LeetCode Day 43: Array Simulation and XOR

🚀 LeetCode Day Problem Solving 🚀 Day-42 📌 Problem: Range Update with Step & XOR You are given an array nums[] and multiple queries. Each query contains: [l, r, k, v] 👉 For every query: • Start from index l • Jump with step k → l, l+k, l+2k ... ≤ r • Update each visited index: nums[i] = (nums[i] * v) % (10⁹ + 7) After processing all queries, return: 👉 XOR of all elements in the array 🛠 Approach: ✔ Direct Simulation 👉 For each query: for i from l to r step k: nums[i] = (nums[i] * v) % MOD ✔ After all updates: 👉 Compute XOR of entire array 🧠 Key Learning: ✔ Handling range with step (k jump) ✔ Modular multiplication (10^9 + 7) ✔ Combining simulation + bitwise XOR 📊 Complexity: ⏱ Time: O(q * (n/k)) → worst case O(n * q) 📦 Space: O(1) 🧪 Example: Input: nums = [2,3,1,5,4] queries = [[1,4,2,3],[0,2,1,2]] After operations → [4,18,2,15,4] Output: 4 ^ 18 ^ 2 ^ 15 ^ 4 = 31 ✅ Day 42 Completed 🚀 Practicing Simulation + Bit Manipulation 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 #LeetCode Day Problem Solving 🚀 Day-23 📌 Problem: Range Update with Step & XOR You are given an array nums[] and multiple queries. Each query contains: [l, r, k, v] 👉 For every query: • Start from index l • Jump with step k → l, l+k, l+2k ... ≤ r • Update each visited index: nums[i] = (nums[i] * v) % (10⁹ + 7) After processing all queries, return: 👉 XOR of all elements in the array 🛠 Approach: ✔ Direct Simulation 👉 For each query: for i from l to r step k: nums[i] = (nums[i] * v) % MOD ✔ After all updates: 👉 Compute XOR of entire array 🧠 Key Learning: ✔ Handling range with step (k jump) ✔ Modular multiplication (10^9 + 7) ✔ Combining simulation + bitwise XOR 📊 Complexity: ⏱ Time: O(q * (n/k)) → worst case O(n * q) 📦 Space: O(1) 🧪 Example: Input: nums = [2,3,1,5,4] queries = [[1,4,2,3],[0,2,1,2]] After operations → [4,18,2,15,4] Output: 4 ^ 18 ^ 2 ^ 15 ^ 4 = 31 ✅ Day 23 Completed 🚀 Practicing Simulation + Bit Manipulation 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 #LeetCode Day Problem Solving 🚀 Day-24 📌 Problem: Range Update with Step & XOR You are given an array nums[] and multiple queries. Each query contains: [l, r, k, v] 👉 For every query: • Start from index l • Jump with step k → l, l+k, l+2k ... ≤ r • Update each visited index: nums[i] = (nums[i] * v) % (10⁹ + 7) After processing all queries, return: 👉 XOR of all elements in the array 🛠 Approach: ✔ Direct Simulation 👉 For each query: for i from l to r step k: nums[i] = (nums[i] * v) % MOD ✔ After all updates: 👉 Compute XOR of entire array 🧠 Key Learning: ✔ Handling range with step (k jump) ✔ Modular multiplication (10^9 + 7) ✔ Combining simulation + bitwise XOR 📊 Complexity: ⏱ Time: O(q * (n/k)) → worst case O(n * q) 📦 Space: O(1) 🧪 Example: Input: nums = [2,3,1,5,4] queries = [[1,4,2,3],[0,2,1,2]] After operations → [4,18,2,15,4] Output: 4 ^ 18 ^ 2 ^ 15 ^ 4 = 31 ✅ Day 24 Completed 🚀 Practicing Simulation + Bit Manipulation 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 #LeetCode Day Problem Solving 🚀 Day-32 📌 Problem: Given an integer n, define its mirror distance as: 👉 |n - reverse(n)| Where: ✔ reverse(n) = number formed by reversing digits ✔ Leading zeros are ignored (e.g., 10 → 01 → 1) 🧠 Examples: Input: n = 25 Output: 27 ➡ reverse(25) = 52 → |25 - 52| = 27 Input: n = 10 Output: 9 ➡ reverse(10) = 1 → |10 - 1| = 9 Input: n = 7 Output: 0 💡 Key Insight: ✔ Reverse the digits of the number ✔ Compute absolute difference 👉 Simple math + digit manipulation