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Edit Distance
Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)
You have the following 3 operations permitted on a word:
a) Insert a character
b) Delete a character
c) Replace a character
class Solution {public: int minDistance(string word1, string word2) { int l1 = word1.length(); int l2 = word2.length(); vector<vector<int>> dp(l1 + 1, vector<int>(l2 + 1, 0)); for (int i = 1; i <= l1; i++) { dp[i][0] = i; } for (int i = 1; i <= l2; i++) { dp[0][i] = i; } for (int i = 1; i <= l1; i++) { for (int j = 1; j <= l2; 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 - 1], min(dp[i - 1][j], dp[i][j - 1])) + 1; } } } return dp[l1][l2]; }};class Solution2 {public: int minDistance(string word1, string word2) { int l1 = word1.length(); int l2 = word2.length(); int dp[l2 + 1] { 0 }; for (int i = 1; i <= l2; i++) { dp[i] = i; } for (int i = 1; i <= l1; i++) { int pre = i; for (int j = 1; j <= l2; j++) { int cur; if(word1[i-1] == word2[j-1]){ cur = dp[j-1]; }else{ cur = min(dp[j-1],min(dp[j],pre))+1; } dp[j-1] = pre; pre = cur; } dp[l2] = pre; } return dp[l2]; }};
Edit Distance
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