Solution - #2401 - Mridul/Edited - 23/03/2025

Problems/2401 - Longest Nice Subarray - Medium/Explanation.md

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+# Explanation of the Code
+
+## **Objective**
+The goal of the program is to find the length of the longest "nice" subarray within the input array `nums`. A subarray is considered "nice" if the bitwise AND operation between every pair of elements in the subarray results in `0`. The program efficiently calculates this using a sliding window approach.
+
+---
+
+## **Step-by-Step Explanation**
+
+### **1. Initialize Variables**
+```java
+int maxLength = 1, left = 0, usedBits = 0;
+
+for (int right = 0; right < nums.length; right++) {
+    while ((usedBits & nums[right]) != 0) {
+        usedBits ^= nums[left++];
+    }
+    usedBits |= nums[right];
+    maxLength = Math.max(maxLength, right - left + 1);
+}
+Outer Loop (for): Iterates through each element of the array using the right pointer, which represents the end of the current window.
+
+Inner Loop (while):
+(usedBits & nums[right]) != 0: Checks if adding the current element (nums[right]) to the subarray would result in a bitwise AND operation that is non-zero, which violates the "nice" condition.
+
+usedBits ^= nums[left++]: Removes the element at the left pointer from the subarray by performing a bitwise XOR, effectively shrinking the window from the left until the subarray is "nice" again.
+
+Update Subarray:
+usedBits |= nums[right]: Adds the current element (nums[right]) to the subarray by updating the usedBits variable using a bitwise OR operation.
+
+maxLength = Math.max(maxLength, right - left + 1): Updates the maximum length of the "nice" subarray encountered so far.
+
+return maxLength;
+Returns the length of the longest "nice" subarray.
+
+Time Complexity: O(n)
+Each element is added and removed from the window at most once, making the overall traversal linear.
+
+Space Complexity: O(1)
+We use only a few integer variables.

Problems/2401 - Longest Nice Subarray - Medium/Solution.java

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+class Solution {
+    public int longestNiceSubarray(int[] nums) {
+        int maxLength = 1, left = 0, usedBits = 0;
+        for (int right = 0; right < nums.length; right++) {
+            while ((usedBits & nums[right]) != 0) {
+                usedBits ^= nums[left++];
+            }
+            usedBits |= nums[right];
+            maxLength = Math.max(maxLength, right - left + 1);
+        }
+        return maxLength;
+    }
+}