Minimum Number of Steps to Make Two Strings Anagram - 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.

Input & Output

Example 1 — Basic Case

$ Input: s = "bab", t = "aba"

› Output: 1

💡 Note: String t has 2 'a's but s only has 1 'a'. Replace one 'a' in t with 'b' to make "bab" or "bba"

Example 2 — Multiple Changes

$ Input: s = "leetcode", t = "practice"

› Output: 5

💡 Note: t needs to be transformed: 'p'→'l', 'r'→'e', 'a'→'e', 'i'→'t', 'c' remains, so 5 changes needed

Example 3 — Already Anagram

$ Input: s = "anagram", t = "mangaar"

› Output: 0

💡 Note: Both strings already have same character frequencies, no changes needed

Constraints

1 ≤ s.length ≤ 5 × 10⁴
s.length == t.length
s and t consist of lowercase English letters only

Visualization

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Asked in

f Facebook 15 a Amazon 12 M Microsoft 8

The key insight is to count character frequencies in both strings and find excess characters in t that need replacement. Best approach uses single-pass frequency counting with O(n) time and O(1) space complexity.

Common Approaches

✓ Character Frequency Counting

⏱️ Time: O(n) Space: O(1)

Count the frequency of each character in both strings. For each character, if t has more occurrences than s, those excess characters need to be replaced. The total number of excess characters is the minimum steps needed.

Character-by-Character Comparison

⏱️ Time: O(n) Space: O(1)

Iterate through both strings simultaneously, comparing characters at each position. Count how many positions have different characters. This naive approach doesn't account for the fact that characters can be rearranged.

Single Pass Optimized

⏱️ Time: O(n) Space: O(1)

Instead of two separate counting passes, we can count characters and calculate the required steps in a more efficient way. We increment count for characters in s and decrement for characters in t. The final result is half the sum of absolute differences.

Character Frequency Counting — Algorithm Steps

Step 1: Count frequency of each character in string s
Step 2: Count frequency of each character in string t
Step 3: For each character, calculate excess in t compared to s
Step 4: Sum all excess characters to get minimum steps

Visualization

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Step-by-Step Walkthrough

Count Frequencies

Count each character in both strings

Find Excess

Identify characters where t has more than s

Sum Excess

Total excess characters = minimum steps

Code -

solution.c — C

#include <stdio.h>
#include <string.h>

int solution(char* s, char* t) {
    // Use arrays to count frequencies (assuming lowercase letters only)
    int countS[26] = {0};
    int countT[26] = {0};
    
    int len = strlen(s);
    
    // Count frequencies
    for (int i = 0; i < len; i++) {
        countS[s[i] - 'a']++;
        countT[t[i] - 'a']++;
    }
    
    int steps = 0;
    // For each character, if t has more than s, add the excess to steps
    for (int i = 0; i < 26; i++) {
        if (countT[i] > countS[i]) {
            steps += countT[i] - countS[i];
        }
    }
    
    return steps;
}

int main() {
    char s[100001], t[100001];
    fgets(s, sizeof(s), stdin);
    fgets(t, sizeof(t), stdin);
    
    // Remove newlines
    s[strcspn(s, "\n")] = 0;
    t[strcspn(t, "\n")] = 0;
    
    int result = solution(s, t);
    printf("%d\n", result);
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

Two passes through strings of length n to count frequencies

✓ Linear Growth

Space Complexity

O(1)

Hash map stores at most 26 lowercase letters (constant space)

✓ Linear Space

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Medium Frequency

~15 min Avg. Time

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Minimum Number of Steps to Make Two Strings Anagram - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Character Frequency Counting — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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