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1898. Maximum Number of Removable Characters

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Description

You are given two strings s and p where p is a subsequence of s. You are also given a distinct 0-indexed integer array removable containing a subset of indices of s (s is also 0-indexed).

You want to choose an integer k (0 <= k <= removable.length) such that, after removing k characters from s using the first k indices in removable, p is still a subsequence of s. More formally, you will mark the character at s[removable[i]] for each 0 <= i < k, then remove all marked characters and check if p is still a subsequence.

Return the maximum k you can choose such that p is still a subsequence of s after the removals.

A subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

 

Example 1:

Input: s = "abcacb", p = "ab", removable = [3,1,0]
Output: 2
Explanation: After removing the characters at indices 3 and 1, "abcacb" becomes "accb".
"ab" is a subsequence of "accb".
If we remove the characters at indices 3, 1, and 0, "abcacb" becomes "ccb", and "ab" is no longer a subsequence.
Hence, the maximum k is 2.

Example 2:

Input: s = "abcbddddd", p = "abcd", removable = [3,2,1,4,5,6]
Output: 1
Explanation: After removing the character at index 3, "abcbddddd" becomes "abcddddd".
"abcd" is a subsequence of "abcddddd".

Example 3:

Input: s = "abcab", p = "abc", removable = [0,1,2,3,4]
Output: 0
Explanation: If you remove the first index in the array removable, "abc" is no longer a subsequence.

 

Constraints:

• 1 <= p.length <= s.length <= 105
• 0 <= removable.length < s.length
• 0 <= removable[i] < s.length
• p is a subsequence of s.
• s and p both consist of lowercase English letters.
• The elements in removable are distinct.

Solutions

Python3

class Solution:
    def maximumRemovals(self, s: str, p: str, removable: List[int]) -> int:
        def check(mid):
            m, n, i, j = len(s), len(p), 0, 0
            ids = set(removable[:mid])
            while i < m and j < n:
                if i not in ids and s[i] == p[j]:
                    j += 1
                i += 1
            return j == n

        left, right = 0, len(removable)
        while left < right:
            mid = (left + right + 1) >> 1
            if check(mid):
                left = mid
            else:
                right = mid - 1
        return left

Java

class Solution {
    public int maximumRemovals(String s, String p, int[] removable) {
        int left = 0, right = removable.length;
        while (left < right) {
            int mid = (left + right + 1) >> 1;
            if (check(s, p, removable, mid)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    private boolean check(String s, String p, int[] removable, int mid) {
        int m = s.length(), n = p.length(), i = 0, j = 0;
        Set<Integer> ids = new HashSet<>();
        for (int k = 0; k < mid; ++k) {
            ids.add(removable[k]);
        }
        while (i < m && j < n) {
            if (!ids.contains(i) && s.charAt(i) == p.charAt(j)) {
                ++j;
            }
            ++i;
        }
        return j == n;
    }
}

TypeScript

function maximumRemovals(s: string, p: string, removable: number[]): number {
    let left = 0,
        right = removable.length;
    while (left < right) {
        let mid = (left + right + 1) >> 1;
        if (isSub(s, p, new Set(removable.slice(0, mid)))) {
            left = mid;
        } else {
            right = mid - 1;
        }
    }
    return left;
}

function isSub(str: string, sub: string, idxes: Set<number>): boolean {
    let m = str.length,
        n = sub.length;
    let i = 0,
        j = 0;
    while (i < m && j < n) {
        if (!idxes.has(i) && str.charAt(i) == sub.charAt(j)) {
            ++j;
        }
        ++i;
    }
    return j == n;
}

C++

class Solution {
public:
    int maximumRemovals(string s, string p, vector<int>& removable) {
        int left = 0, right = removable.size();
        while (left < right) {
            int mid = left + right + 1 >> 1;
            if (check(s, p, removable, mid)) {
                left = mid;
            } else {
                right = mid - 1;
            }
        }
        return left;
    }

    bool check(string s, string p, vector<int>& removable, int mid) {
        int m = s.size(), n = p.size(), i = 0, j = 0;
        unordered_set<int> ids;
        for (int k = 0; k < mid; ++k) {
            ids.insert(removable[k]);
        }
        while (i < m && j < n) {
            if (ids.count(i) == 0 && s[i] == p[j]) {
                ++j;
            }
            ++i;
        }
        return j == n;
    }
};

Go

func maximumRemovals(s string, p string, removable []int) int {
	left, right := 0, len(removable)
	for left < right {
		mid := (left + right + 1) >> 1
		if check(s, p, removable, mid) {
			left = mid
		} else {
			right = mid - 1
		}
	}
	return left
}

func check(s string, p string, removable []int, mid int) bool {
	m, n, i, j := len(s), len(p), 0, 0
	ids := make(map[int]bool)
	for k := 0; k < mid; k++ {
		ids[removable[k]] = true
	}
	for i < m && j < n {
		if !ids[i] && s[i] == p[j] {
			j++
		}
		i++
	}
	return j == n
}

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