Check if One String Swap Can Make Strings Equal

Check if One String Swap Can Make Strings Equal - Problem

You are given two strings s1 and s2 of equal length. A string swap is an operation where you choose two indices in a string (not necessarily different) and swap the characters at these indices.

Return true if it is possible to make both strings equal by performing at most one string swap on exactly one of the strings. Otherwise, return false.

Input & Output

Example 1 — Valid Single Swap

$ Input: s1 = "bank", s2 = "kanb"

› Output: true

💡 Note: Swap s1[0] and s1[3]: "bank" → "kanb". Characters differ at positions 0 and 3, and swapping them makes strings equal.

Example 2 — Already Equal

$ Input: s1 = "ab", s2 = "ab"

› Output: true

💡 Note: Strings are already equal, no swap needed. Zero differences allow for equality.

Example 3 — Too Many Differences

$ Input: s1 = "aa", s2 = "bc"

› Output: false

💡 Note: Characters differ at both positions, but swapping any two characters in s1 cannot produce "bc" since frequencies don't match.

Constraints

1 ≤ s1.length, s2.length ≤ 100
s1.length == s2.length
s1 and s2 consist of only lowercase English letters

Visualization

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Asked in

G Google 15 f Facebook 12 a Amazon 8

The key insight is that two strings can be made equal with at most one swap if and only if they differ in exactly 0 or 2 positions. Best approach uses single pass difference detection with early termination. Time: O(n), Space: O(1)

Common Approaches

✓ Brute Force - Try All Swaps

⏱️ Time: O(n³) Space: O(n)

For each string, try swapping every pair of characters and check if the result equals the other string. This explores all possible single swaps.

Character Frequency Analysis

⏱️ Time: O(n) Space: O(1)

First check if both strings have the same character frequencies. If yes, find positions where characters differ and verify if exactly 0 or 2 differences exist.

Single Pass Optimal Solution

⏱️ Time: O(n) Space: O(1)

Traverse both strings once, collecting differing positions. Check if exactly 0 or 2 differences exist, and validate swap possibility on the fly.

Brute Force - Try All Swaps — Algorithm Steps

Compare strings - if equal, return true
Try all possible swaps in s1, check if result equals s2
Try all possible swaps in s2, check if result equals s1
If no swap works, return false

Visualization

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Step-by-Step Walkthrough

Compare Original

Check if strings are already equal

Try All Swaps

Test every possible character pair swap

Find Match

Return true if any swap creates equality

Code -

solution.c — C

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

bool solution(char* s1, char* s2) {
    if (strcmp(s1, s2) == 0) {
        return true;
    }
    
    int n = strlen(s1);
    char temp1[1001], temp2[1001];
    
    // Try all swaps in s1
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            strcpy(temp1, s1);
            char ch = temp1[i];
            temp1[i] = temp1[j];
            temp1[j] = ch;
            if (strcmp(temp1, s2) == 0) {
                return true;
            }
        }
    }
    
    // Try all swaps in s2
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            strcpy(temp2, s2);
            char ch = temp2[i];
            temp2[i] = temp2[j];
            temp2[j] = ch;
            if (strcmp(temp2, s1) == 0) {
                return true;
            }
        }
    }
    
    return false;
}

int main() {
    char s1[1001], s2[1001];
    fgets(s1, sizeof(s1), stdin);
    fgets(s2, sizeof(s2), stdin);
    s1[strcspn(s1, "\n")] = 0;
    s2[strcspn(s2, "\n")] = 0;
    
    bool result = solution(s1, s2);
    printf(result ? "true\n" : "false\n");
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n³)

For each string (2), we try n² possible swaps, and each comparison takes O(n)

✓ Linear Growth

Space Complexity

O(n)

Need to create new strings for each swap attempt

✓ Linear Space

32.4K Views

Medium Frequency

~15 min Avg. Time

987 Likes

Ln 1, Col 1

Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force - Try All Swaps — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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