Rearrange Characters To Make Target String - Leetcode Solution

The problem "Rearrange Characters To Make Target String" on LeetCode requires us to determine whether it is possible to rearrange the characters of a given string s to form a target string t. If it is possible, we need to return true; otherwise, we return false.

The approach to solving this problem is to count the frequency of each character in s and t and compare them. If the frequency of every character in t is less than or equal to the frequency of that character in s, we can rearrange the characters of s to form t. Otherwise, it is impossible to form t from s.

Let's break down the solution into the following steps:

Step 1: Initialize two frequency arrays freq1 and freq2 of size 26, representing the frequency of characters in s and t respectively. Initially, all elements in these arrays would be 0.

Step 2: Traverse the string s and increase the frequency of each character by 1 in array freq1.

Step 3: Traverse the string t and increase the frequency of each character by 1 in array freq2.

Step 4: Traverse the frequency arrays freq1 and freq2 and compare the frequency of each character. If the frequency of every character in t is less than or equal to the frequency of that character in s, return true, meaning it is possible to rearrange the characters of s to form t. Otherwise, return false.

Here is the Python code to implement the above approach:

def canConvertString(s: str, t: str, k: int) -> bool:
    freq1 = [0] * 26
    freq2 = [0] * 26
    
    if len(s) != len(t):
        return False
    
    for i in range(len(s)):
        diff = (ord(t[i]) - ord(s[i])) % 26
        if diff != 0:
            if freq1[diff] + 26 * freq2[diff] > k - diff:
                return False
            freq1[diff] += 1
            
    return True

As you can see, in addition to the above approach, we also check whether we can form t from s using only k operations, where an operation is defined as adding or subtracting 26 from a character's ASCII value. If we cannot form t using exactly k operations, we return false.

In the code above, we compute the difference between the ASCII values of the corresponding characters in s and t. Then, we use this difference to update freq1 and freq2 arrays. Finally, we check whether we can rearrange the characters of s to form t using k operations. If it is possible, we return true; otherwise, we return false.

Overall, the time complexity of this approach is O(n), where n is the length of the string s. The space complexity is O(1) because the frequency arrays have a fixed size of 26.

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