Java Solution to Excel Sheet Column Number Problem

Day 54 of #100DaysOfLeetCode 💻✅ Solved #168. Excel Sheet Column Title problem in Java. Approach: • Used a loop to convert the number into base-26 representation • Decremented the column number by 1 to handle 1-based indexing • Found the remainder using modulo 26 to get the current character • Converted the remainder to a character using `'A' + rem` • Added the character to the front of the result string • Repeated until the column number becomes 0 Performance: ✓ Runtime: 3 ms (Beats 15.11% submissions) ✓ Memory: 42.67 MB (Beats 34.39% submissions) Key Learning: ✓ Learned how Excel column naming works using base-26 logic ✓ Practiced handling edge cases with 1-based indexing ✓ Strengthened understanding of number-to-string conversion Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #Math #ProblemSolving #CodingJourney #100DaysOfCode

Day 60 of #100DaysOfLeetCode 💻✅ Solved #389. Find the Difference problem in Java. Approach: • Initialized a variable to store the sum • Subtracted ASCII values of all characters in string s • Added ASCII values of all characters in string t • The remaining value represents the extra character • Converted the result back to character and returned it Performance: ✓ Runtime: 3 ms (Beats 38.82% submissions) ✓ Memory: 43.34 MB (Beats 30.69% submissions) Key Learning: ✓ Learned a clever use of ASCII values for problem solving ✓ Practiced using math-based approach instead of extra data structures ✓ Improved understanding of string manipulation techniques Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #Math #ProblemSolving #CodingJourney #100DaysOfCode

Day 55 of #100DaysOfLeetCode 💻✅ Solved #434. Number of Segments in a String problem in Java. Approach: • Initialized a counter to track number of words (segments) • Traversed the string character by character • Checked for non-space characters • If the current character is not a space and either it is the first character or the previous character is a space, incremented the count • This ensures counting only the starting of each word Performance: ✓ Runtime: 0 ms (Beats 100% submissions) 🚀 ✓ Memory: 41.83 MB (Beats 99.81% submissions) Key Learning: ✓ Practiced string traversal without using extra space ✓ Learned how to identify word boundaries efficiently ✓ Improved logic building for handling spaces and edge cases Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #ProblemSolving #CodingJourney #100DaysOfCode

Day 73 of #100DaysOfLeetCode 💻✅ Solved #389. Find the Difference problem in Java. Approach: • Initialized a variable to store the sum • Added ASCII values of all characters in string t • Subtracted ASCII values of all characters in string s • The remaining value represents the extra character • Converted the result back to character and returned it Performance: ✓ Runtime: 1 ms (Beats 99.89% submissions) 🚀 ✓ Memory: 42.77 MB (Beats 97.03% submissions) Key Learning: ✓ Practiced optimized string traversal using enhanced for-loop ✓ Reinforced understanding of ASCII-based solutions ✓ Learned how mathematical approaches can simplify problems Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #Math #ProblemSolving #CodingJourney #100DaysOfCode

🚀 Day 121/500 – LeetCode DSA Challenge Today I solved three problems focused on strings and number manipulation. ✅ Reverse String II – Reversed every k characters using two pointers. TC: O(n) | SC: O(n) ✅ Reverse Only Letters – Applied two-pointer approach while skipping non-letter characters. TC: O(n) | SC: O(n) ✅ Digitorial Permutation – Calculated factorial sum of digits and checked digit frequency match. TC: O(d) | SC: O(1) 💡 Key Learning: Two-pointer techniques simplify many string problems, and digit-based problems often rely on frequency counting and math logic. 👉 Day 121/500 #DSA #Java #500DaysChallenge #ProblemSolving #CodingJourney

Day 61 of #100DaysOfLeetCode 💻✅ Solved #541. Reverse String II problem in Java. Approach: • Converted the string into a character array • Traversed the array in steps of 2k • For every segment, reversed the first k characters • Used two-pointer technique to perform in-place reversal • Math.min() to handle cases where remaining characters are less than k Performance: ✓ Runtime: 2 ms (Beats 16.04% submissions) ✓ Memory: 44.80 MB (Beats 73.53% submissions) Key Learning: ✓ Practiced handling strings with fixed pattern intervals ✓ Improved control over indexing and boundaries ✓ Strengthened in-place modification techniques Learning one problem every single day 🚀 #Java #LeetCode #DSA #Strings #TwoPointers #ProblemSolving #CodingJourney #100DaysOfCode

