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I used the priority queue for this : 

- Create a priority queue to store integers. Priority queues ensure that the smallest element is always at the front.
- Iterate through each integer in the input array nums.
- Add each integer to the priority queue pq.
- If the size of the priority queue exceeds the value of k, remove the smallest element from the priority queue.
- After iterating through all elements, return the top element of the priority queue, which represents the kth largest element in the array.

- Initialize variables i, j, sum, and ans.
- Start a while loop with condition j less than the length of the array nums.
- Inside the loop, add the value at index j to the sum.
- Start an inner while loop with the condition sum greater than or equal to the target, and i less than the length of the array nums.
- Inside the inner loop, update ans to the minimum of (j - i + 1) and ans.
- Subtract the value at index i from sum and increment i.
- Increment j.
- After exiting the loops, return ans if it's greater than the length of the array nums, otherwise return 0.

This was a tough sliding window question :
We have a pattern as most patterns in sliding windows which helped solve it.

- Initialize the variable and to -1.
- Start a loop to iterate through each element of the array nums.
- Check if the current element is not equal to 0.
- If the element is not 0, increment ind and set the element at index ind to the current element.
- After the loop, increment ind to point to the first index after the last non-zero element.
- Start another loop from ind to the end of the array.
- Inside this loop, set each element to 0.
- After the loop, all non-zero elements will be at the beginning of the array, and all zeros will be at the end.

1. Initialize a HashMap inDegree to store the indegree of each class. Initialize an empty ArrayList result to store the order of execution.
2. Iterate through the input list of class dependencies and populate the inDegree map.
3. Create a Queue queue and enqueue all classes with an indegree of 0.
4. While the queue is not empty, do the following:
- Dequeue a class currentClass from the queue.
Add currentClass to the result list.
Iterate through the input list of class dependencies:
-If currentClass is found as a prerequisite of another class dependentClass, decrement the indegree of dependentClass by 1.
- If the indegree of dependentClass becomes 0, enqueue dependentClass.
5. After the loop, if the size of the result list is equal to the total number of classes, return the result list as the order of execution. Otherwise, return an empty list, indicating that there's a cycle in the class dependencies.

I was able to provide the naive solution like iterating through values for x and checking if the expression is satisfying.
Later after so many hints I used bit manipulation.

1. Initialize an integer variable x to 0. This will hold the value of x that satisfies the equation.
2. Start a loop from 30 to 0 (inclusive) to iterate through each bit position. 
3. Within the loop, extract the bits of a,b,c at current bit position i using right shift and bitwise and operations.
4. Check if the XOR of the corresponding bits of a and b is not equal to the corresponding bit of c. If the condition is true, it implies that changing the bit of x at position i affects the equation.
5. Inside the if statement, check two cases :
- if both a and b are 1 at bit position i return -1 which means no solution exists.
-if the corresponding bit of c is 1 at the current bit position i set the corresponding bit of x to 1 using bitwise OR operation with 1<6. After the loop completes, return the value of x which represents the integer x satisfying the equation.