Split Two Strings to Make Palindrome

Split Two Strings to Make Palindrome - Problem

You are given two strings a and b of the same length. Choose an index and split both strings at the same index, splitting a into two strings: aprefix and asuffix where a = aprefix + asuffix, and splitting b into two strings: bprefix and bsuffix where b = bprefix + bsuffix. Check if aprefix + bsuffix or bprefix + asuffix forms a palindrome.

When you split a string s into sprefix and ssuffix, either ssuffix or sprefix is allowed to be empty. For example, if s = "abc", then "" + "abc", "a" + "bc", "ab" + "c", and "abc" + "" are valid splits.

Return true if it is possible to form a palindrome string, otherwise return false.

Input & Output

Example 1 — Basic Valid Case

$ Input: a = "x", b = "y"

› Output: true

💡 Note: Split at position 0: aprefix="" + bsuffix="y" = "y" (palindrome), or split at position 1: aprefix="x" + bsuffix="" = "x" (palindrome)

Example 2 — Valid Combination

$ Input: a = "xbdef", b = "xecab"

› Output: false

💡 Note: No split position creates a palindrome: splitting anywhere produces non-palindromic combinations

Example 3 — Another Valid Case

$ Input: a = "ulacfd", b = "jizalu"

› Output: true

💡 Note: Split at position 3: aprefix="ula" + bsuffix="alu" = "ulaalu" (palindrome)

Constraints

1 ≤ a.length, b.length ≤ 10⁵
a.length == b.length
a and b consist of lowercase English letters

Visualization

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Asked in

F Facebook 25 G Google 20

The key insight is to use two pointers from both ends to efficiently find where strings can be split to form palindromes. Best approach uses two pointers optimization with O(n) time and O(1) space complexity.

Common Approaches

✓ Brute Force

⏱️ Time: O(n²) Space: O(n)

For each possible split position i (0 to n), create aprefix + bsuffix and bprefix + asuffix, then check if either forms a palindrome using a helper function.

Two Pointers Optimization

⏱️ Time: O(n) Space: O(1)

Start with pointers at both ends of the strings. Move inward while characters match. When they don't match, try two scenarios: check if remaining part forms palindrome with opposite string.

Brute Force — Algorithm Steps

Step 1: Try each split position from 0 to string length
Step 2: For each split, form two possible combinations
Step 3: Check if either combination is a palindrome

Visualization

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Step-by-Step Walkthrough

Split at position i

Divide both strings at same index

Form combinations

Create aprefix+bsuffix and bprefix+asuffix

Check palindrome

Test if either combination reads same forwards and backwards

Code -

solution.c — C

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

bool isPalindrome(char* s) {
    int len = strlen(s);
    for (int i = 0; i < len / 2; i++) {
        if (s[i] != s[len - 1 - i]) {
            return false;
        }
    }
    return true;
}

bool solution(char* a, char* b) {
    int n = strlen(a);
    for (int i = 0; i <= n; i++) {
        // aprefix + bsuffix
        char combo1[1001];
        strncpy(combo1, a, i);
        combo1[i] = '\0';
        strcat(combo1, b + i);
        if (isPalindrome(combo1)) {
            return true;
        }
        
        // bprefix + asuffix
        char combo2[1001];
        strncpy(combo2, b, i);
        combo2[i] = '\0';
        strcat(combo2, a + i);
        if (isPalindrome(combo2)) {
            return true;
        }
    }
    return false;
}

int main() {
    char a[501], b[501];
    fgets(a, sizeof(a), stdin);
    fgets(b, sizeof(b), stdin);
    a[strcspn(a, "\n")] = 0;
    b[strcspn(b, "\n")] = 0;
    bool result = solution(a, b);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Loop through n split positions, each palindrome check takes O(n) time

⚠ Quadratic Growth

Space Complexity

O(n)

String concatenation creates new strings of length n

⚡ Linearithmic Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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