LeetCode #647 — MEDIUM

Palindromic Substrings

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

Solve on LeetCode

The Problem

Problem Statement

Given a string s, return the number of palindromic substrings in it.

A string is a palindrome when it reads the same backward as forward.

A substring is a contiguous sequence of characters within the string.

Example 1:

Input: s = "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: s = "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Constraints:

1 <= s.length <= 1000
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, return the number of palindromic substrings in it. A string is a palindrome when it reads the same backward as forward. A substring is a contiguous sequence of characters within the string.

Baseline thinking

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

Pattern signal: Two Pointers · Dynamic Programming

Example 1

"abc"

Example 2

"aaa"

Core Insight

What unlocks the optimal approach

How can we reuse a previously computed palindrome to compute a larger palindrome?
If “aba” is a palindrome, is “xabax” a palindrome? Similarly is “xabay” a palindrome?
Complexity based hint:</br> If we use brute force and check whether for every start and end position a substring is a palindrome we have O(n^2) start - end pairs and O(n) palindromic checks. Can we reduce the time for palindromic checks to O(1) by reusing some previous computation?

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 #647: Palindromic Substrings
class Solution {
    public int countSubstrings(String s) {
        int ans = 0;
        int n = s.length();
        for (int k = 0; k < n * 2 - 1; ++k) {
            int i = k / 2, j = (k + 1) / 2;
            while (i >= 0 && j < n && s.charAt(i) == s.charAt(j)) {
                ++ans;
                --i;
                ++j;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #647: Palindromic Substrings
func countSubstrings(s string) int {
	ans, n := 0, len(s)
	for k := 0; k < n*2-1; k++ {
		i, j := k/2, (k+1)/2
		for i >= 0 && j < n && s[i] == s[j] {
			ans++
			i, j = i-1, j+1
		}
	}
	return ans
}

# Accepted solution for LeetCode #647: Palindromic Substrings
class Solution:
    def countSubstrings(self, s: str) -> int:
        ans, n = 0, len(s)
        for k in range(n * 2 - 1):
            i, j = k // 2, (k + 1) // 2
            while ~i and j < n and s[i] == s[j]:
                ans += 1
                i, j = i - 1, j + 1
        return ans

// Accepted solution for LeetCode #647: Palindromic Substrings
impl Solution {
    pub fn count_substrings(s: String) -> i32 {
        let s = s.chars().collect::<Vec<char>>();
        let (mut count, length): (i32, i32) = (0, s.len() as i32);

        for i in 0..length {
            // odd length

            let (mut l, mut r) = (i, i);
            while l >= 0 && r < length && s[l as usize] == s[r as usize] {
                count += 1;
                l -= 1;
                r += 1;
            }

            // even length

            let (mut l, mut r) = (i, i + 1);

            while l >= 0 && r < length && s[l as usize] == s[r as usize] {
                count += 1;
                l -= 1;
                r += 1;
            }
        }

        count
    }
}

// Accepted solution for LeetCode #647: Palindromic Substrings
function countSubstrings(s: string): number {
    let res = 0;

    for (let i = 0; i < s.length; i++) {
        let l = i;
        let r = i;

        while (l >= 0 && r < s.length && s[l] === s[r]) {
            res += 1;
            l -= 1;
            r += 1;
        }

        l = i;
        r = i + 1;

        while (l >= 0 && r < s.length && s[l] === s[r]) {
            res += 1;
            l -= 1;
            r += 1;
        }
    }

    return res;
}

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)

Space

O(1)

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.

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.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.