LeetCode #2839 — EASY

Check if Strings Can be Made Equal With Operations I

Build confidence with an intuition-first walkthrough focused on core interview patterns fundamentals.

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The Problem

Problem Statement

You are given two strings s1 and s2, both of length 4, consisting of lowercase English letters.

You can apply the following operation on any of the two strings any number of times:

Choose any two indices i and j such that j - i = 2, then swap the two characters at those indices in the string.

Return true if you can make the strings s1 and s2 equal, and false otherwise.

Example 1:

Input: s1 = "abcd", s2 = "cdab"
Output: true
Explanation: We can do the following operations on s1:
- Choose the indices i = 0, j = 2. The resulting string is s1 = "cbad".
- Choose the indices i = 1, j = 3. The resulting string is s1 = "cdab" = s2.

Example 2:

Input: s1 = "abcd", s2 = "dacb"
Output: false
Explanation: It is not possible to make the two strings equal.

Constraints:

s1.length == s2.length == 4
s1 and s2 consist only of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given two strings s1 and s2, both of length 4, consisting of lowercase English letters. You can apply the following operation on any of the two strings any number of times: Choose any two indices i and j such that j - i = 2, then swap the two characters at those indices in the string. Return true if you can make the strings s1 and s2 equal, and false otherwise.

Baseline thinking

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

Pattern signal: General problem-solving

Example 1

"abcd"
"cdab"

Example 2

"abcd"
"dacb"

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">Since the strings are very small you can try a brute-force approach.</div>
<div class="_1l1MA">There are only <code>2</code> different swaps that are possible in a string.</div>

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 #2839: Check if Strings Can be Made Equal With Operations I
class Solution {
    public boolean canBeEqual(String s1, String s2) {
        int[][] cnt = new int[2][26];
        for (int i = 0; i < s1.length(); ++i) {
            ++cnt[i & 1][s1.charAt(i) - 'a'];
            --cnt[i & 1][s2.charAt(i) - 'a'];
        }
        for (int i = 0; i < 26; ++i) {
            if (cnt[0][i] != 0 || cnt[1][i] != 0) {
                return false;
            }
        }
        return true;
    }
}

// Accepted solution for LeetCode #2839: Check if Strings Can be Made Equal With Operations I
func canBeEqual(s1 string, s2 string) bool {
	cnt := [2][26]int{}
	for i := 0; i < len(s1); i++ {
		cnt[i&1][s1[i]-'a']++
		cnt[i&1][s2[i]-'a']--
	}
	for i := 0; i < 26; i++ {
		if cnt[0][i] != 0 || cnt[1][i] != 0 {
			return false
		}
	}
	return true
}

# Accepted solution for LeetCode #2839: Check if Strings Can be Made Equal With Operations I
class Solution:
    def canBeEqual(self, s1: str, s2: str) -> bool:
        return sorted(s1[::2]) == sorted(s2[::2]) and sorted(s1[1::2]) == sorted(
            s2[1::2]
        )

// Accepted solution for LeetCode #2839: Check if Strings Can be Made Equal With Operations I
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2839: Check if Strings Can be Made Equal With Operations I
// class Solution {
//     public boolean canBeEqual(String s1, String s2) {
//         int[][] cnt = new int[2][26];
//         for (int i = 0; i < s1.length(); ++i) {
//             ++cnt[i & 1][s1.charAt(i) - 'a'];
//             --cnt[i & 1][s2.charAt(i) - 'a'];
//         }
//         for (int i = 0; i < 26; ++i) {
//             if (cnt[0][i] != 0 || cnt[1][i] != 0) {
//                 return false;
//             }
//         }
//         return true;
//     }
// }

// Accepted solution for LeetCode #2839: Check if Strings Can be Made Equal With Operations I
function canBeEqual(s1: string, s2: string): boolean {
    const cnt: number[][] = Array.from({ length: 2 }, () => Array.from({ length: 26 }, () => 0));
    for (let i = 0; i < s1.length; ++i) {
        ++cnt[i & 1][s1.charCodeAt(i) - 97];
        --cnt[i & 1][s2.charCodeAt(i) - 97];
    }
    for (let i = 0; i < 26; ++i) {
        if (cnt[0][i] || cnt[1][i]) {
            return false;
        }
    }
    return true;
}

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 + |\Sigma|)

Space

O(|\Sigma|)

Approach Breakdown

BRUTE FORCE

O(n²) time

O(1) space

Two nested loops check every pair or subarray. The outer loop fixes a starting point, the inner loop extends or searches. For n elements this gives up to n²/2 operations. No extra space, but the quadratic time is prohibitive for large inputs.

OPTIMIZED

O(n) time

O(1) space

Most array problems have an O(n²) brute force (nested loops) and an O(n) optimal (single pass with clever state tracking). The key is identifying what information to maintain as you scan: a running max, a prefix sum, a hash map of seen values, or two pointers.

Shortcut: If you are using nested loops on an array, there is almost always an O(n) solution. Look for the right auxiliary state.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.