383. Ransom Note
Problem Description
The problem presents us with two strings: ransomNote
and magazine
. Our task is to determine if it's possible to construct the ransomNote
string using the letters from the magazine
string. There are certain rules to follow for this construction:
- Each letter in the
magazine
can be used only once. - The order of the letters in the
ransomNote
does not have to match the order in themagazine
. - If all letters in the
ransomNote
can be matched to letters inmagazine
, taking into account the frequency of each letter, then the output should betrue
. - Conversely, if there are any letters in the
ransomNote
which appear more times than they do in themagazine
, or if there are any letters in theransomNote
that don't appear in themagazine
at all, the output should befalse
.
Considering the above points, an immediate intuitive answer might seem complex since it implicitly involves tracking the count of each character in both ransomNote
and magazine
. The heart of the problem lies in efficiently checking the availability of each character needed to form the ransomNote
.
Intuition
When considering the constraints of the problem, we naturally lean towards solutions that involve counting the frequency of each letter. The key, intuitive insight here is the realization that if magazine
contains all the letters required by ransomNote
in the necessary quantities, then we should be able to "use up" these letters from the magazine
without any shortage.
To arrive at the solution, we think of keeping a tally of all the letters in magazine
. A sensible way to achieve this is by using a hash table or a fixed-size array (since the alphabet is of fixed size - 26 letters for English). For each character in magazine
, we increase its count in our tally. We then decrement these counts for each letter in ransomNote
. If at any point the count for a particular letter drops below zero, it means ransomNote
requires more of that letter than magazine
can supply, and hence, we cannot construct ransomNote
with the given magazine
. The final result hinges on this check. If the entire ransomNote
is processed without any count going negative, it's possible to create ransomNote
from magazine
; otherwise, it is not.
Solution Approach
The provided solution approach utilizes a hash table data structure, more specifically a Counter
, which comes from Python's standard library collections
. The hash table is particularly suited for this problem as it allows us to efficiently keep track of the counts of individual characters.
Here is how the algorithm works step by step:
-
First, we initialize the hash table by passing the
magazine
string to theCounter
, which will count the occurrences of each character inmagazine
and store them in thecnt
hash table. -
Next, we iterate through each character
c
in theransomNote
string. For each such character, we perform two operations:- We decrement the count of the character
c
in the hash tablecnt
. - We check if the count of the character
c
falls below0
after the decrement. If it does, it indicates thatmagazine
does not have enough occurrences of characterc
to match the requirements ofransomNote
, and we returnFalse
.
- We decrement the count of the character
-
If we can iterate through all characters in
ransomNote
without the count of any character falling below0
, it implies thatmagazine
contains sufficient characters to constructransomNote
. Therefore, we returnTrue
.
Using the hash table for frequency counting is efficient as both insertion and lookup operations are on average O(1) complexity. The solution's overall time complexity is O(n+m), where n
is the length of ransomNote
and m
is the length of magazine
, since we have to iterate over both strings entirely. The space complexity is O(1) - even though we are using extra space for the hash table, it's size depends on the size of the character set (which is fixed), not the size of the input. In this case, it is the 26 letters of the English alphabet, hence it is a constant space overhead.
Using a Counter
is particularly nifty here because it wraps up the necessary operations of initializing the frequency counts from magazine
, decrementing with each character from ransomNote
, and performing the check for a negative amount in a concise and readable manner. This indicates a great example of choosing the right data structures to simplify the implementation of an algorithm.
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Start EvaluatorExample Walkthrough
Let's walk through a small example to illustrate the solution approach:
Imagine the ransomNote
is "aabbcc"
and the magazine
is "abccbaac"
. We want to know if we can construct the ransom note using the letters from the magazine.
Following the solution approach:
-
We initialize the hash table
cnt
usingCounter
with themagazine
string:1cnt = Counter("abccbaac") 2// The counter will have counts for letters as follows: {'a': 3, 'b': 2, 'c': 3}
-
Next, we iterate through each character
c
in theransomNote
string"aabbcc"
and update the hash tablecnt
:- We take the first character 'a' from the
ransomNote
:- Decrement the count of 'a' in
cnt
by 1:Counter({'a': 2, 'b': 2, 'c': 3})
- Count is not below 0, we move to the next character.
- Decrement the count of 'a' in
- We take the second character 'a':
- Decrement the count again:
Counter({'a': 1, 'b': 2, 'c': 3})
- Count is not below 0, we move to the next character.
- Decrement the count again:
- We take the first 'b':
- Decrement the count of 'b':
Counter({'a': 1, 'b': 1, 'c': 3})
- Count is not below 0, we move on.
- Decrement the count of 'b':
- We take the second 'b':
- Decrement the count:
Counter({'a': 1, 'b': 0, 'c': 3})
- Count is not below 0, so we continue.
- Decrement the count:
- We take the first 'c':
- Decrement the count of 'c':
Counter({'a': 1, 'b': 0, 'c': 2})
- Count is not below 0, move to the next character.
