Reverse and Print Common Characters in Two Strings Using Python

Find common characters between two strings at the same index positions after reversing the second string

Understand the Problem

Problem Statement

Given two strings, S1 and S2, reverse S2 and find the characters that are common between S1 and the reversed S2 at the same index positions. Both strings will contain only lowercase letters.

Constraints

The length of S1 should be between 2 and 500.
The length of S2 should also be between 2 and 500.
Both strings contain only lowercase letters (a-z).
Comparison is case-sensitive.

Examples

Example 1

Input

energy
genuine

Output

en

Explanation

After reversing 'genuine' to 'eniuneg', comparing with 'energy' shows matches at index 0 ('e') and index 1 ('n'), resulting in 'en'.

Example 2

Input

growth
thrown

Output

Explanation

After reversing 'thrown' to 'nworht', comparing with 'growth' shows only one match at index 2 ('o'), resulting in 'o'.

Example 3

Input

abc
xyz

Output

Explanation

After reversing 'xyz' to 'zyx', comparing with 'abc' shows no matches at any index position, resulting in an empty string.

Solution

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

int main() {
    char s1[501], s2[501];
    char result[501];
    int i, j, len1, len2, result_len = 0;
    
    // Read input strings
    fgets(s1, sizeof(s1), stdin);
    fgets(s2, sizeof(s2), stdin);
    
    // Remove newlines
    s1[strcspn(s1, "\n")] = 0;
    s2[strcspn(s2, "\n")] = 0;
    
    len1 = strlen(s1);
    len2 = strlen(s2);
    
    // Compare characters from end of s2 to beginning of s1
    for (i = 0; i < len1 && i < len2; i++) {
        if (s1[i] == s2[len2 - 1 - i]) {
            result[result_len++] = s1[i];
        }
    }
    result[result_len] = '\0';
    
    printf("%s\n", result);
    return 0;
}

Time:O(n) where n is the length of the shorter string

Space:O(n) for storing the result string

Approach:

The C solution reads both strings using fgets() to handle spaces. It removes newlines, then iterates through both strings up to the length of the shorter one. For each position i, it compares s1[i] with s2[len2-1-i] (which effectively reverses s2 during comparison). Matching characters are stored in the result array, which is null-terminated and printed at the end.

Visual Explanation

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