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Documentation and renaming variable

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Ashish Bhanu Daulatabad
2021-03-08 10:18:00 +05:30
parent 738c6afe77
commit 8b732a8729
1 changed files with 84 additions and 62 deletions
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146
dynamic_programming/abbreviation.cpp
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@@ -14,103 +14,117 @@
* string a and b (for character indexes i and j respectively):
* 1. If a[i] and b[j] are equal, then move to next position
* 2. If a[i] is lowercase of b[j], then explore two possibilities:
* a. Capitalize or
* a. Skip a[i]
* a. Capitalize or
* b. Skip a[i]
* 3. If the a[i] is not uppercase, just discard the char, else return false
*
* Time Complexity (O(|a|*|b|)) where |a| => length of string
* @author Ashish Daulatabad (https://github.com/AshishYUO)
*/
#include <cassert>
#include <iostream>
#include <string>
#include <vector>
#include <cassert> /// assert function
#include <iostream> /// input/output operations
#include <string> /// std::string library
#include <vector> /// std::vector STL library
/**
* @namespace dynamic_programming
* @brief Dynamic Programming Algorithms
*/
namespace dynamic_programming {
/**
* @namespace Knapsack
* @namespace abbreviation
* @brief Implementation of Abbreivation problem
*/
namespace abbreviation {
/**
* (recursive dp function) https://www.hackerrank.com/challenges/abbr/problem
* Returns whether s can be converted to t with following rules:
* a. Capitalize zero or more of a's lowercase letters from string s
* b. remove all other lowercase letters from string s
* @param dp: memo as parameter to store the result
* @param v: visited boolean to check if the result is already computed
* @param s: given string
* @param t: resultant abbreivated string
* @param i: start of string s
* @param j: start of string j
* @returns bool whether s can be converted to t
* @brief
* Recursive Dynamic Programming function
* @details
* Returns whether `s` can be converted to `t` with following rules:
* a. Capitalize zero or more of a's lowercase letters from string `s`
* b. remove all other lowercase letters from string `s`
* @param memo memo as parameter to store the result
* @param visited visited boolean to check if the result is already computed
* @param str given string, which might not be abbreivated
* @param result resultant abbreivated string
* @param str_idx pointer for string `str`, helpful for transitions
* @param result_idx pointer for string `result`, helpful for transitions
* @returns boolean (`true` or `false`) whether string `str` can be converted to
* `result`
*/
bool abbreviation_recursion(std::vector<std::vector<bool>> *memo,
std::vector<std::vector<bool>> *visited,
const std::string &s, const std::string &t,
int i = 0, int j = 0) {
bool ans = memo->at(i).at(j);
if (i == s.size() && j == t.size()) {
const std::string &str, const std::string &result,
int str_idx = 0, int result_idx = 0) {
bool ans = memo->at(str_idx).at(result_idx);
if (str_idx == str.size() && result_idx == result.size()) {
return true;
} else if (i == s.size() && j != t.size()) {
// result t is not converted, return false
} else if (str_idx == str.size() && result_idx != result.size()) {
// result `t` is not converted, return false
return false;
} else if (!visited->at(i).at(j)) {
} else if (!visited->at(str_idx).at(result_idx)) {
/**
* (s[i] == t[j]): if s char at position i is equal to t char at
* position j, then s character is a capitalized one, move on to next
* character
* s[i] - 32 == t[j]: if s[i] character is lowercase of t[j] then
* explore two possibilites:
* 1. convert it to capitalized and move both to next pointer (i + 1, j
* + 1)
* 2. Discard the character (s[i]) and move to next char (i + 1, j)
* `(str[i] == result[j])`: if str char at position i is equal to
* `result` char at position j, then s character is a capitalized one,
* move on to next character `str[i] - 32 == result[j]`:
* if `str[i]` character is lowercase of `result[j]` then explore two
