LeetCode #3084 — MEDIUM

Count Substrings Starting and Ending with Given Character

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

The Problem

Problem Statement

You are given a string s and a character c. Return the total number of substrings of s that start and end with c.

Example 1:

Input: s = "abada", c = "a"

Output: 6

Explanation: Substrings starting and ending with "a" are: "abada", "abada", "abada", "abada", "abada", "abada".

Example 2:

Input: s = "zzz", c = "z"

Output: 6

Explanation: There are a total of 6 substrings in s and all start and end with "z".

Constraints:

1 <= s.length <= 10⁵
s and c consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s and a character c. Return the total number of substrings of s that start and end with c.

Baseline thinking

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

Pattern signal: Math

Example 1

"abada"
"a"

Example 2

"zzz"
"z"

Step 02

Core Insight

What unlocks the optimal approach

Count the number of characters <code>'c'</code> in string <code>s</code>, let’s call it <code>m</code>.
We can select <code>2</code> numbers <code>i</code> and <code>j</code> such that <code>i <= j</code> are the start and end indices of substring. Note that <code>i</code> and <code>j</code> can be the same.
The answer is <code>m * (m + 1) / 2</code>.

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 #3084: Count Substrings Starting and Ending with Given Character
class Solution {
    public long countSubstrings(String s, char c) {
        long cnt = s.chars().filter(ch -> ch == c).count();
        return cnt + cnt * (cnt - 1) / 2;
    }
}

// Accepted solution for LeetCode #3084: Count Substrings Starting and Ending with Given Character
func countSubstrings(s string, c byte) int64 {
	cnt := int64(strings.Count(s, string(c)))
	return cnt + cnt*(cnt-1)/2
}

# Accepted solution for LeetCode #3084: Count Substrings Starting and Ending with Given Character
class Solution:
    def countSubstrings(self, s: str, c: str) -> int:
        cnt = s.count(c)
        return cnt + cnt * (cnt - 1) // 2

// Accepted solution for LeetCode #3084: Count Substrings Starting and Ending with Given Character
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3084: Count Substrings Starting and Ending with Given Character
// class Solution {
//     public long countSubstrings(String s, char c) {
//         long cnt = s.chars().filter(ch -> ch == c).count();
//         return cnt + cnt * (cnt - 1) / 2;
//     }
// }

// Accepted solution for LeetCode #3084: Count Substrings Starting and Ending with Given Character
function countSubstrings(s: string, c: string): number {
    const cnt = s.split('').filter(ch => ch === c).length;
    return cnt + Math.floor((cnt * (cnt - 1)) / 2);
}

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(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.