2186. Minimum Number of Steps to Make Two Strings Anagram II
Problem Description
In this problem, we are given two strings s
and t
. The task is to perform a series of steps with the goal of making these two strings anagrams of each other. An anagram is a rearrangement of letters to form a different word or phrase, typically using all the original letters exactly once; for example, 'listen' is an anagram of 'silent'.
The only allowed action in each step we can take is to append any character we want to the end of either string s
or string t
. The main objective is to determine the fewest number of such steps required to make the strings anagrams of each other. There is no restriction on how many times a character can be added, and there is no need to add the same character to both strings.
Intuition
The intuition behind the solution comes from the definition of an anagram. Since both s
and t
must have the same characters in the same amounts to be anagrams of each other, what we ultimately need to do is to balance out the character counts in both strings.
To find the minimum number of steps required, we should:
-
Count the frequency of each character in string
s
. This is because we need to know how many of each character we have to begin with and what we may need to add tot
to balance the two strings. -
Subtract the frequency of each character that appears in
t
from the frequency count obtained froms
. When we do this, the count for a character will increase ift
has fewer instances of that character thans
, decrease ift
has more of that character, or remain the same if both have an equal number. This gives us a measure of the deficit or surplus of characters int
relative tos
. -
To make
s
andt
anagrams of each other, we need to make up for the character deficits between the two strings - in other words, we need to convert the negative counts to zero by appending characters tot
, and reduce the positive counts to zero by appending characters tos
. We don't actually append the characters; we just need to count how many characters would need to be appended. -
We sum the absolute values of these counts, which gives us the total number of characters that need to be added to either string; in other words, the minimum number of steps required to make the strings anagrams.
The given solution achieves exactly this with an efficient approach utilizing the Counter
from Python's collections module to tally the characters quickly and a loop to adjust the counts based on the second string t
.
Solution Approach
The solution uses the Counter
class from Python's collections
module, which is a subclass of dict. It is designed to count objects, a special data structure that is ideal for tallying a hashable object (in our case, the characters in the string s
).
Here's how the implementation works step by step:
-
cnt = Counter(s)
: We instantiate aCounter
object for the first strings
. This counts the frequency of each character ins
and stores it as key-value pairs, where the key is the character, and the value is the number of times that character occurs ins
. -
for c in t: cnt[c] -= 1
: We iterate over each characterc
in the second stringt
. For every character, we decrement its count in theCounter
by one. Ift
has more of a particular character thans
, the count for that character will become negative, representing a surplus of that character int
. Ift
has fewer of that character, the count for that character will either decrease towards zero (ifs
had more initially) or stay negative. -
return sum(abs(v) for v in cnt.values())
: After adjusting the counts based on both strings, we compute the sum of the absolute values of these counts. This is because we want the total number of characters that are in surplus or deficit—regardless of the string to which they need to be added—to makes
andt
anagrams. As explained in the intuition, negative counts indicate a surplus int
(and thus a need to add characters tos
), while positive counts indicate a surplus ins
(and a need to add characters tot
). The absolute value treats both cases the same way, as simply characters to be added.
The use of abs()
function in step 3 is crucial because we want to ignore whether the final count is positive or negative for our purpose. We are only interested in how many characters in total are out of balance.
This approach offers an efficient and straightforward solution to the problem: it leverages Python's Counter
to handle the calculations in a condensed manner without the need for lots of loops or conditionals, and the final result is simply the sum of absolute differences in character counts between the two strings.
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Start EvaluatorExample Walkthrough
Let's walk through a small example using the solution approach described above. Suppose we have s = "abc"
and t = "xabc"
as our strings, and we want to make them anagrams of each other.
Using the solution approach:
-
Count the characters in
s
: First, we will create aCounter
object for strings
. After this step,cnt
will look like{'a': 1, 'b': 1, 'c': 1}
since these are the frequencies of the characters ins
. -
Adjust counts based on
t
: Next, we'll iterate through the second stringt
and update the counts in ourCounter
object. For each character int
, we will decrease the corresponding count incnt
. After processingt
, which hasx
that doesn't exist ins
and all the other characters exactly once,cnt
will be updated like this:cnt['a']
will be1 - 1 = 0
cnt['b']
will be1 - 1 = 0
cnt['c']
will be1 - 1 = 0
cnt['x']
will be0 - 1 = -1
(since 'x' was not ins
to begin with, it starts at zero and gets decreased)
-
Compute the number of steps needed to make
s
andt
anagrams: After the second step, our countercnt
shows thats
andt
are already anagrams for charactersa
,b
, andc
(because their counts are 0), but we have an extra 'x' in stringt
that needs to be balanced. Thus,cnt
is{'a': 0, 'b': 0, 'c': 0, 'x': -1}
now. We are interested in the sum of the absolute values of the counts:abs(0) + abs(0) + abs(0) + abs(-1) = 0 + 0 + 0 + 1 = 1
So, in this example, only 1 step is required to make s
and t
anagrams of each other, and that step is to append an 'x' to s
.
