Power Programmer

Here is a step-by-step solution for the problem described:

1. **Input**: Obtain the input values, including the number of guests (N) and their happiness values (Ci).

2. **Sort the happiness values**: Arrange the happiness values in ascending order. This step helps minimize the total unhappiness.

3. **Initialize variables**: Set up variables such as total unhappiness (totalUnhappiness) and current time (currentTime) to keep track of the serving process.

4. **Iterate through guests**:
  - Start a loop to iterate through each guest, starting from the first (smallest happiness value).
  - For each guest:
    - Calculate their unhappiness as |Ci - currentTime|.
    - Add this unhappiness to the totalUnhappiness.
    - Increment currentTime by 1 unit as each guest takes exactly one unit of time to be served.

5. **Output**: Once all guests are served, output the totalUnhappiness, which represents the minimum total unhappiness achieved by serving all guests in an optimized order.

Let's illustrate this with an example:
Input: 4 2 2 3 3

1. Sort the happiness values: 2 2 3 3
2. Initialize totalUnhappiness = 0 and currentTime = 0
3. Iterate through guests:
  - Guest 1 (happiness 2): |2 - 0| = 2, totalUnhappiness = 2, currentTime = 1
  - Guest 2 (happiness 2): |2 - 1| = 1, totalUnhappiness = 2 + 1 = 3, currentTime = 2
  - Guest 3 (happiness 3): |3 - 2| = 1, totalUnhappiness = 3 + 1 = 4, currentTime = 3
  - Guest 4 (happiness 3): |3 - 3| = 0, totalUnhappiness = 4 + 0 = 4, currentTime = 4
4. Output: Total unhappiness = 4

This step-wise solution ensures an optimized order of serving guests to minimize total unhappiness.

We declare a HashMap to store the frequency map.
We store the character of string 1 in the map and then while traversing string 2, we erase the characters if the map is empty at the end that means the characters in both strings are the same and we can continue, or else we return -1.
We make a variable res and point two pointers i and j to the last of both strings and start traversing from back.
As soon as see an ith character that doesn’t match with the jth character, we start increasing res by 1 until again both the characters are the same.
At last, we return the res.

Here's a stepwise solution for finding the sum of the XOR values of all nodes in a tree:

1. Create Tree Structure:
  - Initialize a list of lists or a dictionary to represent the tree structure based on the parent-child relationships defined in array P.

3. Calculate XOR Values Recursively:
  - Implement a recursive function to calculate the XOR value for each node and its subtree.
  - Start with the root node (node 0) and recursively process each node:
    - For each node, calculate its XOR value by XORing its value (A[i]) with the XOR values of its children.
    - Recursively call the function for each child node to calculate their XOR values.

4. Sum XOR Values:
  - Traverse the tree and sum up the XOR values of all nodes to get the total sum of XOR values in the tree.

5. Modulo Operation:
  - After computing the sum of XOR values, perform a modulo operation with 10^9+7 to ensure the result stays within the specified range.

6. Return Result:
  - Return the final sum of XOR values modulo 10^9+7 as the result.

Explanation:

In the given array [1, 2, 3, 4, 5, 6], the elements from index 1 to index 4 are [2, 3, 4, 5]. The sum of these elements is 2 + 3 + 4 + 5 = 14.

Code Snippet:

def calculateSumInRange(arr, start, end):
   if start < 0 or end >= len(arr) or start > end:
       return "Invalid range"

   return sum(arr[start:end+1])

if __name__ == "__main__":
   arr = [1, 2, 3, 4, 5, 6]
   start = 1
   end = 4

   result = calculateSumInRange(arr, start, end)
   print("Sum:", result)

This code calculates the sum of elements in the array within the specified range [start, end] and handles invalid ranges by returning an appropriate message. Adjust the input array and range values based on the actual problem requirements.

Here are the steps to solve the Two Sum problem efficiently:

1. **Input Preparation**:
  - Given an array of integers `nums` and a target sum `target`.

2. **Create a HashMap**:
  - Initialize an empty HashMap to store the indices of elements.

3. **Iterate Through Array**:
  - Iterate through each element in the array `nums` with index `i`.

4. **Check Complement**:
  - Calculate the complement as `complement = target - nums[i]`.
  - Check if the complement exists in the HashMap:
    - If yes, return the indices `[hashMap[complement], i]`.
    - If not, add the current element and its index to the HashMap.

5. **Return "No Solution"**:
  - If the loop completes without finding a valid pair, return "No solution".

6. **Code Implementation** (Python Example):

```python
def twoSum(nums, target):
   # Initialize HashMap to store indices
   hashMap = {}
   
   # Iterate through array
   for i in range(len(nums)):
       complement = target - nums[i]
       # Check if complement exists in HashMap
       if complement in hashMap:
           return [hashMap[complement], i]
       # Add current element and index to HashMap
       hashMap[nums[i]] = i
   
   # No solution found
   return "No solution"

# Example usage
nums = [2, 7, 11, 15]
target = 9
result = twoSum(nums, target)
print("Indices:", result)  # Output: Indices: [0, 1] (nums[0] + nums[1] = 9)
```

This implementation uses a HashMap to efficiently find the two numbers that add up to the target sum by checking for their complements. Adjust the input array `nums` and target sum `target` based on the actual problem requirements.

Here are the steps to solve the Three Sum problem:

1. **Input Preparation**:
  - Given an array of integers `nums` and a target sum `target`.

2. **Sort the Array**:
  - Sort the array `nums` in non-decreasing order. This step is crucial for optimizing the algorithm.

3. **Initialize Result List**:
  - Initialize an empty list `result` to store the triplets that sum up to the target.

4. **Iterate Through Array**:
  - Iterate through each element in the sorted array `nums` up to the third-to-last element (to leave room for two pointers).

5. **Two Pointers Approach**:
  - For each element `nums[i]`, initialize two pointers `left = i + 1` and `right = len(nums) - 1`.
  - While `left < right`, perform the following steps:
    - Calculate the current sum as `current_sum = nums[i] + nums[left] + nums[right]`.
    - If `current_sum` equals `target`, add the triplet `[nums[i], nums[left], nums[right]]` to the `result` list.
    - If `current_sum` is less than `target`, increment `left` to move towards larger values.
    - If `current_sum` is greater than `target`, decrement `right` to move towards smaller values.

6. **Avoid Duplicates**:
  - While iterating through `nums`, skip duplicate elements to avoid duplicate triplets in the result.

7. **Return Result**:
  - After completing the iterations, return the `result` list containing all valid triplets that sum up to the target.

This implementation uses the two-pointers approach along with sorting and skipping duplicates to efficiently find all triplets in the array `nums` that sum up to the target sum `target`. Adjust the input array `nums` and target sum `target` based on the actual problem requirements.