LeetCode #1898 — MEDIUM

Maximum Number of Removable Characters

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Two Pointers · Binary Search

Example 1

"abcacb"
"ab"
[3,1,0]

Example 2

"abcbddddd"
"abcd"
[3,2,1,4,5,6]

Example 3

"abcab"
"abc"
[0,1,2,3,4]

Core Insight

What unlocks the optimal approach

First, we need to think about solving an easier problem, If we remove a set of indices from the string does P exist in S as a subsequence
We can binary search the K and check by solving the above problem.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1898: Maximum Number of Removable Characters
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;
    }
}

// Accepted solution for LeetCode #1898: Maximum Number of Removable Characters
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
}

# Accepted solution for LeetCode #1898: Maximum Number of Removable Characters
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

// Accepted solution for LeetCode #1898: Maximum Number of Removable Characters
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
    }
}

// Accepted solution for LeetCode #1898: Maximum Number of Removable Characters
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;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(k × log k)

Space

O(n)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair of elements. The outer loop picks one element, the inner loop scans the rest. For n elements that is n × (n−1)/2 comparisons = O(n²). No extra memory — just two loop variables.

TWO POINTERS

O(n) time

O(1) space

Each pointer traverses the array at most once. With two pointers moving inward (or both moving right), the total number of steps is bounded by n. Each comparison is O(1), giving O(n) overall. No auxiliary data structures are needed — just two index variables.

Shortcut: Two converging pointers on sorted data → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Moving both pointers on every comparison

Wrong move: Advancing both pointers shrinks the search space too aggressively and skips candidates.

Usually fails on: A valid pair can be skipped when only one side should move.

Fix: Move exactly one pointer per decision branch based on invariant.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.