This approach highlights the power of using a Counter
object to efficiently tackle this problem and provides a simple and clear Example Walkthrough showing how one would apply the solution to a real-world example.
Solution Implementation
1from collections import Counter
2
3class Solution:
4 def minSteps(self, s: str, t: str) -> int:
5 # Create a Counter object for the string s to keep track of character frequencies
6 char_count = Counter(s)
7
8 # Decrease the count in the Counter for every character in string t
9 for char in t:
10 char_count[char] -= 1
11
12 # Sum the absolute values of the counts in the Counter.
13 # This gives the total number of characters that are different
14 # between s and t, which is the number of steps required.
15 return sum(abs(value) for value in char_count.values())
16
1class Solution {
2 public int minSteps(String s, String t) {
3 // Initialize an array to store the character counts for the alphabet
4 int[] charCounts = new int[26];
5
6 // Increment count for each character in string s
7 for (char c : s.toCharArray()) {
8 charCounts[c - 'a']++;
9 }
10
11 // Decrement count for each character in string t
12 for (char c : t.toCharArray()) {
13 charCounts[c - 'a']--;
14 }
15
16 // Initialize a variable to store the minimum number of steps required
17 int minSteps = 0;
18
19 // Sum the absolute values of the differences in character counts
20 for (int count : charCounts) {
21 minSteps += Math.abs(count);
22 }
23
24 // Return the total number of steps to make t an anagram of s
25 return minSteps;
26 }
27}
28
1#include <vector>
2#include <string>
3#include <cmath> // Include cmath for std::abs
4
5class Solution {
6public:
7 // Function to find the minimum number of steps to make two strings anagrams
8 int minSteps(string s, string t) {
9
10 // Create a vector to store the count of each character in the alphabet
11 vector<int> charCount(26, 0);
12
13 // Increment the count for each character in string 's'
14 for (char& c : s) {
15 ++charCount[c - 'a'];
16 }
17
18 // Decrement the count for each character in string 't'
19 for (char& c : t) {
20 --charCount[c - 'a'];
21 }
22
23 // Initialize the answer variable to count the number of steps
24 int steps = 0;
25
26 // Sum the absolute values of the counts
27 // This represents the minimum number of character changes needed
28 for (int count : charCount) {
29 steps += std::abs(count);
30 }
31
32 // Return the total number of steps required
33 return steps;
34 }
35};
36
1// Function to determine the minimum number of steps to make two strings anagrams of each other
2function minSteps(s: string, t: string): number {
3 // Initialize an array to count character occurrences
4 const charCounts: number[] = new Array(128).fill(0);
5
6 // Count the occurrences of each character in string `s`
7 for (const char of s) {
8 charCounts[char.charCodeAt(0)]++;
9 }
10
11 // Subtract the occurrences of each character in string `t`
12 for (const char of t) {
13 charCounts[char.charCodeAt(0)]--;
14 }
15
16 // Calculate the total number of steps required to make the strings anagrams
17 let totalSteps = 0;
18 for (const count of charCounts) {
19 totalSteps += Math.abs(count);
20 }
21
22 // Return the total number of steps
23 return totalSteps;
24}
25
Time and Space Complexity
Time complexity
The time complexity of the provided code is derived from iterating over both strings s
and t
and the summation of the values in the counter cnt
.
cnt = Counter(s)
: Creating a counter for strings
has a time complexity ofO(n)
wheren
is the length of strings
.- The for-loop iterating over string
t
: The loop runs for the length of stringt
, which ism
. Therefore, the time complexity for this part isO(m)
. - The summation
sum(abs(v) for v in cnt.values())
: This operation is dependent on the number of unique characters ins
. In the worst case, each character is unique, and the complexity of this summation would beO(u)
, whereu
is the number of unique characters, which is at mostn
.
Summing up, if n
is the length of string s
and m
is the length of string t
, the total time complexity is O(n + m + u)
. Since u <= n
, we can simplify this to O(n + m)
.
Space complexity
The space complexity is mainly due to the storage required for the counter cnt
.
cnt = Counter(s)
: The counter for strings
needs space for each unique character. In the worst case, this isO(u)
whereu
is the number of unique characters ins
, which is at mostn
characters.
Thus, the space complexity of the algorithm is O(u)
, and considering u <= n
, it simplifies to O(n)
.
Learn more about how to find time and space complexity quickly using problem constraints.
What are the two properties the problem needs to have for dynamic programming to be applicable? (Select 2)
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