LeetCode #1790 — EASY

Check if One String Swap Can Make Strings Equal

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

The Problem

Problem Statement

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.

Example 1:

Input: s1 = "bank", s2 = "kanb"
Output: true
Explanation: For example, swap the first character with the last character of s2 to make "bank".

Example 2:

Input: s1 = "attack", s2 = "defend"
Output: false
Explanation: It is impossible to make them equal with one string swap.

Example 3:

Input: s1 = "kelb", s2 = "kelb"
Output: true
Explanation: The two strings are already equal, so no string swap operation is required.

Constraints:

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"bank"
"kanb"

Example 2

"attack"
"defend"

Example 3

"kelb"
"kelb"

Core Insight

What unlocks the optimal approach

The answer is false if the number of nonequal positions in the strings is not equal to 0 or 2.
Check that these positions have the same set of characters.

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 #1790: Check if One String Swap Can Make Strings Equal
class Solution {
    public boolean areAlmostEqual(String s1, String s2) {
        int cnt = 0;
        char c1 = 0, c2 = 0;
        for (int i = 0; i < s1.length(); ++i) {
            char a = s1.charAt(i), b = s2.charAt(i);
            if (a != b) {
                if (++cnt > 2 || (cnt == 2 && (a != c2 || b != c1))) {
                    return false;
                }
                c1 = a;
                c2 = b;
            }
        }
        return cnt != 1;
    }
}

// Accepted solution for LeetCode #1790: Check if One String Swap Can Make Strings Equal
func areAlmostEqual(s1 string, s2 string) bool {
	cnt := 0
	var c1, c2 byte
	for i := range s1 {
		a, b := s1[i], s2[i]
		if a != b {
			cnt++
			if cnt > 2 || (cnt == 2 && (a != c2 || b != c1)) {
				return false
			}
			c1, c2 = a, b
		}
	}
	return cnt != 1
}

# Accepted solution for LeetCode #1790: Check if One String Swap Can Make Strings Equal
class Solution:
    def areAlmostEqual(self, s1: str, s2: str) -> bool:
        cnt = 0
        c1 = c2 = None
        for a, b in zip(s1, s2):
            if a != b:
                cnt += 1
                if cnt > 2 or (cnt == 2 and (a != c2 or b != c1)):
                    return False
                c1, c2 = a, b
        return cnt != 1

// Accepted solution for LeetCode #1790: Check if One String Swap Can Make Strings Equal
impl Solution {
    pub fn are_almost_equal(s1: String, s2: String) -> bool {
        if s1 == s2 {
            return true;
        }
        let (s1, s2) = (s1.as_bytes(), s2.as_bytes());
        let mut idxs = vec![];
        for i in 0..s1.len() {
            if s1[i] != s2[i] {
                idxs.push(i);
            }
        }
        if idxs.len() != 2 {
            return false;
        }
        s1[idxs[0]] == s2[idxs[1]] && s2[idxs[0]] == s1[idxs[1]]
    }
}

// Accepted solution for LeetCode #1790: Check if One String Swap Can Make Strings Equal
function areAlmostEqual(s1: string, s2: string): boolean {
    let c1, c2;
    let cnt = 0;
    for (let i = 0; i < s1.length; ++i) {
        const a = s1.charAt(i);
        const b = s2.charAt(i);
        if (a != b) {
            if (++cnt > 2 || (cnt == 2 && (a != c2 || b != c1))) {
                return false;
            }
            c1 = a;
            c2 = b;
        }
    }
    return cnt != 1;
}

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(n)

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.