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| 1 | +// Time: O(m * n) |
| 2 | +// Space: O(n) |
| 3 | + |
| 4 | +// DP with rolling window. |
| 5 | +class Solution { |
| 6 | +public: |
| 7 | + int minimumDeleteSum(string s1, string s2) { |
| 8 | + vector<vector<int>> dp(2, vector<int>(s2.length() + 1)); |
| 9 | + for (int j = 0; j < s2.length(); ++j) { |
| 10 | + dp[0][j + 1] = dp[0][j] + s2[j]; |
| 11 | + } |
| 12 | + |
| 13 | + for (int i = 0; i < s1.length(); ++i) { |
| 14 | + dp[(i + 1) % 2][0] = dp[i % 2][0] + s1[i]; |
| 15 | + for (int j = 0; j < s2.length(); ++j) { |
| 16 | + if (s1[i] == s2[j]) { |
| 17 | + dp[(i + 1) % 2][j + 1] = dp[i % 2][j]; |
| 18 | + } else { |
| 19 | + dp[(i + 1) % 2][j + 1] = min(dp[i % 2][j + 1] + s1[i], |
| 20 | + dp[(i + 1) % 2][j] + s2[j]); |
| 21 | + } |
| 22 | + } |
| 23 | + } |
| 24 | + |
| 25 | + return dp[s1.length() % 2][s2.length()]; |
| 26 | + } |
| 27 | +}; |
| 28 | + |
| 29 | +// Time: O(m * n) |
| 30 | +// Space: O(m * n) |
| 31 | +class Solution2 { |
| 32 | +public: |
| 33 | + int minimumDeleteSum(string s1, string s2) { |
| 34 | + vector<vector<int>> dp(s1.length() + 1, vector<int>(s2.length() + 1)); |
| 35 | + for (int i = 0; i < s1.length(); ++i) { |
| 36 | + dp[i + 1][0] = dp[i][0] + s1[i]; |
| 37 | + } |
| 38 | + for (int j = 0; j < s2.length(); ++j) { |
| 39 | + dp[0][j + 1] = dp[0][j] + s2[j]; |
| 40 | + } |
| 41 | + |
| 42 | + for (int i = 0; i < s1.length(); ++i) { |
| 43 | + for (int j = 0; j < s2.length(); ++j) { |
| 44 | + if (s1[i] == s2[j]) { |
| 45 | + dp[i + 1][j + 1] = dp[i][j]; |
| 46 | + } else { |
| 47 | + dp[i + 1][j + 1] = min(dp[i][j + 1] + s1[i], |
| 48 | + dp[i + 1][j] + s2[j]); |
| 49 | + } |
| 50 | + } |
| 51 | + } |
| 52 | + |
| 53 | + return dp[s1.length()][s2.length()]; |
| 54 | + } |
| 55 | +}; |
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