Solution Leetcode and Explaination Every single line

Author: Github-Saurabh0

62. Letter Combinations of a Phone Number/Program Explaination.txt

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+✅ Java Code (Backtracking Approach)
+
+class Solution {
+ private static final String[] KEYPAD = {
+ "", "", "abc", "def", "ghi",
+ "jkl", "mno", "pqrs", "tuv", "wxyz"
+ };
+
+ public List<String> letterCombinations(String digits) {
+ List<String> result = new ArrayList<>();
+ if (digits.isEmpty()) return result;
+ backtrack(digits, 0, new StringBuilder(), result);
+ return result;
+ }
+
+ private void backtrack(String digits, int index, StringBuilder current, List<String> result) {
+ if (index == digits.length()) {
+ result.add(current.toString());
+ return;
+ }
+
+ String letters = KEYPAD[digits.charAt(index) - '0'];
+ for (char c : letters.toCharArray()) {
+ current.append(c);
+ backtrack(digits, index + 1, current, result);
+ current.deleteCharAt(current.length() - 1); // Backtrack step
+ }
+ }
+}
+
+📘 Hinglish Explanation (Line-by-Line)
+
+private static final String[] KEYPAD = {...};
+🔹 Mobile keypad ke letters ko array me store kiya (2 = abc, 3 = def ...).
+
+public List<String> letterCombinations(String digits) {
+ List<String> result = new ArrayList<>();
+ 🔹 Final result store karne ke liye list banayi.
+
+ if (digits.isEmpty()) return result;
+ 🔹 Agar input empty ho toh khali list return kar do.
+
+ backtrack(digits, 0, new StringBuilder(), result);
+ 🔹 Backtracking start kiya from index 0.
+}
+
+private void backtrack(String digits, int index, StringBuilder current, List<String> result) {
+ if (index == digits.length()) {
+ result.add(current.toString());
+ return;
+ 🔹 Base case: agar sab digits cover ho gaye, toh combination result me daal do.
+ }
+
+ String letters = KEYPAD[digits.charAt(index) - '0'];
+ 🔹 Current digit ke corresponding letters uthaye (eg. '2' = "abc").
+
+ for (char c : letters.toCharArray()) {
+ current.append(c);
+ 🔹 Letter ko current combination me add kiya.
+
+ backtrack(digits, index + 1, current, result);
+ 🔹 Next digit ke liye recursive call.
+
+ current.deleteCharAt(current.length() - 1);
+ 🔹 Backtrack → last added letter ko hata diya.
+ }
+}
+
+
+🧠 Example Input:
+
+Input: digits = "23"
+Output: ["ad","ae","af","bd","be","bf","cd","ce","cf"]
+🚀 Time & Space Complexity:
+
+Time: O(3ⁿ × 4ᵐ), where n = number of digits like 2-6-8 (3 letters), m = number of digits like 7 or 9 (4 letters)
+
+Space: O(n) → recursive call stack
+
+🔗 Need more help or Java tricks?
+https://www.linkedin.com/in/saurabh884095/

62. Letter Combinations of a Phone Number/Solution.java

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+class Solution {
+ private static final String[] KEYPAD = {
+ "", "", "abc", "def", "ghi",
+ "jkl", "mno", "pqrs", "tuv", "wxyz"
+ };
+
+ public List<String> letterCombinations(String digits) {
+ List<String> result = new ArrayList<>();
+ if (digits.isEmpty()) return result;
+ backtrack(digits, 0, new StringBuilder(), result);
+ return result;
+ }
+
+ private void backtrack(String digits, int index, StringBuilder current, List<String> result) {
+ if (index == digits.length()) {
+ result.add(current.toString());
+ return;
+ }
+
+ String letters = KEYPAD[digits.charAt(index) - '0'];
+ for (char c : letters.toCharArray()) {
+ current.append(c);
+ backtrack(digits, index + 1, current, result);
+ current.deleteCharAt(current.length() - 1); // Backtrack step
+ }
+ }
+}