LeetCode #3083 — EASY

Existence of a Substring in a String and Its Reverse

Build confidence with an intuition-first walkthrough focused on hash map fundamentals.

Solve on LeetCode

The Problem

Problem Statement

Given a string s, find any substring of length 2 which is also present in the reverse of s.

Return true if such a substring exists, and false otherwise.

Example 1:

Input: s = "leetcode"

Output: true

Explanation: Substring "ee" is of length 2 which is also present in reverse(s) == "edocteel".

Example 2:

Input: s = "abcba"

Output: true

Explanation: All of the substrings of length 2 "ab", "bc", "cb", "ba" are also present in reverse(s) == "abcba".

Example 3:

Input: s = "abcd"

Output: false

Explanation: There is no substring of length 2 in s, which is also present in the reverse of s.

Constraints:

1 <= s.length <= 100
s consists only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: Given a string s, find any substring of length 2 which is also present in the reverse of s. Return true if such a substring exists, and false otherwise.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"leetcode"

Example 2

"abcba"

Example 3

"abcd"

Step 02

Core Insight

What unlocks the optimal approach

Make a new string by reversing the string <code>s</code>.
For every substring of length <code>2</code> in <code>s</code>, check if there is a corresponding substring in the reverse of <code>s</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 #3083: Existence of a Substring in a String and Its Reverse
class Solution {
    public boolean isSubstringPresent(String s) {
        boolean[][] st = new boolean[26][26];
        int n = s.length();
        for (int i = 0; i < n - 1; ++i) {
            st[s.charAt(i + 1) - 'a'][s.charAt(i) - 'a'] = true;
        }
        for (int i = 0; i < n - 1; ++i) {
            if (st[s.charAt(i) - 'a'][s.charAt(i + 1) - 'a']) {
                return true;
            }
        }
        return false;
    }
}

// Accepted solution for LeetCode #3083: Existence of a Substring in a String and Its Reverse
func isSubstringPresent(s string) bool {
	st := [26][26]bool{}
	for i := 0; i < len(s)-1; i++ {
		st[s[i+1]-'a'][s[i]-'a'] = true
	}
	for i := 0; i < len(s)-1; i++ {
		if st[s[i]-'a'][s[i+1]-'a'] {
			return true
		}
	}
	return false
}

# Accepted solution for LeetCode #3083: Existence of a Substring in a String and Its Reverse
class Solution:
    def isSubstringPresent(self, s: str) -> bool:
        st = {(a, b) for a, b in pairwise(s[::-1])}
        return any((a, b) in st for a, b in pairwise(s))

// Accepted solution for LeetCode #3083: Existence of a Substring in a String and Its Reverse
fn is_substring_present(s: String) -> bool {
    let cs1: Vec<char> = s.chars().collect();
    let cs2: Vec<char> = s.chars().rev().collect();

    let cs1: Vec<_> = cs1.windows(2).collect();
    let cs2: Vec<_> = cs2.windows(2).collect();

    for &s1 in &cs1 {
        for &s2 in &cs2 {
            if *s1 == *s2 {
                return true;
            }
        }
    }

    false
}

fn main() {
    println!("ret={}", is_substring_present("leetcode".to_string()));
}

#[test]
fn test() {
    assert!(is_substring_present("leetcode".to_string()));
    assert!(is_substring_present("abcba".to_string()));
    assert!(!is_substring_present("abcd".to_string()));
}

// Accepted solution for LeetCode #3083: Existence of a Substring in a String and Its Reverse
function isSubstringPresent(s: string): boolean {
    const st: boolean[][] = Array.from({ length: 26 }, () => Array(26).fill(false));
    for (let i = 0; i < s.length - 1; ++i) {
        st[s.charCodeAt(i + 1) - 97][s.charCodeAt(i) - 97] = true;
    }
    for (let i = 0; i < s.length - 1; ++i) {
        if (st[s.charCodeAt(i) - 97][s.charCodeAt(i + 1) - 97]) {
            return true;
        }
    }
    return false;
}

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(|\Sigma|^2)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

For each element, scan the rest of the array looking for a match. Two nested loops give n × (n−1)/2 comparisons = O(n²). No extra space since we only use loop indices.

HASH MAP

O(n) time

O(n) space

One pass through the input, performing O(1) hash map lookups and insertions at each step. The hash map may store up to n entries in the worst case. This is the classic space-for-time tradeoff: O(n) extra memory eliminates an inner loop.

Shortcut: Need to check “have I seen X before?” → hash map → O(n) time, O(n) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.