problem ⚙ Approach: 1️⃣ Extract digits using % 10 2️⃣ Build reversed number 3️⃣ Compute abs(n - reversed) 📊 Complexity Analysis: ⏱ Time Complexity: O(d) (d = number of digits) 📦 Space Complexity: O(1) 🧠 What I Learned: ✔ Number reversal logic ✔ Handling leading zeros properly ✔ Clean mathematical problem solving ✅ Day 32 Completed 🚀 Improving skills in Math + Implementation Problems 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 #LeetCode Day Problem Solving 🚀 Day-30 📌 Problem: You are given a circular array nums and an array queries. For each query index i, find the minimum circular distance to another index j such that: ✔ nums[j] == nums[queries[i]] ✔ If no such index exists → return -1 🧠 Example: Input: nums = [1,3,1,4,1,3,2] queries = [0,3,5] ✅ Output: [2, -1, 3] 📖 Explanation: Query 0 → value = 1 → nearest same value at index 2 → distance = 2 Query 1 → value = 4 → no duplicate → -1 Query 2 → value = 3 → nearest at index 1 (circular path) → distance = 3 💡 Key Insight: ✔ Since array is circular, distance = 👉 min(|i - j|, n - |i - j|) ✔ Store indices of each value using a map (value → list of indices) ✔ For each query: 👉 Use binary search to find nearest index efficiently 📊 Complexity Analysis: ⏱ Time Complexity: O(n + q log n) 📦 Space Complexity: O(n) 🧠 What I Learned: ✔ Handling circular arrays smartly ✔ Using binary search on index lists ✔ Optimizing brute force to efficient lookup ✅ Day 30 Completed 🚀 Leveling up in Arrays + Binary Search + Optimization 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 #LeetCode Day Problem Solving 🚀 Day-25 📌 Problem: You are given an integer array nums. A tuple (i, j, k) is called good if: ✔ i, j, k are distinct ✔ nums[i] == nums[j] == nums[k] 👉 Distance of a good tuple: |i - j| + |j - k| + |k - i| 🎯 Goal: Find the minimum possible distance among all good tuples. If none exist → return -1. 🧠 Example: Input: nums = [1,2,1,1,3] ✅ Output: 6 📖 Explanation: Good tuple → (0,2,3) Distance = 2 + 1 + 3 = 6 💡 Key Insight: ✔ For same values, we only care about indices ✔ Sort indices for each number ✔ Try triplets of closest indices 👉 Important formula simplification: For sorted i < j < k Distance = (j-i) + (k-j) + (k-i) = 2*(k - i) ✔ So minimize → difference between farthest indices ⚡ Approach: ✔ Group indices by value (HashMap) ✔ For each value: - If size ≥ 3 → check consecutive triplets - Compute minimum distance 📊 Complexity Analysis: ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 🧠 What I Learned: ✔ Grouping using HashMap ✔ Optimizing triplet computation ✔ Using math to simplify distance formula ✅ Day 25 Completed 🚀 Leveling up in Arrays + Hashing + Optimization 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency

🚀 LeetCode Day Problem Solving 🚀 Day-52 📌 Problem: Given an integer n, define its mirror distance as: 👉 |n - reverse(n)| Where: ✔ reverse(n) = number formed by reversing digits ✔ Leading zeros are ignored (e.g., 10 → 01 → 1) 🧠 Examples: Input: n = 25 Output: 27 ➡ reverse(25) = 52 → |25 - 52| = 27 Input: n = 10 Output: 9 ➡ reverse(10) = 1 → |10 - 1| = 9 Input: n = 7 Output: 0 💡 Key Insight: ✔ Reverse the digits of the number ✔ Compute absolute difference 👉 Simple math + digit manipulation problem ⚙ Approach: 1️⃣ Extract digits using % 10 2️⃣ Build reversed number 3️⃣ Compute abs(n - reversed) 📊 Complexity Analysis: ⏱ Time Complexity: O(d) (d = number of digits) 📦 Space Complexity: O(1) 🧠 What I Learned: ✔ Number reversal logic ✔ Handling leading zeros properly ✔ Clean