Problem

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.

Examples

Example 1:

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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:

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

Solution

Method 1 - Using LCS DP

To make them identical, just find the longest common subsequence. The rest of the characters have to be deleted from the both the strings, which does not belong to longest common subsequence.

Code

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public int minDistance(String word1, String word2) {
	int m = word1.length();
	int n = word2.length();
	int[][] dp = new int[m + 1][n + 1];
	for (int i = 0; i <= m; i++) {
		for (int j = 0; j <= n; j++) {
			if (i == 0 || j == 0) {
				continue;
			}
			if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
				dp[i][j] = 1 + dp[i - 1][j - 1];
			} else {
				dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]);
			}
		}
	}
	int val =  dp[m][n];
	return m - val + n - val; // m + n - 2 * val
}

Complexity

  • ⏰ Time complexity: O(m * n)
  • 🧺 Space complexity: O(m * n)