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Formatted question description: https://leetcode.ca/all/1420.html

1420. Build Array Where You Can Find The Maximum Exactly K Comparisons (Hard)

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 10^9 + 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 satisify 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]

Example 4:

Input: n = 50, m = 100, k = 25
Output: 34549172
Explanation: Don't forget to compute the answer modulo 1000000007

Example 5:

Input: n = 37, m = 17, k = 7
Output: 418930126

 

Constraints:

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

Related Topics:
Dynamic Programming

Solution 1. DP

// OJ: https://leetcode.com/problems/build-array-where-you-can-find-the-maximum-exactly-k-comparisons/
// Time: O(M^2 * NK)
// Space: O(MNK)
// Ref: https://leetcode.com/problems/build-array-where-you-can-find-the-maximum-exactly-k-comparisons/discuss/586576/C%2B%2B-Bottom-Up-Dynamic-Programming-with-Explanation
class Solution {
    typedef long long LL;
    LL dp[51][101][51] = {};
public:
    int numOfArrays(int n, int m, int k) {
        int mod = 1e9+7;
        for (int i = 0; i <= m; ++i) dp[1][i][1] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                for (int t = 1; t <= k; ++t) {
                    LL s = 0;
                    s = (s + j * dp[i - 1][j][t]) % mod;
                    for (int x = 1; x < j; ++x) s = (s + dp[i - 1][x][t - 1]) %mod;
                    dp[i][j][t] = (dp[i][j][t] + s) % mod;
                }
            }
        }
        LL ans = 0;
        for (int i = 1; i <= m; ++i) ans = (ans + dp[n][i][k]) % mod;
        return ans;
    }
};

  • class Solution {
    public int numOfArrays(int n, int m, int k) {
        final int MODULO = 1000000007;
        int[][][] dp = new int[n + 1][m + 1][k + 2];
        dp[0][0][0] = 1;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= m; j++) {
                for (int p = 0; p <= k; p++) {
                    for (int q = 1; q <= m; q++) {
                        if (q > j)
                            dp[i + 1][q][p + 1] = (dp[i + 1][q][p + 1] + dp[i][j][p]) % MODULO;
                        else
                            dp[i + 1][j][p] = (dp[i + 1][j][p] + dp[i][j][p]) % MODULO;
                    }
                }
            }
        }
        int count = 0;
        for (int i = 1; i <= m; i++)
            count = (count + dp[n][i][k]) % MODULO;
        return count;
    }
}

############

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;
    }
}

  • // OJ: https://leetcode.com/problems/build-array-where-you-can-find-the-maximum-exactly-k-comparisons/
// Time: O(M^2 * NK)
// Space: O(MNK)
// Ref: https://leetcode.com/problems/build-array-where-you-can-find-the-maximum-exactly-k-comparisons/discuss/586576/C%2B%2B-Bottom-Up-Dynamic-Programming-with-Explanation
class Solution {
    typedef long long LL;
    LL dp[51][101][51] = {};
public:
    int numOfArrays(int n, int m, int k) {
        int mod = 1e9+7;
        for (int i = 0; i <= m; ++i) dp[1][i][1] = 1;
        for (int i = 1; i <= n; ++i) {
            for (int j = 1; j <= m; ++j) {
                for (int t = 1; t <= k; ++t) {
                    LL s = 0;
                    s = (s + j * dp[i - 1][j][t]) % mod;
                    for (int x = 1; x < j; ++x) s = (s + dp[i - 1][x][t - 1]) %mod;
                    dp[i][j][t] = (dp[i][j][t] + s) % mod;
                }
            }
        }
        LL ans = 0;
        for (int i = 1; i <= m; ++i) ans = (ans + dp[n][i][k]) % mod;
        return ans;
    }
};

  • 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



  • 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
}

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