Simple Java Solution


  • 0

    the idea is: if both characters are same, then pass the dp[i-1][j-1] since no more deletion are needed; if not, dp[i][j] should be the minimum of previous two: delete from word1 or delete from word2, plus the current deletion 1.

    public class Solution {
        public int minDistance(String word1, String word2) {
            int m = word1.length(), n = word2.length();
            int[][] dp = new int[m + 1][n + 1];
            for (int i = 1; i <= m; i++) {
                dp[i][0] = i;
            }
            for (int j = 1; j <= n; j++) {
                dp[0][j] = j;
            }
            for (int i = 1; i <= m; i++) {
                for (int j = 1; j <= n; j++) {
                    if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                        dp[i][j] = dp[i - 1][j - 1];
                    } else {
                        dp[i][j] = Math.min(dp[i - 1][j], dp[i][j - 1]) + 1;
                    }
                }
            }
            return dp[m][n];
        }
    }
    

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