Strings

Reverse String (LeetCode 344)

class Solution {
    public void reverseString(char[] s) {
        int left = 0, right = s.length - 1;
        while (left < right) {
            char temp = s[left];
            s[left] = s[right];
            s[right] = temp;
            left++; right--;
        }
    }
}

• Dry run (s = [h,e,l,l,o]):
  • swap 0↔4 → [o,e,l,l,h]
  • swap 1↔3 → [o,l,l,e,h]
  • left=2,right=2 → stop → [o,l,l,e,h]

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Length of Last Word (LeetCode 58)

class Solution {
    public int lengthOfLastWord(String s) {
        String[] words = s.trim().split(" ");
        return words[words.length - 1].length();
    }
}

• Dry run ("Hello World "):
  • trim → "Hello World"
  • split → ["Hello","World"] → last="World" → len=5

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Roman to Integer (LeetCode 13)

class Solution {
    public int romanToInt(String s) {
        int sum = 0;
        int num = 0;
        for (int i = s.length() - 1; i >= 0; i--) {
            switch (s.charAt(i)) {
                case 'I': num = 1; break;
                case 'V': num = 5; break;
                case 'X': num = 10; break;
                case 'L': num = 50; break;
                case 'C': num = 100; break;
                case 'D': num = 500; break;
                case 'M': num = 1000; break;
            }
            if (num * 4 < sum) sum -= num; else sum += num;
        }
        return sum;
    }
}

• Dry run ("MCMXCIV") from right:
  • V(5) → sum=5
  • I(1): 1*4<5 → sum=4
  • C(100): 100*4<4? no → sum=104
  • X(10): 10*4<104 → sum=94
  • M(1000): add → 1094
  • C(100): subtract → 994
  • M(1000): add → 1994

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Anagram (LeetCode 242)

class Solution {
    public boolean isAnagram(String s, String t) {
        if (s.length() != t.length()) return false;
        int[] freq = new int[26];
        for (int i = 0; i < s.length(); i++) {
            freq[s.charAt(i) - 'a']++;
            freq[t.charAt(i) - 'a']--;
        }
        for (int count : freq) if (count != 0) return false;
        return true;
    }
}

• Dry run (s="ab", t="ba"):
  • i=0: freq[a]++ →1; freq[b]-- →-1
  • i=1: freq[b]++ →0; freq[a]-- →0
  • All zeros → true

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Frequency of Character (uppercase)

class Main {
    public static void main(String[] args) {
        String s = "SURYAS";
        int[] freq = new int[26];
        for (int i = 0; i < s.length(); i++) {
            freq[s.charAt(i) - 'A']++;
        }
        for (int i = 0; i < freq.length; i++) {
            if (freq[i] > 0) {
                System.out.println((char)(i + 'A') + " " + freq[i]);
            }
        }
    }
}

• Dry run:
  • S→ index 18 → freq[18]=1
  • U→ 20 → 1
  • R→ 17 → 1
  • Y→ 24 → 1
  • A→ 0 → 1
  • S→ 18 → 2
  • Output in alphabetical order: A 1, R 1, S 2, U 1, Y 1