Existence of a Substring in a String and Its Reverse

Existence of a Substring in a String and Its Reverse - Problem

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.

Note: A substring is a contiguous sequence of characters within a string.

Input & Output

Example 1 — Basic Case

$ Input: s = "abcba"

› Output: true

💡 Note: The 2-character substring "ab" appears at the beginning of the original string and also appears at the beginning of the reversed string "abcba"

Example 2 — No Match

$ Input: s = "abcd"

› Output: false

💡 Note: Original string has substrings "ab", "bc", "cd". Reversed string "dcba" has substrings "dc", "cb", "ba". No common 2-character substring exists.

Example 3 — Single Match

$ Input: s = "leetcode"

› Output: true

💡 Note: The substring "ee" appears in the original string. The reversed string "edocteel" also contains "ee", so return true.

Constraints

1 ≤ s.length ≤ 100
s consists of lowercase English letters only

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that we need to find any 2-character substring that appears in both the original string and its reverse. The most efficient approach uses a hash set to store all 2-character substrings from the original string, then checks if any 2-character substring from the reversed string exists in the set. Time: O(n), Space: O(n)

Common Approaches

✓ Brute Force - Compare All Pairs

⏱️ Time: O(n²) Space: O(1)

Extract all possible 2-character substrings from the original string and all 2-character substrings from the reversed string. Then compare every substring from the original with every substring from the reversed string to find a match.

Hash Set Optimization

⏱️ Time: O(n) Space: O(n)

Use a hash set to store all 2-character substrings from the original string. Then iterate through the reversed string and check if any of its 2-character substrings exist in the set. This eliminates the need for nested loops.

Two Pointers from Ends

⏱️ Time: O(n²) Space: O(1)

Start with two pointers, one at the beginning and one at the end of the string. Check if 2-character substrings starting from these positions create matching patterns when one is from the original and one is from the conceptual reverse direction.

Brute Force - Compare All Pairs — Algorithm Steps

Step 1: Generate all 2-character substrings from original string
Step 2: Generate all 2-character substrings from reversed string
Step 3: Compare each original substring with each reversed substring

Visualization

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Step-by-Step Walkthrough

Extract Substrings

Get all 2-char substrings from original and reversed string

Compare Pairs

Check each original substring against each reversed substring

Find Match

Return true when first match is found

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(char* s) {
    int len = strlen(s);
    if (len < 2) {
        return 0;
    }
    
    // Create reversed string
    char* reversedS = (char*)malloc((len + 1) * sizeof(char));
    for (int i = 0; i < len; i++) {
        reversedS[i] = s[len - 1 - i];
    }
    reversedS[len] = '\0';
    
    // Generate all 2-char substrings from original string
    for (int i = 0; i < len - 1; i++) {
        // Check against all 2-char substrings from reversed string
        for (int j = 0; j < len - 1; j++) {
            if (s[i] == reversedS[j] && s[i + 1] == reversedS[j + 1]) {
                free(reversedS);
                return 1;
            }
        }
    }
    
    free(reversedS);
    return 0;
}

int main() {
    char s[1001];
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0; // Remove newline
    
    int result = solution(s);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Two nested loops: outer loop generates n-1 substrings from original, inner loop checks against n-1 substrings from reversed string

⚠ Quadratic Growth

Space Complexity

O(1)

Only using constant extra variables for iteration and comparison

✓ Linear Space

12.5K Views

Medium Frequency

~15 min Avg. Time

342 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Compare All Pairs — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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