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Difficulty: 🟡 Medium
Topics: Dynamic Programming, Combinatorics
You are given three positive integers:
zero → Number of 0'sone → Number of 1'slimit → Maximum allowed consecutive identical elementsA binary array arr is called stable if:
0 in arr is exactly zero1 in arr is exactly onelimit must contain both 0 and 1Return the total number of stable binary arrays.
Since the answer may be very large, return it modulo:
10^9 + 7
Input:
zero = 1, one = 1, limit = 2
Output:
2
Explanation:
The two possible stable binary arrays are:
[1,0][0,1]Both arrays:
0 and one 1Input:
zero = 1, one = 2, limit = 1
Output:
1
Explanation:
The only stable binary array is:
[1,0,1]The arrays:
[1,1,0][0,1,1]❌ Are NOT stable because they contain subarrays of length 2 with identical elements.
Input:
zero = 3, one = 3, limit = 2
Output:
14
Explanation:
All possible stable arrays:
[0,0,1,0,1,1]
[0,0,1,1,0,1]
[0,1,0,0,1,1]
[0,1,0,1,0,1]
[0,1,0,1,1,0]
[0,1,1,0,0,1]
[0,1,1,0,1,0]
[1,0,0,1,0,1]
[1,0,0,1,1,0]
[1,0,1,0,0,1]
[1,0,1,0,1,0]
[1,0,1,1,0,0]
[1,1,0,0,1,0]
[1,1,0,1,0,0]
1 <= zero, one, limit <= 200
The constraint:
Any subarray of length > limit must contain both 0 and 1
Means:
👉 You cannot have more than limit consecutive 0's
👉 You cannot have more than limit consecutive 1's
This is a Dynamic Programming + Counting problem.
import java.util.Arrays;
public class Main {
static int MOD = 1_000_000_007;
static int[][][][] memo;
static int limit;
public static void main(String[] args) {
// 🔹 Hardcoded Inputs
int zero = 3;
int one = 3;
limit = 2;
memo = new int[zero + 1][one + 1][2][limit + 1];
for (int[][][] a : memo)
for (int[][] b : a)
for (int[] c : b)
Arrays.fill(c, -1);
int result = 0;
// Start with 0
if (zero > 0)
result = (result + dfs(zero - 1, one, 0, 1)) % MOD;
// Start with 1
if (one > 0)
result = (result + dfs(zero, one - 1, 1, 1)) % MOD;
System.out.println("Total Stable Arrays: " + result);
}
static int dfs(int zeroLeft, int oneLeft, int last, int count) {
if (zeroLeft == 0 && oneLeft == 0)
return 1;
if (memo[zeroLeft][oneLeft][last][count] != -1)
return memo[zeroLeft][oneLeft][last][count];
long ways = 0;
// Try placing 0
if (zeroLeft > 0) {
if (last == 0) {
if (count < limit)
ways += dfs(zeroLeft - 1, oneLeft, 0, count + 1);
} else {
ways += dfs(zeroLeft - 1, oneLeft, 0, 1);
}
}
// Try placing 1
if (oneLeft > 0) {
if (last == 1) {
if (count < limit)
ways += dfs(zeroLeft, oneLeft - 1, 1, count + 1);
} else {
ways += dfs(zeroLeft, oneLeft - 1, 1, 1);
}
}
return memo[zeroLeft][oneLeft][last][count] = (int)(ways % MOD);
}
}
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