mathematical problem solving ✅ Day 52 Completed 🚀 Improving skills in Math + Implementation Problems 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 LeetCode Day Problem Solving 🚀 Day-50 📌 Problem: You are given a circular array nums and an array queries. For each query index i, find the minimum circular distance to another index j such that: ✔ nums[j] == nums[queries[i]] ✔ If no such index exists → return -1 🧠 Example: Input: nums = [1,3,1,4,1,3,2] queries = [0,3,5] ✅ Output: [2, -1, 3] 📖 Explanation: Query 0 → value = 1 → nearest same value at index 2 → distance = 2 Query 1 → value = 4 → no duplicate → -1 Query 2 → value = 3 → nearest at index 1 (circular path) → distance = 3 💡 Key Insight: ✔ Since array is circular, distance = 👉 min(|i - j|, n - |i - j|) ✔ Store indices of each value using a map (value → list of indices) ✔ For each query: 👉 Use binary search to find nearest index efficiently 📊 Complexity Analysis: ⏱ Time Complexity: O(n + q log n) 📦 Space Complexity: O(n) 🧠 What I Learned: ✔ Handling circular arrays smartly ✔ Using binary search on index lists ✔ Optimizing brute force to efficient lookup ✅ Day 50 Completed 🚀 Leveling up in Arrays + Binary Search + Optimization 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 LeetCode Day Problem Solving 🚀 Day-57 📌 Problem: For each index i in array nums, compute: 👉 arr[i] = sum of |i - j| for all j such that nums[j] == nums[i] and j ≠ i ✔ If no such j exists → arr[i] = 0 🧠 Example: Input: nums = [1,3,1,1,2] ✅ Output: [5,0,3,4,0] 📖 Explanation: For value 1 → indices = [0,2,3] For index 0: → |0-2| + |0-3| = 5 For index 2: → |2-0| + |2-3| = 3 For index 3: → |3-0| + |3-2| = 4 💡 Key Insight: ✔ Group indices by same values (using HashMap) ✔ For each group, compute distances efficiently 👉 Instead of brute force O(n²) ❌ Use Prefix Sum on indices 🔥 ⚡ Optimized Approach: 1️⃣ Store indices for each value 2️⃣ For each group: Maintain prefix sum of indices Use formula to compute distance in O(1) per index 👉 Left contribution + Right contribution 📊 Complexity Analysis: ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 🧠 What I Learned: ✔ Grouping using HashMap ✔ Using prefix sums on indices ✔ Turning brute force into efficient solution ✅ Day 57 Completed 🚀 Leveling up in Hashing + Prefix Sum Optimization 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency #MilanSahoo 🚀

🚀 #LeetCode Day Problem Solving 🚀 Day-37 📌 Problem: For each index i in array nums, compute: 👉 arr[i] = sum of |i - j| for all j such that nums[j] == nums[i] and j ≠ i ✔ If no such j exists → arr[i] = 0 🧠 Example: Input: nums = [1,3,1,1,2] ✅ Output: [5,0,3,4,0] 📖 Explanation: For value 1 → indices = [0,2,3] For index 0: → |0-2| + |0-3| = 5 For index 2: → |2-0| + |2-3| = 3 For index 3: → |3-0| + |3-2| = 4 💡 Key Insight: ✔ Group indices by same values (using HashMap) ✔ For each group, compute distances efficiently 👉 Instead of brute force O(n²) ❌ Use Prefix Sum on indices 🔥 ⚡ Optimized Approach: 1️⃣ Store indices for each value 2️⃣ For each group: Maintain prefix sum of indices Use formula to compute distance in O(1) per index 👉 Left contribution + Right contribution 📊 Complexity Analysis: ⏱ Time Complexity: O(n) 📦 Space Complexity: O(n) 🧠 What I Learned: ✔ Grouping using HashMap ✔ Using prefix sums on indices ✔ Turning brute force into efficient solution ✅ Day 37 Completed 🚀 Leveling up in Hashing + Prefix Sum Optimization 💪 #Leetcode #DSA #ProblemSolving #BitManipulation #CodingJourney #InterviewPreparation #Consistency