problemmediumalgorithmsleetcode-1347leetcode 1347leetcode1347

Minimum Number of Steps to Make Two Strings Anagram

MediumUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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.

Examples

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 * 10^4
  • s.length == t.length
  • s and t consist of lowercase English letters only.

Solution

Method 1 – Counting Character Differences

Intuition

To make t an anagram of s, we need to match the character counts. For each character, count how many more are needed in t to match s. The sum of positive differences gives the minimum steps.

Approach

  1. Count frequency of each character in s and t.
  2. For each character, if t has fewer than s, add the difference to the answer.
  3. The total is the minimum steps needed.

Code

C++
#include <string>
#include <vector>
using namespace std;
class Solution {
public:
    int minSteps(string s, string t) {
        vector<int> cnt(26, 0);
        for (char c : s) cnt[c - 'a']++;
        for (char c : t) cnt[c - 'a']--;
        int res = 0;
        for (int x : cnt) if (x > 0) res += x;
        return res;
    }
};
Go
func minSteps(s, t string) int {
    cnt := make([]int, 26)
    for _, c := range s {
        cnt[c-'a']++
    }
    for _, c := range t {
        cnt[c-'a']--
    }
    res := 0
    for _, x := range cnt {
        if x > 0 {
            res += x
        }
    }
    return res
}
Java
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 res = 0;
        for (int x : cnt) if (x > 0) res += x;
        return res;
    }
}
Kotlin
class Solution {
    fun minSteps(s: String, t: String): Int {
        val cnt = IntArray(26)
        for (c in s) cnt[c - 'a']++
        for (c in t) cnt[c - 'a']--
        var res = 0
        for (x in cnt) if (x > 0) res += x
        return res
    }
}
Python
class Solution:
    def minSteps(self, s: str, t: str) -> int:
        from collections import Counter
        cs, ct = Counter(s), Counter(t)
        return sum(max(cs[c] - ct[c], 0) for c in cs)
Rust
use std::collections::HashMap;
impl Solution {
    pub fn min_steps(s: String, t: String) -> i32 {
        let mut cs = [0; 26];
        let mut ct = [0; 26];
        for c in s.chars() { cs[(c as u8 - b'a') as usize] += 1; }
        for c in t.chars() { ct[(c as u8 - b'a') as usize] += 1; }
        (0..26).map(|i| (cs[i] - ct[i]).max(0)).sum()
    }
}
TypeScript
class Solution {
    minSteps(s: string, t: string): number {
        const cs = Array(26).fill(0), ct = Array(26).fill(0);
        for (const c of s) cs[c.charCodeAt(0) - 97]++;
        for (const c of t) ct[c.charCodeAt(0) - 97]--;
        let res = 0;
        for (let i = 0; i < 26; ++i) if (cs[i] + ct[i] > 0) res += cs[i] + ct[i];
        return res;
    }
}

Complexity

  • ⏰ Time complexity: O(n) — n = length of s (and t), single pass for counts.
  • 🧺 Space complexity: O(1) — constant extra space.

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