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583. Delete Operation for Two Strings

Description

Given two strings word1 and word2, return the minimum number of steps required to make word1 and word2 the same.

In one step, you can delete exactly one character in either string.

 

Example 1:

Input: word1 = "sea", word2 = "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Example 2:

Input: word1 = "leetcode", word2 = "etco"
Output: 4

 

Constraints:

  • 1 <= word1.length, word2.length <= 500
  • word1 and word2 consist of only lowercase English letters.

Solutions

Dynamic programming.

  • 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] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
}

  • class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vector<vector<int>> dp(m + 1, vector<int>(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[i - 1] == word2[j - 1])
                    dp[i][j] = dp[i - 1][j - 1];
                else
                    dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m][n];
    }
};

  • class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            dp[i][0] = i
        for j in range(1, n + 1):
            dp[0][j] = j
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1])
        return dp[-1][-1]


  • func minDistance(word1 string, word2 string) int {
	m, n := len(word1), len(word2)
	dp := make([][]int, m+1)
	for i := range dp {
		dp[i] = make([]int, n+1)
		dp[i][0] = i
	}
	for j := range dp[0] {
		dp[0][j] = j
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if word1[i-1] == word2[j-1] {
				dp[i][j] = dp[i-1][j-1]
			} else {
				dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])
			}
		}
	}
	return dp[m][n]
}

  • function minDistance(word1: string, word2: string): number {
    const m = word1.length;
    const n = word2.length;
    const dp = Array.from({ length: m + 1 }, () => Array(n + 1).fill(0));
    for (let i = 1; i <= m; i++) {
        for (let j = 1; j <= n; j++) {
            if (word1[i - 1] === word2[j - 1]) {
                dp[i][j] = dp[i - 1][j - 1] + 1;
            } else {
                dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
            }
        }
    }
    const max = dp[m][n];
    return m - max + n - max;
}


  • impl Solution {
    pub fn min_distance(word1: String, word2: String) -> i32 {
        let (m, n) = (word1.len(), word2.len());
        let (word1, word2) = (word1.as_bytes(), word2.as_bytes());
        let mut dp = vec![vec![0; n + 1]; m + 1];
        for i in 1..=m {
            for j in 1..=n {
                dp[i][j] = if word1[i - 1] == word2[j - 1] {
                    dp[i - 1][j - 1] + 1
                } else {
                    dp[i - 1][j].max(dp[i][j - 1])
                };
            }
        }
        let max = dp[m][n];
        (m - max + (n - max)) as i32
    }
}

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