TheAlgorithms-C-Plus-Plus/dynamic_programming/word_break.cpp

/**
 * @file
 * @brief [Word Break Problem](https://leetcode.com/problems/word-break/)
 * @details
 * 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
 *
 * @author [Akshay Anand] (https://github.com/axayjha)
 */

#include <cassert>
#include <climits>
#include <iostream>
#include <string>
#include <unordered_set>
#include <vector>

/**
 * @namespace dynamic_programming
 * @brief Dynamic programming algorithms
 */
namespace dynamic_programming {

/**
 * @namespace word_break
 * @brief Functions for [Word Break](https://leetcode.com/problems/word-break/)
 * problem
 */
namespace word_break {

/**
 * @brief Function that checks if the string passed in param is present in
 * the the unordered_set passed
 *
 * @param str the string to be searched
 * @param strSet unordered set of string, that is to be looked into
 * @returns `true` if str is present in strSet
 * @returns `false` if str is not present in strSet
 */
bool exists(const std::string &str,
            const std::unordered_set<std::string> &strSet) {
    return strSet.find(str) != strSet.end();
}

/**
 * @brief Function that checks if the string passed in param can be
 * segmented from position 'pos', and then correctly go on to segment the
 * rest of the string correctly as well to reach a solution
 *
 * @param s the complete string to be segmented
 * @param strSet unordered set of string, that is to be used as the
 * reference dictionary
 * @param pos the index value at which we will segment string and test
 * further if it is correctly segmented at pos
 * @param dp the vector to memoize solution for each position
 * @returns `true` if a valid solution/segmentation is possible by segmenting at
 * index pos
 * @returns `false` otherwise
 */
bool check(const std::string &s, const std::unordered_set<std::string> &strSet,
           int pos, std::vector<int> *dp) {
    if (pos == s.length()) {
        // if we have reached till the end of the string, means we have
        // segmented throughout correctly hence we have a solution, thus
        // returning true
        return true;
    }

    if (dp->at(pos) != INT_MAX) {
        // if dp[pos] is not INT_MAX, means we must have saved a solution
        // for the position pos; then return if the solution at pos is true
        // or not
        return dp->at(pos) == 1;
    }

    std::string wordTillNow =
        "";  // string to save the prefixes of word till different positons

    for (int i = pos; i < s.length(); i++) {
        // Loop starting from pos to end, to check valid set of
        // segmentations if any
        wordTillNow +=
            std::string(1, s[i]);  // storing the prefix till the position i

        // if the prefix till current position is present in the dictionary
        // and the remaining substring can also be segmented legally, then
        // set solution at position pos in the memo, and return true
        if (exists(wordTillNow, strSet) and check(s, strSet, i + 1, dp)) {
            dp->at(pos) = 1;
            return true;
        }
    }
    // if function has still not returned, then there must be no legal
    // segmentation possible after segmenting at pos
    dp->at(pos) = 0;  // so set solution at pos as false
    return false;     // and return no solution at position pos
}

/**
 * @brief Function that checks if the string passed in param can be
 * segmented into the strings present in the vector.
 * In others words, it checks if any permutation of strings in
 * the vector can be concatenated to form the final string.
 *
 * @param s the complete string to be segmented
 * @param wordDict a vector of words to be used as dictionary to look into
 * @returns `true` if s can be formed by a combination of strings present in
 * wordDict
 * @return `false` otherwise
 */
bool wordBreak(const std::string &s, const std::vector<std::string> &wordDict) {
    // unordered set to store words in the dictionary for constant time
    // search
    std::unordered_set<std::string> strSet;
    for (const auto &s : wordDict) {
        strSet.insert(s);
    }
    // a vector to be used for memoization, whose value at index i will
    // tell if the string s can be segmented (correctly) at position i.
    // initializing it with INT_MAX (which will denote no solution)
    std::vector<int> dp(s.length(), INT_MAX);

    // calling check method with position = 0, to check from left
    // from where can be start segmenting the complete string in correct
    // manner
    return check(s, strSet, 0, &dp);
}

}  // namespace word_break
}  // namespace dynamic_programming

/**
 * @brief Test implementations
 * @returns void
 */
static void test() {
    // the complete string
    const std::string s = "applepenapple";
    // the dictionary to be used
    const std::vector<std::string> wordDict = {"apple", "pen"};

    assert(dynamic_programming::word_break::wordBreak(s, wordDict));

    // should return true, as applepenapple can be segmented as apple + pen +
    // apple
    std::cout << dynamic_programming::word_break::wordBreak(s, wordDict)
              << std::endl;
    std::cout << "Test implementation passed!\n";
}
/**
 * @brief Main function
 * @returns 0 on exit
 */
int main() {
    test();  // call the test function :)

    // the complete string
    const std::string s = "applepenapple";
    // the dictionary to be used
    const std::vector<std::string> wordDict = {"apple", "pen"};

    // should return true, as applepenapple can be segmented as apple + pen +
    // apple
    std::cout << dynamic_programming::word_break::wordBreak(s, wordDict)
              << std::endl;
}