 # 1420. Build Array Where You Can Find The Maximum Exactly K Comparisons

Updated: 2023-10-27 LeetCode Problem 1420

## Problem Statement

In this problem, we have three integers, n, m, and k. We need to construct an array arr having the following properties:

1. It consists of exactly n integers.
2. Each integer in the array is between 1 and m inclusive.
3. After executing a certain algorithm on arr, we get a value known as search_cost. Our goal is to ensure search_cost is equal to k.

The main challenge is determining how many ways we can construct such an array arr.

## Naive Solution

A naive approach might involve generating all possible array combinations, then determining which ones fulfill our criteria. This method, however, would be inefficient due to its exponential time complexity. Given the constraints, this naive method won’t be feasible.

## Hints & Tips

• Utilize dynamic programming to avoid recalculating overlapping subproblems.
• Keeping track of the maximum value encountered so far can help narrow down the possible outcomes.

## Approach / Idea

To tackle this problem efficiently, we use dynamic programming. The main idea is to maintain a three-dimensional dp array, which keeps track of:

1. Current length of the array we’re constructing (i).
2. The maximum value used so far (max_so_far).
3. Remaining comparisons (remain).

With this DP table, we can progressively compute how many ways we can construct an array of length i while meeting our conditions.

## Steps / High-Level Algorithm

1. Initialize the DP Array: Create a three-dimensional dp array filled with zeros.

2. Base Case: When the array length equals n, the possible values for max_so_far are already decided, hence set dp[n][max_so_far] to 1.

3. Fill the DP Table:

• Iterate backwards, starting from the end towards the beginning.
• For each i, determine the number of ways we can construct an array of that length based on max_so_far and remain.

Note: This is where the majority of the dynamic programming logic comes into play.

4. Calculate the Result: Once the DP table is complete, dp[k] contains the number of ways we can construct the array.

## Solution

Here’s the python code:

class Solution:
def numOfArrays(self, n: int, m: int, k: int) -> int:
dp = [[ * (k + 1) for _ in range(m + 1)] for __ in range(n + 1)]
MOD = 10 ** 9 + 7

for num in range(len(dp)):
dp[n][num] = 1

for i in range(n - 1, -1, -1):
for max_so_far in range(m, -1, -1):
for remain in range(k + 1):
ans = (max_so_far * dp[i + 1][max_so_far][remain]) % MOD

if remain > 0:
for num in range(max_so_far + 1, m + 1):
ans = (ans + dp[i + 1][num][remain - 1]) % MOD

dp[i][max_so_far][remain] = ans

return dp[k]