0712 Minimum ASCII Delete Sum for Two Strings

0712 Minimum ASCII Delete Sum for Two Strings#

Problem#

Given two strings s1 and s2, return the lowest ASCII sum of deleted characters to make two strings equal.

examples#

Example 1:

Input: s1 = "sea", s2 = "eat"
Output: 231
Explanation: Deleting "s" from "sea" adds the ASCII value of "s" (115) to the sum.
Deleting "t" from "eat" adds 116 to the sum.
At the end, both strings are equal, and 115 + 116 = 231 is the minimum sum possible to achieve this.

Example 2:

Input: s1 = "delete", s2 = "leet"
Output: 403
Explanation: Deleting "dee" from "delete" to turn the string into "let",
adds 100[d] + 101[e] + 101[e] to the sum.
Deleting "e" from "leet" adds 101[e] to the sum.
At the end, both strings are equal to "let", and the answer is 100+101+101+101 = 403.
If instead we turned both strings into "lee" or "eet", we would get answers of 433 or 417, which are higher.

Constraint#

1 <= s1.length, s2.length <= 1000
s1 and s2 consist of lowercase English letters.

Analysis#

A typical dynamic programming problem:

Define a DP table of size (m+1, n+1) so that dp[i][j] stores the minimum delete sum for s1[:i] and s2[:j]

  • base case: dp[0][:] and dp[:][0]

  • choice: delete s1[i] or delete s2[j] or skip

  • state: minimum delete sum

  • state change:

For example:

for s1 = “sea’, s2 = “eat”, the dp table should be like:

        0     1     2      3
        x     e     a      t
0   x   0     e    e+a   e+a+t
1   s   s    s+e  e+a+s e+a+t+s
2   e  s+e    s    s+a   s+a+e
3   a s+e+a  s+a    s     s+t

We can get state change as:

# if the ending characters are the same, then the skip editing (deletion)
if s1[i-1] == s2[j-1]:
    dp[i][j] = dp[i-1][j-1]
# else select the minimum delete sum of all possible edits
else:
    dp[i][j] = min(
                dp[i-1][j] + s1[i-1], # delete s1[i-1]
                dp[i][j-1] + s2[j-1], # delete s2[j-1]
    )

Solution#

def minimumDeleteSum(s1, s2):
    m, n = len(s1), len(s2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    # base case
    for i in range(1, m + 1):
        dp[i][0] = dp[i - 1][0] + ord(s1[i - 1])
    for j in range(1, n + 1):
        dp[0][j] = dp[0][j - 1] + ord(s2[j - 1])
    # iterative dp table
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if s1[i-1] == s2[j-1]:
                dp[i][j] = dp[i-1][j-1]
            else:
                dp[i][j] = min(dp[i-1][j] + ord(s1[i-1]), dp[i][j-1] + ord(s2[j-1]))
    return dp[m][n]

# test
# 231
s1 = "sea"
s2 = "eat"
print(minimumDeleteSum(s1, s2))

# 403
s1 = "delete"
s2 = "leet"
print(minimumDeleteSum(s1, s2))

231
403
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