LeetCode #2207 — MEDIUM

Maximize Number of Subsequences in a String

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

The Problem

Problem Statement

You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters.

You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text.

Return the maximum number of times pattern can occur as a subsequence of the modified text.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Example 1:

Input: text = "abdcdbc", pattern = "ac"
Output: 4
Explanation:
If we add pattern[0] = 'a' in between text[1] and text[2], we get "abadcdbc". Now, the number of times "ac" occurs as a subsequence is 4.
Some other strings which have 4 subsequences "ac" after adding a character to text are "aabdcdbc" and "abdacdbc".
However, strings such as "abdcadbc", "abdccdbc", and "abdcdbcc", although obtainable, have only 3 subsequences "ac" and are thus suboptimal.
It can be shown that it is not possible to get more than 4 subsequences "ac" by adding only one character.

Example 2:

Input: text = "aabb", pattern = "ab"
Output: 6
Explanation:
Some of the strings which can be obtained from text and have 6 subsequences "ab" are "aaabb", "aaabb", and "aabbb".

Constraints:

1 <= text.length <= 10⁵
pattern.length == 2
text and pattern consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters. You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text. Return the maximum number of times pattern can occur as a subsequence of the modified text. A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.

Baseline thinking

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

Pattern signal: Greedy

Example 1

"abdcdbc"
"ac"

Example 2

"aabb"
"ab"

Core Insight

What unlocks the optimal approach

Find the optimal position to add pattern[0] so that the number of subsequences is maximized. Similarly, find the optimal position to add pattern[1].
For each of the above cases, count the number of times the pattern occurs as a subsequence in text. The larger count is the required answer.

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 #2207: Maximize Number of Subsequences in a String
class Solution {
    public long maximumSubsequenceCount(String text, String pattern) {
        long ans = 0;
        int x = 0, y = 0;
        for (int i = 0; i < text.length(); ++i) {
            if (text.charAt(i) == pattern.charAt(1)) {
                ++y;
                ans += x;
            }
            if (text.charAt(i) == pattern.charAt(0)) {
                ++x;
            }
        }
        ans += Math.max(x, y);
        return ans;
    }
}

// Accepted solution for LeetCode #2207: Maximize Number of Subsequences in a String
func maximumSubsequenceCount(text string, pattern string) (ans int64) {
	x, y := 0, 0
	for _, c := range text {
		if byte(c) == pattern[1] {
			y++
			ans += int64(x)
		}
		if byte(c) == pattern[0] {
			x++
		}
	}
	ans += int64(max(x, y))
	return
}

# Accepted solution for LeetCode #2207: Maximize Number of Subsequences in a String
class Solution:
    def maximumSubsequenceCount(self, text: str, pattern: str) -> int:
        ans = x = y = 0
        for c in text:
            if c == pattern[1]:
                y += 1
                ans += x
            if c == pattern[0]:
                x += 1
        ans += max(x, y)
        return ans

// Accepted solution for LeetCode #2207: Maximize Number of Subsequences in a String
/**
 * [2207] Maximize Number of Subsequences in a String
 *
 * You are given a 0-indexed string text and another 0-indexed string pattern of length 2, both of which consist of only lowercase English letters.
 * You can add either pattern[0] or pattern[1] anywhere in text exactly once. Note that the character can be added even at the beginning or at the end of text.
 * Return the maximum number of times pattern can occur as a subsequence of the modified text.
 * A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters.
 *  
 * Example 1:
 *
 * Input: text = "abdcdbc", pattern = "ac"
 * Output: 4
 * Explanation:
 * If we add pattern[0] = 'a' in between text[1] and text[2], we get "ab<u>a</u>dcdbc". Now, the number of times "ac" occurs as a subsequence is 4.
 * Some other strings which have 4 subsequences "ac" after adding a character to text are "<u>a</u>abdcdbc" and "abd<u>a</u>cdbc".
 * However, strings such as "abdc<u>a</u>dbc", "abd<u>c</u>cdbc", and "abdcdbc<u>c</u>", although obtainable, have only 3 subsequences "ac" and are thus suboptimal.
 * It can be shown that it is not possible to get more than 4 subsequences "ac" by adding only one character.
 *
 * Example 2:
 *
 * Input: text = "aabb", pattern = "ab"
 * Output: 6
 * Explanation:
 * Some of the strings which can be obtained from text and have 6 subsequences "ab" are "<u>a</u>aabb", "aa<u>a</u>bb", and "aab<u>b</u>b".
 *
 *  
 * Constraints:
 *
 * 	1 <= text.length <= 10^5
 * 	pattern.length == 2
 * 	text and pattern consist only of lowercase English letters.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximize-number-of-subsequences-in-a-string/
// discuss: https://leetcode.com/problems/maximize-number-of-subsequences-in-a-string/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn maximum_subsequence_count(text: String, pattern: String) -> i64 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2207_example_1() {
        let text = "abdcdbc".to_string();
        let pattern = "ac".to_string();

        let result = 4;

        assert_eq!(Solution::maximum_subsequence_count(text, pattern), result);
    }

    #[test]
    #[ignore]
    fn test_2207_example_2() {
        let text = "aabb".to_string();
        let pattern = "ab".to_string();

        let result = 6;

        assert_eq!(Solution::maximum_subsequence_count(text, pattern), result);
    }
}

// Accepted solution for LeetCode #2207: Maximize Number of Subsequences in a String
function maximumSubsequenceCount(text: string, pattern: string): number {
    let ans = 0;
    let [x, y] = [0, 0];
    for (const c of text) {
        if (c === pattern[1]) {
            ++y;
            ans += x;
        }
        if (c === pattern[0]) {
            ++x;
        }
    }
    ans += Math.max(x, y);
    return ans;
}

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(n log n)

Space

O(1)

Approach Breakdown

EXHAUSTIVE

O(2ⁿ) time

O(n) space

Try every possible combination of choices. With n items each having two states (include/exclude), the search space is 2ⁿ. Evaluating each combination takes O(n), giving O(n × 2ⁿ). The recursion stack or subset storage uses O(n) space.

GREEDY

O(n log n) time

O(1) space

Greedy algorithms typically sort the input (O(n log n)) then make a single pass (O(n)). The sort dominates. If the input is already sorted or the greedy choice can be computed without sorting, time drops to O(n). Proving greedy correctness (exchange argument) is harder than the implementation.

Shortcut: Sort + single pass → O(n log n). If no sort needed → O(n). The hard part is proving it works.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Using greedy without proof

Wrong move: Locally optimal choices may fail globally.

Usually fails on: Counterexamples appear on crafted input orderings.

Fix: Verify with exchange argument or monotonic objective before committing.