columns is the same as new_dp[j]
                new_dp[j + 1] = new_dp[j];

                // Option 2: Use column j to match target[i]
                // Ways to form target[:i] with j-1 columns * count of target[i] in column j
                long ways_using_col_j = (dp[j] * counts[j][char_code]);
                new_dp[j + 1] = (new_dp[j + 1] + ways_using_col_j) % MOD;
            }
            dp = new_dp;
        }

        return (int) dp[n];
    }
}

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Problem Info

Description

Problem 1639: Number of Ways to Form a Target String Given a Dictionary - Detailed explanation and solution approaches.

Topics

Array String Dynamic Programming

Companies

DunzoSnowflake

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:1639

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

6 steps

1
Step 1: Initialize a 2D array `counts` to store the frequency of each character (a-z) at each column index in the list of words.
2
Step 2: Initialize a 1D DP array `dp` of size `n+1` with all values set to 1. `dp[j]` represents the number of ways to form the target string up to the current character using the first `j` columns.
3
Step 3: Iterate through each character of the target string. For each character, create a new DP array `new_dp`.
4
Step 4: For each column index `j` from 0 to `n-1`, calculate `new_dp[j+1]` based on two options: not using column `j`, and using column `j` to match the current character of the target. If not using column `j`, `new_dp[j+1] = new_dp[j]`. If using column `j`, `new_dp[j+1] = (new_dp[j+1] + dp[j] * counts[j][char_code]) % MOD`, where `char_code` is the index of the target character in the alphabet.
5
Step 5: Update the `dp` array with the `new_dp` array after each iteration through the target string characters.
6
Step 6: Return `dp[n]`, which represents the total number of ways to form the entire target string using all columns of the words.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(m*n)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from the nested loops iterating through the target string and the columns of words. The space complexity comes primarily from the dp array, which stores intermediate results for each column.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach with Memoization

A recursive approach with memoization can also be used to solve this problem. It would involve defining a recursive function that takes the current index in the target string and the current column index as input, and returns the number of ways to form the remaining part of the target string. Memoization would be used to avoid recomputing the same subproblems.

Time: O(m*n)Space: O(m*n)

Trade-offs: The recursive approach with memoization would have the same time complexity as the iterative DP solution. However, it may have a higher space complexity due to the recursion stack and the memoization table, especially for larger input sizes. The iterative DP solution is generally preferred for its better space complexity and reduced overhead.

✨

Key Insights

1Insight 1: Dynamic programming is crucial for efficiently exploring all possible combinations of character selections without redundant calculations.
2Insight 2: The restriction that once we use the k-th character of any word, no character at or to the left of k can be used again, allows us to process each column sequentially.
3Insight 3: Counting the frequencies of characters at each column index of the words is necessary for efficiently determining the number of ways to match a particular character in the target string.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: When there is no way to form the target string, the algorithm will correctly return 0 because `dp` will be initialized to 1 and updated accordingly in the iterations.
Edge case 2: Modulo operation is used to prevent integer overflow, especially since the number of ways can be very large.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not taking the modulo at each step can lead to integer overflow, resulting in incorrect answers.

✗Common mistake 2: Incorrectly initializing the DP array can lead to subtle bugs in the calculation of the number of ways.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: This algorithm is related to string matching and sequence alignment, which are used in bioinformatics for comparing DNA sequences.

🏦Real-world scenario 2: It also finds application in text processing and data mining, where identifying patterns and forming target sequences from given data is crucial.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 13

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 6

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Number of Ways to Form a Target String Given a Dictionary"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether string techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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