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Copy pathQ13_EditDistanceProblem.cpp
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Copy pathQ13_EditDistanceProblem.cpp
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187 lines (160 loc) · 5.88 KB
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#include<iostream>
#include<vector>
using namespace std;
int solveByRecursion(string txt1 ,string txt2, int i , int j){
//base cases
if( i == txt1.length()){
return txt2.length() - j ;
}
if( j == txt2.length()){
return txt1.length() - i;
}
//initializing the ans
int ans = 0;
//logic
if(txt1[i] == txt2[j]){
ans = solveByRecursion(txt1, txt2, i+1 , j+1);
}
else{
// one operation cost 1 unit
int insert = 1+ solveByRecursion(txt1, txt2, i ,j+1);
int deleted = 1+ solveByRecursion(txt1, txt2, i+1, j );
int replaced = 1+ solveByRecursion(txt1, txt2 ,i+1 , j+1);
//we have to give minimum ways (units) ans thats why returning the min of all operations
ans = min(insert , min(deleted , replaced));
}
return ans;
}
int soleByMemoisation(string& txt1 ,string& txt2, int i , int j,vector<vector<int> > &dp ){
//base cases
if( i == txt1.length()){
return txt2.length() - j ;
}
if( j == txt2.length()){
return txt1.length() - i;
}
// step3 : cheking if the ANS IS ALREADY EXIST
if(dp[i][j] != -1){
return dp[i][j];
}
//initializing the ans
int ans = 0;
//logic
if(txt1[i] == txt2[j]){
ans = soleByMemoisation(txt1, txt2, i+1 , j+1,dp);
}
else{
// one operation cost 1 unit
int insert = 1+ soleByMemoisation(txt1, txt2, i ,j+1,dp );
int deleted = 1+ soleByMemoisation(txt1, txt2, i+1, j ,dp );
int replaced = 1+ soleByMemoisation(txt1, txt2 ,i+1 , j+1 ,dp);
//we have to give minimum ways (units) ans thats why returning the min of all operations
ans = min(insert , min(deleted , replaced));
}
// step2: storing the ans to the dp arrat
dp[i][j] = ans;
return dp[i][j];
}
int solveByTabulation(string txt1, string txt2){
//step1 : creating the dpAray
vector<vector<int> > dp(txt1.length()+1 , vector<int>(txt2.length()+1 ,0));
//analizing the base cass
for(int j = 0 ;j<= txt2.length() ; j++){
dp[txt1.length()][j] = txt2.length() -j;
}
for(int i =0; i<= txt1.length();i++){
dp[i][txt2.length()] = txt1.length() -i;
}
//settting up the flow of the code
// NOT FORGET TO START THE i FROM .length()-1
for(int i = txt1.length()-1 ; i>=0;i--){
for(int j = txt2.length()-1 ; j>=0 ;j--){
//logic
//initializing the ans
int ans = 0;
//logic
if(txt1[i] == txt2[j]){
ans = dp[ i+1] [ j+1];
}
else{
// one operation cost 1 unit
int insert = 1+ dp[ i ][j+1];
int deleted = 1+ dp[ i+1][ j];
int replaced = 1+ dp[i+1] [ j+1];
//we have to give minimum ways (units) ans thats why returning the min of all operations
ans = min(insert , min(deleted , replaced));
}
// step2: storing the ans to the dp arrat
dp[i][j] = ans;
}
}
return dp[0][0];
}
int SpaceOptimisedSolution(string txt1, string txt2){
//by two one dimentional dparrays
vector<int>curr(txt2.length()+1 , 0);
vector<int>next(txt2.length()+1 , 0);
//GIVE SPECIAL ATTENTION ON THE BASE CASES
for(int j = 0 ;j<= txt2.length() ; j++){
next[j] = txt2.length() -j;
}
//settting up the flow of the code
// NOT FORGET TO START THE i FROM .length()-1
for(int i = txt1.length()-1 ; i>=0;i--){
//every row starts here
curr[txt2.length()] = txt1.length()-i;
for(int j = txt2.length()-1 ; j>=0 ;j--){
//logic
//initializing the ans
int ans = 0;
//logic
if(txt1[i] == txt2[j]){
ans = next[ j+1];
}
else{
// one operation cost 1 unit
int insert = 1+ curr[j+1];
int deleted = 1+ next[ j];
int replaced = 1+ next [ j+1];
//we have to give minimum ways (units) ans thats why returning the min of all operations
ans = min(insert , min(deleted , replaced));
}
// step2: storing the ans to the dp arrat
curr[j] = ans;
}
//SHIFTING IS IMPORTANT
next = curr;
}
//returning the answer
return next[0];
}
int main(){
// string word1 ="horse";
// string word2 ="ros";
string word1 ="intention";
string word2 ="execution";
//solution
if(word1.length() ==0){
return word2.length();
}
if(word2.length() == 0){
return word1.length();
}
int i = 0 ;
int j = 0;
//methhod 1
int ansByRec = solveByRecursion(word1 ,word2, i , j);
cout<<"The answer of recursive solution is: "<<ansByRec<<endl;
//method 2: solving by memoisation method
//step1: creating the dp array (vector)
vector<vector<int> > dp(word1.length()+1 , vector<int>(word2.length() , -1));
int ansByMemoisation = soleByMemoisation(word1, word2, i , j , dp);
cout<<"The answer of Memoisation solution is: "<<ansByMemoisation<<endl;
//method 3: the tabulaton method
int ansByTabulation = solveByTabulation(word1, word2);
cout<<"the answer of tabulaton solution is : "<<ansByTabulation<<endl;
//Method 4: solving by space optimised
int ansBySpaceOptimisation = SpaceOptimisedSolution(word1, word2);
cout<<"the answer of space optimised solution is : "<< ansByMemoisation<<endl;
return 0;
}