LeetCode #2840 — MEDIUM

Check if Strings Can be Made Equal With Operations II

Move from brute-force thinking to an efficient approach using hash map strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given two strings s1 and s2, both of length n, 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 i < j and the difference j - i is even, 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 = "abcdba", s2 = "cabdab"
Output: true
Explanation: We can apply the following operations on s1:
- Choose the indices i = 0, j = 2. The resulting string is s1 = "cbadba".
- Choose the indices i = 2, j = 4. The resulting string is s1 = "cbbdaa".
- Choose the indices i = 1, j = 5. The resulting string is s1 = "cabdab" = s2.

Example 2:

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

Constraints:

n == s1.length == s2.length
1 <= n <= 10⁵
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 n, 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 i < j and the difference j - i is even, 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: Hash Map

Example 1

"abcdba"
"cabdab"

Example 2

"abe"
"bea"

Step 02

Core Insight

What unlocks the optimal approach

<div class="_1l1MA">Characters in two positions can be swapped if and only if the two positions have the same parity.</div>
<div class="_1l1MA">To be able to make the two strings equal, the characters at even and odd positions in the strings should be the same.</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 #2840: Check if Strings Can be Made Equal With Operations II
class Solution {
    public boolean checkStrings(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 #2840: Check if Strings Can be Made Equal With Operations II
func checkStrings(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 #2840: Check if Strings Can be Made Equal With Operations II
class Solution:
    def checkStrings(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 #2840: Check if Strings Can be Made Equal With Operations II
// 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 #2840: Check if Strings Can be Made Equal With Operations II
// class Solution {
//     public boolean checkStrings(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 #2840: Check if Strings Can be Made Equal With Operations II
function checkStrings(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

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.