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/**
 * @file
 * @brief Implementation of [Minimum Edit Distance](https://en.wikipedia.org/wiki/Edit_distance) using Dynamic Programing
 *
 * @details
 *
 * Given two strings str1 & str2 and we have to calculate the minimum
 * number of operations (Insert, Remove, Replace) required to convert
 * str1 to str2.
 *
 * ### Algorithm
 *
 * We will solve this problem using Naive recursion. But as we are
 * approaching with a DP solution. So, we will take a DP array to
 * store the solution of all sub-problems so that we don't have to
 * perform recursion again and again. Now to solve the problem, We
 * can traverse all characters from either right side of the strings
 * or left side. Suppose we will do it from the right side. So, there
 * are two possibilities for every pair of characters being traversed.
 * 1. If the last characters of two strings are the same, Ignore
 * the characters and get the count for the remaining string.
 * So, we get the solution for lengths m-1 and n-1 in a DP array.
 *
 * 2. Else, (If last characters are not the same), we will consider all
 * three operations (Insert, Remove, Replace) on the last character of
 * the first string and compute the minimum cost for all three operations
 * and take the minimum of three values in the DP array.
 * For Insert: Recur for m and n-1
 * For Remove: Recur for for m-1 and n
 * For Replace: Recur for for m-1 and n-1
 *
 * @author [Nirjas Jakilim](github.com/nirzak)
 */

#include <cassert>     /// for assert
#include <iostream>   /// for IO operations
#include <vector>    /// for std::vector
/**
 * @namespace dynamic_programming
 * @brief Dynamic Programming algorithms
 */

namespace dynamic_programming {

/**
 * @namespace Minimum Edit Distance
 * @brief Implementation of [Minimum Edit Distance](https://en.wikipedia.org/wiki/Edit_distance) algorithm
 */

namespace minimum_edit_distance {

/**
 * @brief Takes input of the cost of
 * three operations: Insert, Replace and Delete
 * and return the minimum cost among them.
 * @param x used to pass minimum cost of Insert operations
 * @param y used to pass minimum cost of Replace operations
 * @param z used to pass minimum cost of Delete operations
 * @returns x if `x` is the minimum value
 * @returns y if `y` is the minimum value
 * @returns z if `z` is the minimum value
 */
uint64_t min(uint64_t x, uint64_t y, uint64_t z) {
  if (x <= y && x <= z) {
    return x; /// returns x, if x is the minimum value
  }
  if (y <= x && y <= z) {
    return y; /// returns y, if y is the minimum value
  }
  else {
    return z; /// returns z if z is the minimum value
  }
}

/**
 * @brief Calculates and stores the result
 * of all the sub-problems, so that we don't have to recur to compute
 * the minimum cost of a particular operation if it is already
 * computed and stored in the `dp` vector.
 * @param dp vector to store the computed minimum costs
 * @param str1 to pass the 1st string
 * @param str2 to pass the 2nd string
 * @param m the length of str1
 * @param n the length of str2
 * @returns dp[m][n] the minimum cost of operations
 * needed to convert str1 to str2
 */
uint64_t editDistDP(std::string str1, std::string str2, uint64_t m, uint64_t n) {
  /// Create a table to store results of subproblems
  std::vector<std::vector<uint64_t>>dp(m+1, std::vector<uint64_t>(n+1)); /// creasting 2D vector dp to store the results of subproblems

  /// Fill d[][] in bottom up manner
  for (uint64_t i = 0; i <= m; i++) {
    for (uint64_t j = 0; j <= n; j++) {
      /// If first string is empty, only option is to
      /// insert all characters of second string
      if (i == 0) {
        dp[i][j] = j; /// Minimum operations = j
      }

      /// If second string is empty, only option is to
      /// remove all characters of second string
      else if (j == 0) {
        dp[i][j] = i; /// Minimum operations = i
      }

      /// If last characters are same, ignore last char
      /// and recur for remaining string
      else if (str1[i - 1] == str2[j - 1]) {
        dp[i][j] = dp[i - 1][j - 1];
      }

      /// If the last character is different, consider all
      /// possibilities and find the minimum
      else {
        dp[i][j] = 1 + min(dp[i][j - 1],      // Insert
                           dp[i - 1][j],      // Remove
                           dp[i - 1][j - 1]); // Replace
      }
    }
  }

  return dp[m][n]; /// returning the minimum cost of operations needed to convert str1 to str2
}
}  // namespace minimum_edit_distance
}  // namespace dynamic_programming

/**
 * @brief Self-test implementations
 * @returns void
 */
static void test() {
  // 1st test
  std::string str1 = "INTENTION"; // Sample input of 1st string
  std::string str2 = "EXECUTION"; // Sample input of 2nd string
  uint64_t expected_output1 = 5; // Expected minimum cost
  uint64_t output1 = dynamic_programming::minimum_edit_distance::editDistDP(
      str1, str2, str1.length(), str2.length()); // calling the editDistDP function and storing the result on output1
  assert(output1 == expected_output1); // comparing the output with the expected output
  std::cout << "Minimum Number of Operations Required: " << output1
            << std::endl;

  // 2nd test
  std::string str3 = "SATURDAY";
  std::string str4 = "SUNDAY";
  uint64_t expected_output2 = 3;
  uint64_t output2 = dynamic_programming::minimum_edit_distance::editDistDP(
      str3, str4, str3.length(), str4.length());
  assert(output2 == expected_output2);
  std::cout << "Minimum Number of Operations Required: " << output2
            << std::endl;
}

/**
 * @brief main function
 * @param argc commandline argument count (ignored)
 * @param argv commandline array of arguments (ignored)
 * @returns 0 on exit
 */
int main(int argc, char *argv[]) {
  test();  // run self-test implementations
  return 0;
}

Minimum Edit Distance