LeetCode #1347 — MEDIUM

Minimum Number of Steps to Make Two Strings Anagram

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

The Problem

Problem Statement

You are given two strings of the same length s and t. In one step you can choose any character of t and replace it with another character.

Return the minimum number of steps to make t an anagram of s.

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

Example 1:

Input: s = "bab", t = "aba"
Output: 1
Explanation: Replace the first 'a' in t with b, t = "bba" which is anagram of s.

Example 2:

Input: s = "leetcode", t = "practice"
Output: 5
Explanation: Replace 'p', 'r', 'a', 'i' and 'c' from t with proper characters to make t anagram of s.

Example 3:

Input: s = "anagram", t = "mangaar"
Output: 0
Explanation: "anagram" and "mangaar" are anagrams.

Constraints:

1 <= s.length <= 5 * 10⁴
s.length == t.length
s and t consist of lowercase English letters only.

Step 01

Brute Force Baseline

Problem summary: You are given two strings of the same length s and t. In one step you can choose any character of t and replace it with another character. Return the minimum number of steps to make t an anagram of s. 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

"bab"
"aba"

Example 2

"leetcode"
"practice"

Example 3

"anagram"
"mangaar"

Core Insight

What unlocks the optimal approach

Count the frequency of characters of each string.
Loop over all characters if the frequency of a character in t is less than the frequency of the same character in s then add the difference between the frequencies to the answer.

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 #1347: Minimum Number of Steps to Make Two Strings Anagram
class Solution {
    public int minSteps(String s, String t) {
        int[] cnt = new int[26];
        for (char c : s.toCharArray()) {
            cnt[c - 'a']++;
        }
        int ans = 0;
        for (char c : t.toCharArray()) {
            if (--cnt[c - 'a'] < 0) {
                ans++;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1347: Minimum Number of Steps to Make Two Strings Anagram
func minSteps(s string, t string) (ans int) {
	cnt := [26]int{}
	for _, c := range s {
		cnt[c-'a']++
	}
	for _, c := range t {
		cnt[c-'a']--
		if cnt[c-'a'] < 0 {
			ans++
		}
	}
	return
}

# Accepted solution for LeetCode #1347: Minimum Number of Steps to Make Two Strings Anagram
class Solution:
    def minSteps(self, s: str, t: str) -> int:
        cnt = Counter(s)
        ans = 0
        for c in t:
            cnt[c] -= 1
            ans += cnt[c] < 0
        return ans

// Accepted solution for LeetCode #1347: Minimum Number of Steps to Make Two Strings Anagram
struct Solution;

impl Solution {
    fn min_steps(s: String, t: String) -> i32 {
        let mut count: Vec<i32> = vec![0; 26];
        for c in s.chars() {
            let i = (c as u32 - 'a' as u32) as usize;
            count[i] += 1;
        }
        for c in t.chars() {
            let i = (c as u32 - 'a' as u32) as usize;
            count[i] -= 1;
        }
        count.iter().map(|x| x.abs()).sum::<i32>() / 2
    }
}

#[test]
fn test() {
    let s = "bab".to_string();
    let t = "aba".to_string();
    let res = 1;
    assert_eq!(Solution::min_steps(s, t), res);
    let s = "leetcode".to_string();
    let t = "practice".to_string();
    let res = 5;
    assert_eq!(Solution::min_steps(s, t), res);
    let s = "anagram".to_string();
    let t = "mangaar".to_string();
    let res = 0;
    assert_eq!(Solution::min_steps(s, t), res);
    let s = "xxyyzz".to_string();
    let t = "xxyyzz".to_string();
    let res = 0;
    assert_eq!(Solution::min_steps(s, t), res);
    let s = "friend".to_string();
    let t = "family".to_string();
    let res = 4;
    assert_eq!(Solution::min_steps(s, t), res);
}

// Accepted solution for LeetCode #1347: Minimum Number of Steps to Make Two Strings Anagram
function minSteps(s: string, t: string): number {
    const cnt: number[] = Array(26).fill(0);
    for (const c of s) {
        ++cnt[c.charCodeAt(0) - 97];
    }
    let ans = 0;
    for (const c of t) {
        if (--cnt[c.charCodeAt(0) - 97] < 0) {
            ++ans;
        }
    }
    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(m + n)

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.