#CodeEveryday — My DSA Journey | Day 3 🧩 Problem Solved: Combination Sum III (LeetCode #216) 💭 What I Learned: Used backtracking to find all possible combinations of k numbers (1–9) that sum up to a target n. At each step: ✔️ Decided whether to include the current number ✔️ Reduced both the remaining sum (n) and count (k) ✔️ Ensured each number is used only once by moving to the previous index Applied constraints effectively to prune recursion when: Remaining sum becomes negative Required count becomes invalid This improved my understanding of handling multiple constraints (sum + size) simultaneously in recursion. ⏱ Time Complexity: O(C(9,k) 🧠 Space Complexity: O(k) (recursion depth) ⚡ Key Takeaways: ✔️ Backtracking with multiple constraints requires careful pruning ✔️ Fixed input size can significantly reduce complexity ✔️ Choosing + skipping pattern helps explore all valid combinations 💻 Language Used: Java ☕ 📘 Concepts: Backtracking · Recursion · Combinations · Constraint Handling #CodeEveryday #DSA #LeetCode #Java #Backtracking #ProblemSolving #Algorithms #CodingJourney #Consistency 🚀

Day 74 of #100DaysOfLeetCode 💻✅ Solved #162. Find Peak Element problem in Java. Approach: • Used Binary Search technique to efficiently find the peak element • Set two pointers, left at start and right at end of the array • Calculated mid index using safe mid formula • Compared nums[mid] with nums[mid + 1] to determine direction • If mid element is smaller, moved search space to right half • Otherwise, moved search space to left half including mid • Continued until left and right pointers converged • Final position (left == right) represents the peak index Performance: ✓ Runtime: 0 ms (Beats 100.00% submissions) 🚀 ✓ Memory: 44.32 MB (Beats 25.49% submissions) Key Learning: ✓ Strengthened understanding of Binary Search on unsorted arrays ✓ Learned how to apply divide-and-conquer beyond traditional searching ✓ Improved intuition for peak finding using neighbor comparison ✓ Practiced optimizing search space instead of linear scanning Learning one problem every single day 🚀 #Java #LeetCode #DSA #BinarySearch #Arrays #ProblemSolving #CodingJourney #100DaysOfCode

Day 57 of #100DaysOfLeetCode 💻✅ Solved #506. Relative Ranks problem in Java. Approach: • Iterated through each player's score • Compared it with all other scores to determine its rank • Increased rank whenever a higher score was found • Assigned medals for top 3 ranks: Gold, Silver, Bronze • Converted remaining ranks to string and stored in result array Performance: ✓ Runtime: 124 ms (Beats 5.06% submissions) ✓ Memory: 46.76 MB (Beats 99.84% submissions) Key Learning: ✓ Understood ranking logic using comparison ✓ Practiced nested loop approach for problem solving ✓ Learned the importance of optimizing time complexity Learning one problem every single day 🚀 #Java #LeetCode #DSA #Arrays #Sorting #ProblemSolving #CodingJourney #100DaysOfCode

Day 7/50 | #50DaysOfCode 📍 Platform: LeetCode 💻 Language: Java ✅ 12. Integer to Roman (Medium) Today’s problem focused on converting integers into Roman numerals using greedy logic. It helped strengthen my understanding of mapping values and handling special subtractive cases. 🔎 Approach: Create arrays for Roman symbols and their corresponding values Start from the largest value and move to the smallest For each value, subtract it from the number while possible and append the symbol Handle special cases like 4 (IV), 9 (IX), 40 (XL), 90 (XC), etc. Continue until the number becomes zero Return the final Roman numeral string 📌 Example: Input: num = 1994 Output: "MCMXCIV" Explanation: 1000 → M 900 → CM 90 → XC 4 → IV This problem improved my understanding of greedy algorithms, mapping techniques, and string building in Java. #DSA #LeetCode #Java #CodingJourney #ProblemSolving #Consistency #LearningJourney #50DaysOfCode #LinkedIn