- Decrement the count of 'c':
- We take the second 'c':
- Decrement the count:
Counter({'a': 1, 'b': 0, 'c': 1})
- Count is not below 0.
- Decrement the count:
- We take the first character 'a' from the
-
Since we have gone through all the characters in
ransomNote
without any count falling below 0, we can conclude that it's possible to constructransomNote
from themagazine
.
Thus, the result for the example would be True
.
This example illustrates how the solution approach uses Counter
to efficiently manage character counts from the magazine, and then check against the characters required to form the ransom note.
Solution Implementation
1from collections import Counter
2
3class Solution:
4 def canConstruct(self, ransom_note: str, magazine: str) -> bool:
5 # Create a counter object for all characters in the magazine
6 char_count = Counter(magazine)
7
8 # Check each character in the ransom note
9 for char in ransom_note:
10 # Decrement the count for this character in the counter
11 char_count[char] -= 1
12
13 # If count goes below zero, we cannot construct the note from the magazine
14 if char_count[char] < 0:
15 return False
16
17 # If we haven't returned False, we can construct the ransom note
18 return True
19
1class Solution {
2
3 public boolean canConstruct(String ransomNote, String magazine) {
4 // Array to count occurrences of each letter in the magazine.
5 int[] letterCounts = new int[26];
6
7 // Populate the letterCounts array with the count of each character in the magazine.
8 for (int i = 0; i < magazine.length(); i++) {
9 // Increment the count of the current character.
10 letterCounts[magazine.charAt(i) - 'a']++;
11 }
12
13 // Check if the ransom note can be constructed using the letters in the magazine.
14 for (int i = 0; i < ransomNote.length(); i++) {
15 // Decrement the count of the current character, as it is used in the ransom note.
16 if (--letterCounts[ransomNote.charAt(i) - 'a'] < 0) {
17 // If any letter in the ransom note is in deficit, return false.
18 return false;
19 }
20 }
21
22 // If all letters are accounted for, return true.
23 return true;
24 }
25}
26
1class Solution {
2public:
3 // Function to check if ransomNote can be constructed from magazine
4 bool canConstruct(string ransomNote, string magazine) {
5 // Create a frequency array for each letter in the alphabet
6 int letterCount[26] = {};
7
8 // Iterate through each character in the magazine
9 for (char& letter : magazine) {
10 // Increase the count for the corresponding letter
11 // 'a' maps to index 0, 'b' to 1, ..., 'z' to 25
12 ++letterCount[letter - 'a'];
13 }
14
15 // Iterate through each character in the ransomNote
16 for (char& letter : ransomNote) {
17 // Decrease the count for the corresponding letter
18 if (--letterCount[letter - 'a'] < 0) {
19 // If the count goes negative, magazine doesn't have enough of this letter
20 return false;
21 }
22 }
23
24 // If we've gone through the entire note without issues, the note can be constructed
25 return true;
26 }
27};
28
1function canConstruct(ransomNote: string, magazine: string): boolean {
2 // Initialize an array to keep track of the frequency of each letter in the magazine
3 const letterCount = new Array(26).fill(0);
4
5 // Iterate over each character in the 'magazine' string
6 for (const char of magazine) {
7 // Increment the count corresponding to the current character
8 letterCount[char.charCodeAt(0) - 'a'.charCodeAt(0)]++;
9 }
10
11 // Iterate over each character in the 'ransomNote' string
12 for (const char of ransomNote) {
13 // Get the index of the current character in 'letterCount' array
14 // and decrement the frequency count of the corresponding character
15 const index = char.charCodeAt(0) - 'a'.charCodeAt(0);
16 letterCount[index]--;
17
18 // If the frequency count goes negative, we do not have enough of this character in the magazine
19 if (letterCount[index] < 0) {
20 return false;
21 }
22 }
23
24 // If we have not returned false by this point, it means we have enough characters for the ransomNote
25 return true;
26}
27
Time and Space Complexity
The time complexity of the function canConstruct
is O(m + n)
, where m
is the length of the ransomNote
string and n
is the length of the magazine
string. This is because the function first counts the occurrences of each character in the magazine string, which takes O(n)
time, and then iterates through each character in the ransom note, which takes O(m)
time. Each character decrement and comparison is an O(1)
operation, thus the total time for the loop is O(m)
. Combined, it leads to O(m + n)
time complexity.
The space complexity of the function is O(C)
, where C
is the size of the character set involved in the magazine and ransom note. Given that these are likely to be letters from the English alphabet, the size of the character set C
is 26. This fixed size of the character set means the space taken to store the counts is independent of the lengths of the input strings and is hence constant.
Learn more about how to find time and space complexity quickly using problem constraints.
Consider the classic dynamic programming of longest increasing subsequence:
Find the length of the longest subsequence of a given sequence such that all elements of the subsequence are sorted in increasing order.
For example, the length of LIS for [50, 3, 10, 7, 40, 80]
is 4
and LIS is
[3, 7, 40, 80]
.
What is the recurrence relation?
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