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edit_distance.cpp
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edit_distance.cpp
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/*
72. Edit Distance
Hard
Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.
You have the following 3 operations permitted on a word:
Insert a character
Delete a character
Replace a character
Example 1:
Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation:
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')
Example 2:
Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation:
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')
*/
//We use Dynamic Programming to find the minimum number of operations to convert word1 to word2
//Operations : 1) Insert a character 2) Delete character 3) Replace character
//We try to find out some pattern while filling in the DP 2D array
class Solution {
public:
int minDistance(string word1, string word2) {
int m=word1.length();
int n = word2.length();
int dp[m+1][n+1];
for(int i=0;i<=m;i++)
dp[i][0]=i;
for(int i=0;i<=n;i++)
dp[0][i]=i;
for(int i=1;i<=m;i++){
for(int j=1;j<=n;j++){
if(word1[i-1] == word2[j-1])
dp[i][j] = dp[i-1][j-1];
else
dp[i][j] = min(dp[i-1][j], min(dp[i][j-1], dp[i-1][j-1]))+1;
}
}
return dp[m][n];
}
};
/*
Ex : word1 : "horse" and word2 : "ros"
This is the 2D DP array we use :
r o s
0 1 2 3 --> (0,0) : Null -> Null - 0 opn; (0,1) : Null -> r - 1 opn; (0,2) : Null -> ro - 2 opn; (0,3) : Null -> ros : 3 opn
h 1 1 2 3 --> (1,0) : h -> Null - 1 opn; (1,1) : h->r : 1 opn; (1,2) : h->ro - 2 opn;
o 2 2 1 2
r 3 2 2 2
s 4 3 3 2
e 5 4 4 3
Replace (i-1,j-1) Remove (i-1,j) (downwards)
insert (i,j-1) (---->)
*/