* possibilites:
* 1. convert it to capitalized letter and move both to next pointer
* `(i + 1, j + 1)`
* 2. Discard the character `(str[i])` and move to next char `(i + 1,
* j)`
*/
if (s[i] == t[j]) {
ans = abbreviation_recursion(memo, visited, s, t, i + 1, j + 1);
} else if (s[i] - 32 == t[j]) {
ans = abbreviation_recursion(memo, visited, s, t, i + 1, j + 1) ||
abbreviation_recursion(memo, visited, s, t, i + 1, j);
if (str[str_idx] == result[result_idx]) {
ans = abbreviation_recursion(memo, visited, str, result,
str_idx + 1, result_idx + 1);
} else if (str[str_idx] - 32 == result[result_idx]) {
ans = abbreviation_recursion(memo, visited, str, result,
str_idx + 1, result_idx + 1) ||
abbreviation_recursion(memo, visited, str, result,
str_idx + 1, result_idx);
} else {
// if s[i] is uppercase, then cannot be converted, return false
// else s[i] is lowercase, only option is to discard this character
if (s[i] >= 'A' && s[i] <= 'Z') {
// if `str[i]` is uppercase, then cannot be converted, return
// `false`
// else `str[i]` is lowercase, only option is to discard this
// character
if (str[str_idx] >= 'A' && str[str_idx] <= 'Z') {
ans = false;
} else {
ans = abbreviation_recursion(memo, visited, s, t, i + 1, j);
ans = abbreviation_recursion(memo, visited, str, result,
str_idx + 1, result_idx);
}
}
}
(*memo)[i][j] = ans;
(*visited)[i][j] = true;
return (*memo)[i][j];
(*memo)[str_idx][result_idx] = ans;
(*visited)[str_idx][result_idx] = true;
return (*memo)[str_idx][result_idx];
}
/**
* (iterative dp function) https://www.hackerrank.com/challenges/abbr/problem:
* Returns whether s can be converted to t with following rules:
* a. Capitalize zero or more of a's lowercase letters from string s
* b. remove all other lowercase letters from string s
* @param s: string
* @param t: resultant string
* @returns boolean (true or false) whether s can be converted to t
* @brief
* Iterative Dynamic Programming function
* @details
* Returns whether `s` can be converted to `t` with following rules:
* a. Capitalize zero or more of s's lowercase letters from string `s`
* b. remove all other lowercase letters from string `s`
* Note: The transition states for iterative is similar to recursive as well
* @param str given string, which might not be abbreivated
* @param result resultant abbreivated string
* @returns boolean (`true` or `false`) whether string `str` can be converted to
* `result`
*/
bool abbreviation(const std::string &s, const std::string &t) {
std::vector<std::vector<bool>> memo(s.size() + 1,
std::vector<bool>(t.size() + 1, false));
for (int i = 0; i <= s.size(); ++i) memo[i][0] = true;
for (int i = 1; i <= t.size(); ++i) memo[0][i] = false;
for (int i = 1; i <= s.size(); ++i) {
for (int j = 1; j <= t.size(); ++j) {
if (s[i - 1] == t[j - 1]) {
bool abbreviation(const std::string &str, const std::string &result) {
std::vector<std::vector<bool>> memo(
str.size() + 1, std::vector<bool>(result.size() + 1, false));
for (int i = 0; i <= str.size(); ++i) memo[i][0] = true;
for (int i = 1; i <= result.size(); ++i) memo[0][i] = false;
for (int i = 1; i <= str.size(); ++i) {
for (int j = 1; j <= result.size(); ++j) {
if (str[i - 1] == result[j - 1]) {
memo[i][j] = memo[i - 1][j - 1];
} else if (s[i - 1] - 32 == t[j - 1]) {
} else if (str[i - 1] - 32 == result[j - 1]) {
memo[i][j] = (memo[i - 1][j - 1] || memo[i - 1][j]);
} else {
if (s[i - 1] >= 'A' && s[i - 1] <= 'Z') {
if (str[i - 1] >= 'A' && str[i - 1] <= 'Z') {
memo[i][j] = false;
} else {
memo[i][j] = memo[i - 1][j];
@@ -123,6 +137,10 @@ bool abbreviation(const std::string &s, const std::string &t) {
} // namespace abbreviation
} // namespace dynamic_programming
/**
* @brief Self test-implementations
* @returns void
*/
static void test() {
std::string s = "daBcd", t = "ABC";
std::vector<std::vector<bool>> memo(s.size() + 1,
@@ -158,6 +176,10 @@ static void test() {
assert(dynamic_programming::abbreviation::abbreviation(s, t) == true);
}
/**
* @brief Main function
* @returns 0 on exit
*/
int main() {
test();
return 0;