3316. Find Maximum Removals From Source String

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3316. Find Maximum Removals From Source String

Description

You are given a string source of size n, a string pattern that is a subsequence of source, and a sorted integer array targetIndices that contains distinct numbers in the range [0, n - 1].

We define an operation as removing a character at an index idx from source such that:

  • idx is an element of targetIndices.
  • pattern remains a subsequence of source after removing the character.

Performing an operation does not change the indices of the other characters in source. For example, if you remove 'c' from "acb", the character at index 2 would still be 'b'.

Return the maximum number of operations that can be performed.

Example 1:

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Input: source = "abbaa", pattern = "aba", targetIndices = [0,1,2]

Output: 1

Explanation:

We can’t remove source[0] but we can do either of these two operations:

  • Remove source[1], so that source becomes "a_baa".
  • Remove source[2], so that source becomes "ab_aa".

Example 2:

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Input: source = "bcda", pattern = "d", targetIndices = [0,3]

Output: 2

Explanation:

We can remove source[0] and source[3] in two operations.

Example 3:

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Input: source = "dda", pattern = "dda", targetIndices = [0,1,2]

Output: 0

Explanation:

We can’t remove any character from source.

Example 4:

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Input: source = "yeyeykyded", pattern = "yeyyd", targetIndices = [0,2,3,4]

Output: 2

Explanation:

We can remove source[2] and source[3] in two operations.

Constraints:

  • 1 <= n == source.length <= 3 * 10^3
  • 1 <= pattern.length <= n
  • 1 <= targetIndices.length <= n
  • targetIndices is sorted in ascending order.
  • The input is generated such that targetIndices contains distinct elements in the range [0, n - 1].
  • source and pattern consist only of lowercase English letters.
  • The input is generated such that pattern appears as a subsequence in source.

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    // the meaning of dp[i][j]:
    //  when we are at ith index of source and jth index of pattern
    //  the maximum number of removals we can achieve later
    vector<vector<int>> dp;
    string source_, pattern_;

    int maxRemovals(string source, string pattern, vector<int>& targetIndices) {
        int m = source.size(), n = pattern.size();
        set<int> targets(targetIndices.begin(), targetIndices.end());
        source_ = source; pattern_ = pattern;
        dp.resize(m, vector<int>(n, -1));
        int res = dfs(0, 0, targets);
        return res;
    }

    int dfs(int idx1, int idx2, set<int>& targets) {
        // all pattern matched, all deletions afterwards can be applied
        if (idx2 == pattern_.size()) {
            return distance(targets.lower_bound(idx1), targets.end());
        }
        // we cannot match the pattern, return INT_MIN as a sign of failure
        // any value larger or equal to target size works here
        if (idx1 == source_.size()) {
            return INT_MIN;
        }
        if (dp[idx1][idx2] != -1) {
            return dp[idx1][idx2];
        }
        int res = 0;
        // delete or just skip
        if (targets.contains(idx1)) {
            res = 1 + dfs(idx1 + 1, idx2, targets);
        } else {
            res = dfs(idx1 + 1, idx2, targets);
        }
        // match
        if (source_[idx1] == pattern_[idx2]) {
            res = max(res, dfs(idx1 + 1, idx2 + 1, targets));
        }
        dp[idx1][idx2] = res;
        return res;
    }
};