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

Solution 1: Binary Search

We notice that if removing the characters at the first $k$ indices in $\textit{removable}$ still makes $p$ a subsequence of $s$, then removing the characters at $k \lt k' \leq \textit{removable.length}$ indices will also satisfy the condition. This monotonicity allows us to use binary search to find the maximum $k$.

We define the left boundary of the binary search as $l = 0$ and the right boundary as $r = \textit{removable.length}$. Then we perform binary search. In each search, we take the middle value $mid = \left\lfloor \frac{l + r + 1}{2} \right\rfloor$ and check if removing the characters at the first $mid$ indices in $\textit{removable}$ still makes $p$ a subsequence of $s$. If it does, we update the left boundary $l = mid$; otherwise, we update the right boundary $r = mid - 1$.

After the binary search ends, we return the left boundary $l$.

The time complexity is $O(k \times \log k)$, and the space complexity is $O(n)$. Here, $n$ is the length of the string $s$, and $k$ is the length of $\textit{removable}$.

Python3

class Solution:
    def maximumRemovals(self, s: str, p: str, removable: List[int]) -> int:
        def check(k: int) -> bool:
            rem = [False] * len(s)
            for i in removable[:k]:
                rem[i] = True
            i = j = 0
            while i < len(s) and j < len(p):
                if not rem[i] and p[j] == s[i]:
                    j += 1
                i += 1
            return j == len(p)

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

Java

class Solution {
    private char[] s;
    private char[] p;
    private int[] removable;

    public int maximumRemovals(String s, String p, int[] removable) {
        int l = 0, r = removable.length;
        this.s = s.toCharArray();
        this.p = p.toCharArray();
        this.removable = removable;
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }

    private boolean check(int k) {
        boolean[] rem = new boolean[s.length];
        for (int i = 0; i < k; ++i) {
            rem[removable[i]] = true;
        }
        int i = 0, j = 0;
        while (i < s.length && j < p.length) {
            if (!rem[i] && p[j] == s[i]) {
                ++j;
            }
            ++i;
        }
        return j == p.length;
    }
}

C++

class Solution {
public:
    int maximumRemovals(string s, string p, vector<int>& removable) {
        int m = s.size(), n = p.size();
        int l = 0, r = removable.size();
        bool rem[m];

        auto check = [&](int k) {
            memset(rem, false, sizeof(rem));
            for (int i = 0; i < k; i++) {
                rem[removable[i]] = true;
            }
            int i = 0, j = 0;
            while (i < m && j < n) {
                if (!rem[i] && s[i] == p[j]) {
                    ++j;
                }
                ++i;
            }
            return j == n;
        };
        while (l < r) {
            int mid = (l + r + 1) >> 1;
            if (check(mid)) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }
        return l;
    }
};

Go

func maximumRemovals(s string, p string, removable []int) int {
	m, n := len(s), len(p)
	l, r := 0, len(removable)
	check := func(k int) bool {
		rem := make([]bool, m)
		for i := 0; i < k; i++ {
			rem[removable[i]] = true
		}
		i, j := 0, 0
		for i < m && j < n {
			if !rem[i] && s[i] == p[j] {
				j++
			}
			i++
		}
		return j == n
	}
	for l < r {
		mid := (l + r + 1) >> 1
		if check(mid) {
			l = mid
		} else {
			r = mid - 1
		}
	}
	return l
}

TypeScript

function maximumRemovals(s: string, p: string, removable: number[]): number {
    const [m, n] = [s.length, p.length];
    let [l, r] = [0, removable.length];
    const rem: boolean[] = Array(m);

    const check = (k: number): boolean => {
        rem.fill(false);
        for (let i = 0; i < k; i++) {
            rem[removable[i]] = true;
        }

        let i = 0,
            j = 0;
        while (i < m && j < n) {
            if (!rem[i] && s[i] === p[j]) {
                j++;
            }
            i++;
        }
        return j === n;
    };

    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    return l;
}

Rust

impl Solution {
    pub fn maximum_removals(s: String, p: String, removable: Vec<i32>) -> i32 {
        let m = s.len();
        let n = p.len();
        let s: Vec<char> = s.chars().collect();
        let p: Vec<char> = p.chars().collect();
        let mut l = 0;
        let mut r = removable.len();

        let check = |k: usize| -> bool {
            let mut rem = vec![false; m];
            for i in 0..k {
                rem[removable[i] as usize] = true;
            }
            let mut i = 0;
            let mut j = 0;
            while i < m && j < n {
                if !rem[i] && s[i] == p[j] {
                    j += 1;
                }
                i += 1;
            }
            j == n
        };

        while l < r {
            let mid = (l + r + 1) / 2;
            if check(mid) {
                l = mid;
            } else {
                r = mid - 1;
            }
        }

        l as i32
    }
}

JavaScript

/**
 * @param {string} s
 * @param {string} p
 * @param {number[]} removable
 * @return {number}
 */
var maximumRemovals = function (s, p, removable) {
    const [m, n] = [s.length, p.length];
    let [l, r] = [0, removable.length];
    const rem = Array(m);

    const check = k => {
        rem.fill(false);
        for (let i = 0; i < k; i++) {
            rem[removable[i]] = true;
        }

        let i = 0,
            j = 0;
        while (i < m && j < n) {
            if (!rem[i] && s[i] === p[j]) {
                j++;
            }
            i++;
        }
        return j === n;
    };

    while (l < r) {
        const mid = (l + r + 1) >> 1;
        if (check(mid)) {
            l = mid;
        } else {
            r = mid - 1;
        }
    }

    return l;
};

Kotlin

class Solution {
    fun maximumRemovals(s: String, p: String, removable: IntArray): Int {
        val m = s.length
        val n = p.length
        var l = 0
        var r = removable.size

        fun check(k: Int): Boolean {
            val rem = BooleanArray(m)
            for (i in 0 until k) {
                rem[removable[i]] = true
            }
            var i = 0
            var j = 0
            while (i < m && j < n) {
                if (!rem[i] && s[i] == p[j]) {
                    j++
                }
                i++
            }
            return j == n
        }

        while (l < r) {
            val mid = (l + r + 1) / 2
            if (check(mid)) {
                l = mid
            } else {
                r = mid - 1
            }
        }

        return l
    }
}