LeetCode #1420 — HARD

Build Array Where You Can Find The Maximum Exactly K Comparisons

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

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.
1 <= arr[i] <= m where (0 <= i < n).
After applying the mentioned algorithm to arr, the value search_cost is equal to k.

Return the number of ways to build the array arr under the mentioned conditions. As the answer may grow large, the answer must be computed modulo 10⁹ + 7.

Example 1:

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]

Example 2:

Input: n = 5, m = 2, k = 3
Output: 0
Explanation: There are no possible arrays that satisfy the mentioned conditions.

Example 3:

Input: n = 9, m = 1, k = 1
Output: 1
Explanation: The only possible array is [1, 1, 1, 1, 1, 1, 1, 1, 1]

Constraints:

1 <= n <= 50
1 <= m <= 100
0 <= k <= n

Step 01

Brute Force Baseline

Problem summary: 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. 1 <= arr[i] <= m where (0 <= i < n). After applying the mentioned algorithm to arr, the value search_cost is equal to k. Return the number of ways to build the array arr under the mentioned conditions. As the answer may grow large, the answer must be computed modulo 109 + 7.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Dynamic Programming

Example 1

2
3
1

Example 2

5
2
3

Example 3

9
1
1

Step 02

Core Insight

What unlocks the optimal approach

Use dynamic programming approach. Build dp table where dp[a][b][c] is the number of ways you can start building the array starting from index a where the search_cost = c and the maximum used integer was b.
Recursively, solve the small sub-problems first. Optimize your answer by stopping the search if you exceeded k changes.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
class Solution {
    private static final int MOD = (int) 1e9 + 7;

    public int numOfArrays(int n, int m, int k) {
        if (k == 0) {
            return 0;
        }
        long[][][] dp = new long[n + 1][k + 1][m + 1];
        for (int i = 1; i <= m; ++i) {
            dp[1][1][i] = 1;
        }
        for (int i = 2; i <= n; ++i) {
            for (int c = 1; c <= Math.min(i, k); ++c) {
                for (int j = 1; j <= m; ++j) {
                    dp[i][c][j] = (dp[i - 1][c][j] * j) % MOD;
                    for (int j0 = 1; j0 < j; ++j0) {
                        dp[i][c][j] = (dp[i][c][j] + dp[i - 1][c - 1][j0]) % MOD;
                    }
                }
            }
        }
        long ans = 0;
        for (int i = 1; i <= m; ++i) {
            ans = (ans + dp[n][k][i]) % MOD;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
func numOfArrays(n int, m int, k int) int {
	if k == 0 {
		return 0
	}
	mod := int(1e9) + 7
	dp := make([][][]int, n+1)
	for i := range dp {
		dp[i] = make([][]int, k+1)
		for j := range dp[i] {
			dp[i][j] = make([]int, m+1)
		}
	}
	for i := 1; i <= m; i++ {
		dp[1][1][i] = 1
	}
	for i := 2; i <= n; i++ {
		for c := 1; c <= k && c <= i; c++ {
			for j := 1; j <= m; j++ {
				dp[i][c][j] = (dp[i-1][c][j] * j) % mod
				for j0 := 1; j0 < j; j0++ {
					dp[i][c][j] = (dp[i][c][j] + dp[i-1][c-1][j0]) % mod
				}
			}
		}
	}
	ans := 0
	for i := 1; i <= m; i++ {
		ans = (ans + dp[n][k][i]) % mod
	}
	return ans
}

# Accepted solution for LeetCode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
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

// Accepted solution for LeetCode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
struct Solution;
use std::collections::HashMap;

const MOD: i64 = 1_000_000_007;

impl Solution {
    fn num_of_arrays(n: i32, m: i32, k: i32) -> i32 {
        let mut memo: HashMap<(i32, i32, i32), i64> = HashMap::new();
        Self::dp(n, m, k, &mut memo) as i32
    }

    fn dp(n: i32, m: i32, k: i32, memo: &mut HashMap<(i32, i32, i32), i64>) -> i64 {
        if let Some(&res) = memo.get(&(n, m, k)) {
            return res;
        }
        let res = if k == 0 || m < 1 {
            0
        } else {
            if n == 1 {
                if k == 1 {
                    m as i64
                } else {
                    0
                }
            } else {
                (Self::dp(n, m - 1, k, memo)
                    + Self::dp(n - 1, m - 1, k - 1, memo)
                    + (Self::dp(n - 1, m, k, memo) + MOD - Self::dp(n - 1, m - 1, k, memo))
                        * m as i64)
                    % MOD
            }
        };
        memo.insert((n, m, k), res);
        res
    }
}

#[test]
fn test() {
    let n = 2;
    let m = 3;
    let k = 1;
    let res = 6;
    assert_eq!(Solution::num_of_arrays(n, m, k), res);
    let n = 5;
    let m = 2;
    let k = 3;
    let res = 0;
    assert_eq!(Solution::num_of_arrays(n, m, k), res);
    let n = 9;
    let m = 1;
    let k = 1;
    let res = 1;
    assert_eq!(Solution::num_of_arrays(n, m, k), res);
    let n = 50;
    let m = 100;
    let k = 25;
    let res = 34549172;
    assert_eq!(Solution::num_of_arrays(n, m, k), res);
    let n = 37;
    let m = 17;
    let k = 7;
    let res = 418930126;
    assert_eq!(Solution::num_of_arrays(n, m, k), res);
}

// Accepted solution for LeetCode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1420: Build Array Where You Can Find The Maximum Exactly K Comparisons
// class Solution {
//     private static final int MOD = (int) 1e9 + 7;
// 
//     public int numOfArrays(int n, int m, int k) {
//         if (k == 0) {
//             return 0;
//         }
//         long[][][] dp = new long[n + 1][k + 1][m + 1];
//         for (int i = 1; i <= m; ++i) {
//             dp[1][1][i] = 1;
//         }
//         for (int i = 2; i <= n; ++i) {
//             for (int c = 1; c <= Math.min(i, k); ++c) {
//                 for (int j = 1; j <= m; ++j) {
//                     dp[i][c][j] = (dp[i - 1][c][j] * j) % MOD;
//                     for (int j0 = 1; j0 < j; ++j0) {
//                         dp[i][c][j] = (dp[i][c][j] + dp[i - 1][c - 1][j0]) % MOD;
//                     }
//                 }
//             }
//         }
//         long ans = 0;
//         for (int i = 1; i <= m; ++i) {
//             ans = (ans + dp[n][k][i]) % MOD;
//         }
//         return (int) ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × m)

Space

O(n × m)

Approach Breakdown

RECURSIVE

O(2ⁿ) time

O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING

O(n × m) time

O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.