2

The problem asks us to find which how many times can the target string be formed from the source string. This is clearly a frequency problem. We keep track of frequency of all characters in a map for both strings. Then the minimum number of characters of string $target$ present in string $source$ is the answer. This is because if any character falls short, we can’t make a complete target string: thus won’t be considered for a valid answer.

Code:

Java

class Solution {
    public int rearrangeCharacters(String s, String target) {
        int[] freq = new int[26];
        int[] freq2 = new int[26];
 
        Arrays.fill(freq,0);
        Arrays.fill(freq2,0);
 
        for (Character c: s.toCharArray()) {
            freq[c-'a']++;
        }
 
        for (Character c: target.toCharArray()) {
            freq2[c-'a']++;
        }
 
        int res = Integer.MAX_VALUE;
        for (Character c: target.toCharArray()) {
            res = Math.min(res, freq[c-'a']/freq2[c-'a']);
        }
 
        return res == Integer.MAX_VALUE ? 0 : res;
 
    }
}

Big O Analysis

• Runtime The runtime complexity here is O(N) as there are no nested loops and we have visited the characters of source and target strings once in a loop.
• Memory The memory usage is O(26) since we have arrays of size 26: so, constant space!

— A

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