LeetCode #2186 — MEDIUM

Minimum Number of Steps to Make Two Strings Anagram 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 s and t. In one step, you can append any character to either s or t.

Return the minimum number of steps to make s and t anagrams of each other.

An anagram of a string is a string that contains the same characters with a different (or the same) ordering.

Example 1:

Input: s = "leetcode", t = "coats"
Output: 7
Explanation: 
- In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcodeas".
- In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coatsleede".
"leetcodeas" and "coatsleede" are now anagrams of each other.
We used a total of 2 + 5 = 7 steps.
It can be shown that there is no way to make them anagrams of each other with less than 7 steps.

Example 2:

Input: s = "night", t = "thing"
Output: 0
Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps.

Constraints:

1 <= s.length, t.length <= 2 * 10⁵
s and t consist of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given two strings s and t. In one step, you can append any character to either s or t. Return the minimum number of steps to make s and t anagrams of each other. An anagram of a string is a string that contains the same characters with a different (or the same) ordering.

Baseline thinking

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

Pattern signal: Hash Map

Example 1

"leetcode"
"coats"

Example 2

"night"
"thing"

Core Insight

What unlocks the optimal approach

Notice that for anagrams, the order of the letters is irrelevant.
For each letter, we can count its frequency in s and t.
For each letter, its contribution to the answer is the absolute difference between its frequency in s and t.

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 #2186: Minimum Number of Steps to Make Two Strings Anagram II
class Solution {
    public int minSteps(String s, String t) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            ++cnt[c - 'a'];
        }
        for (char c : t.toCharArray()) {
            --cnt[c - 'a'];
        }
        int ans = 0;
        for (int v : cnt) {
            ans += Math.abs(v);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2186: Minimum Number of Steps to Make Two Strings Anagram II
func minSteps(s string, t string) int {
	cnt := make([]int, 26)
	for _, c := range s {
		cnt[c-'a']++
	}
	for _, c := range t {
		cnt[c-'a']--
	}
	ans := 0
	for _, v := range cnt {
		ans += abs(v)
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}

# Accepted solution for LeetCode #2186: Minimum Number of Steps to Make Two Strings Anagram II
class Solution:
    def minSteps(self, s: str, t: str) -> int:
        cnt = Counter(s)
        for c in t:
            cnt[c] -= 1
        return sum(abs(v) for v in cnt.values())

// Accepted solution for LeetCode #2186: Minimum Number of Steps to Make Two Strings Anagram II
/**
 * [2186] Minimum Number of Steps to Make Two Strings Anagram II
 *
 * You are given two strings s and t. In one step, you can append any character to either s or t.
 * Return the minimum number of steps to make s and t anagrams of each other.
 * An anagram of a string is a string that contains the same characters with a different (or the same) ordering.
 *  
 * Example 1:
 *
 * Input: s = "<u>lee</u>tco<u>de</u>", t = "co<u>a</u>t<u>s</u>"
 * Output: 7
 * Explanation:
 * - In 2 steps, we can append the letters in "as" onto s = "leetcode", forming s = "leetcode<u>as</u>".
 * - In 5 steps, we can append the letters in "leede" onto t = "coats", forming t = "coats<u>leede</u>".
 * "leetcodeas" and "coatsleede" are now anagrams of each other.
 * We used a total of 2 + 5 = 7 steps.
 * It can be shown that there is no way to make them anagrams of each other with less than 7 steps.
 *
 * Example 2:
 *
 * Input: s = "night", t = "thing"
 * Output: 0
 * Explanation: The given strings are already anagrams of each other. Thus, we do not need any further steps.
 *
 *  
 * Constraints:
 *
 * 	1 <= s.length, t.length <= 2 * 10^5
 * 	s and t consist of lowercase English letters.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii/
// discuss: https://leetcode.com/problems/minimum-number-of-steps-to-make-two-strings-anagram-ii/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    pub fn min_steps(s: String, t: String) -> i32 {
        0
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    #[ignore]
    fn test_2186_example_1() {
        let s = "leetcode".to_string();
        let t = "coats".to_string();

        let result = 7;

        assert_eq!(Solution::min_steps(s, t), result);
    }

    #[test]
    #[ignore]
    fn test_2186_example_2() {
        let s = "night".to_string();
        let t = "thing".to_string();

        let result = 0;

        assert_eq!(Solution::min_steps(s, t), result);
    }
}

// Accepted solution for LeetCode #2186: Minimum Number of Steps to Make Two Strings Anagram II
function minSteps(s: string, t: string): number {
    let cnt = new Array(128).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0)];
    }
    for (const c of t) {
        --cnt[c.charCodeAt(0)];
    }
    let ans = 0;
    for (const v of cnt) {
        ans += Math.abs(v);
    }
    return ans;
}

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.