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Build Array Where You Can Find The Maximum Exactly K Comparisons

Difficulty: Hard

Problem Description

You are given three integers n, m, and k. You need to build an array arr of exactly n integers, where each integer is between 1 and m, such that the total number of comparisons needed to find the maximum value in arr is exactly k. You should return the number of valid arrays modulo 10^9 + 7.

Key Insights

  • The maximum element in the array can only be compared to other elements in specific ways to achieve exactly k comparisons.
  • The number of elements equal to the maximum affects how many comparisons can be made.
  • If the maximum element occurs x times, the number of comparisons made will depend on the positions of these maximums relative to other elements in the array.
  • The combination of choosing positions for maximum elements and filling the remaining positions with elements less than that maximum is crucial to solve the problem.

Space and Time Complexity

Time Complexity: O(n * m * k)
Space Complexity: O(n * k)

Solution

To solve the problem, we can use dynamic programming to keep track of the number of ways to fill the array based on the number of maximums and the number of comparisons allowed. We define a DP table where dp[i][j][x] represents the number of ways to create an array of size i with j comparisons and the maximum element appearing x times.

  1. Initialize the DP table with base cases.
  2. Iterate through possible sizes of the array and the number of comparisons.
  3. For each possible maximum element, calculate the number of valid arrays based on how many times it appears.
  4. Sum the valid configurations to get the result.

Code Solutions

def countArr(n, m, k):
    MOD = 10**9 + 7
    dp = [[[0] * (n + 1) for _ in range(k + 1)] for _ in range(n + 1)]
    dp[0][0][0] = 1  # Base case: one way to create an empty array
    
    for i in range(1, n + 1):  # For each size of the array
        for j in range(k + 1):  # For each number of comparisons
            for x in range(1, min(i, m) + 1):  # For each possible maximum
                # Fill the dp table according to the rules
                for num_max in range(1, i + 1):  # Number of max elements
                    if j >= num_max - 1:  # Only if we can make those comparisons
                        dp[i][j][num_max] += dp[i - num_max][j - (num_max - 1)][0] * pow(m - 1, i - num_max, MOD)
                        dp[i][j][num_max] %= MOD
                        
    result = 0
    for x in range(1, n + 1):
        result += dp[n][k][x]
        result %= MOD
        
    return result
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