How to Attempt? Monica and Joey Monica is making N food items for Joey. Each of these food items has some nutritional value stored in an array V. Help Monica to divide these food items into separate meals while making sure of following points

Question

How to Attempt? Monica and Joey Monica is making N food items for Joey. Each of these food items has some nutritional value stored in an array V. Help Monica to divide these food items into separate meals while making sure of following points: – Each meal should contain at least K number of food items. – The difference between the maximum and the minimum nutritional value of food items in the meals should be at most M. Your task is to find and return the minimum number of Meals in which Monica can divide the food items. If the food items cannot be divided into meals by following the above criteria, return −1. Input Specification input 1: An integer value N, representing number of food items. input2: An integer value M, representing maximum allowed nutritional value in a meal. input3: An integer value K, representing minimum number of food items required in one meal. input4: An integer array V, representing nutritional values of food items. Output Specification Return the minimum number of meals in which Monica can divide the food items. Examole 1 one meal. input4: An integer array V, representing nutritional values of food items. Output Specification Return the minimum number of meals in which Monica can divide the food items: Example 1 input1: 3 input2: 1 input3: 1 input4: {1,1,1} input 1: 3 input2: 1 input3: 1 input4: {1,1,1} Output: 1 Explanation: Here all 3 food items can be a single meal as the difference between maximum and minimum nutritional value of items is 0 while maximum allowed difference is M=1. Hence, 1 is returned as output. Example 2  input1: 2 input2: 1 input3: 2 input4: {1,4}​ Output: -1 Explanation: Here we must put both items in one meal as the number of food items is 2 and

Answer

Monica’s food item division into meals can be tackled with a dynamic programming approach. Here’s how you can solve the problem:

1. State Definition:

Define the state dp[i][j] where:

  • i represents the index of the food item currently being considered.
  • j represents the maximum difference in nutritional value allowed within the current meal.

This state signifies the minimum number of meals needed to distribute the first i food items, ensuring each meal has at least K items and a maximum difference of j in nutritional value.

2. Base Cases:

  • dp[0][0] = 0: No food items, no meals needed.
  • dp[i][j] = INF for i < K or j < 0: Not enough items or exceeding allowed difference, invalid state.

3. Recurrence Relation:

dp[i][j] = min(dp[i - 1][j], 1 + dp[i - K][min(j, V[i] - V[i - K + 1])])

This equation computes the minimum number of meals for the current state:

  • dp[i - 1][j]: Represents reusing the previous meal (no change in difference).
  • 1 + dp[i - K][min(j, V[i] - V[i - K + 1])]: Creates a new meal with the latest K items (updating difference).
  • min(j, V[i] - V[i - K + 1]): Clamps the difference to the allowed maximum between the current item and the last item in the new meal.

4. Iteration:

Iterate over i from K to N and j from 0 to M. At each state, compute the minimum using the equation and store it in dp[i][j].

5. Answer:

Finally, after iterating through all states, dp[N][0] provides the minimum number of meals required to distribute all food items, adhering to the constraints. If dp[N][0] remains INF, it signifies an impossible division, and you should return -1.

Complexity:

This approach has a time complexity of O(NM) and a space complexity of O(NM).

Example Implementations:

You can implement this solution in various programming languages like Python, Java, C++, etc. Here’s a Python example:

Python

def minMeals(N, M, K, V):
    dp = [[float('inf')] * (M + 1) for _ in range(N + 1)]
    dp[0][0] = 0
    for i in range(K, N + 1):
        for j in range(M + 1):
            dp[i][j] = min(dp[i - 1][j], 1 + dp[i - K][min(j, V[i] - V[i - K + 1])])
    return dp[N][0] if dp[N][0] != float('inf') else -1

# Example usage
N, M, K, V = 3, 1, 1, [1, 1, 1]
min_meals = minMeals(N, M, K, V)
print(f"Minimum meals needed: {min_meals}")

This example shows how to calculate the minimum meals for the given input in Example 1. You can adjust the parameters and V array to test different scenarios.

Remember that this is an optimal solution with efficient complexity. Depending on the problem constraints and desired trade-offs, alternative approaches might also be possible.

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