Leetcode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
In this guide, we solve Leetcode #1420 Build Array Where You Can Find The Maximum Exactly K Comparisons in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
You are given three integers n, m and k. Consider the following algorithm to find the maximum element of an array of positive integers: You should build the array arr which has the following properties: arr has exactly n integers.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Dynamic Programming, Prefix Sum
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
Input: n = 2, m = 3, k = 1
Output: 6
Explanation: The possible arrays are [1, 1], [2, 1], [2, 2], [3, 1], [3, 2] [3, 3]
Python Solution
class Solution:
def numOfArrays(self, n: int, m: int, k: int) -> int:
if k == 0:
return 0
dp = [[[0] * (m + 1) for _ in range(k + 1)] for _ in range(n + 1)]
mod = 10**9 + 7
for i in range(1, m + 1):
dp[1][1][i] = 1
for i in range(2, n + 1):
for c in range(1, min(k + 1, i + 1)):
for j in range(1, m + 1):
dp[i][c][j] = dp[i - 1][c][j] * j
for j0 in range(1, j):
dp[i][c][j] += dp[i - 1][c - 1][j0]
dp[i][c][j] %= mod
ans = 0
for i in range(1, m + 1):
ans += dp[n][k][i]
ans %= mod
return ans
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.