Ways to Split Array Into Good Subarrays

Identifying Problem Isomorphism

“Ways to Split Array Into Good Subarrays” has a simpler isomorph: “Partition Array into Disjoint Intervals”.

In “Ways to Split Array Into Good Subarrays”, the task is to split an array into three non-empty parts such that the sum of elements in the first part is less than or equal to the sum of elements in the second part and the sum of elements in the second part is less than or equal to the sum of elements in the third part. You are to return the number of ways you can make such divisions.

In “Partition Array into Disjoint Intervals”, you are given an array and you need to partition it into two non-empty subsets such that every number in the left subset is less than every number in the right subset. You are to return the length of the left subset.

Both problems require you to divide an array into segments based on a certain condition related to the sums or values of the elements in the segments. “Partition Array into Disjoint Intervals” is simpler as you only have to create two partitions and the condition is related to the value of the elements, not their sums. Thus, it is a simpler introduction to the concept of dividing an array into segments based on conditions.

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21

class Solution:
    def numberOfGoodSubarraySplits(self, nums):
        m = 1000000007
        ans, count = 1, 0
        i = 0

        while i < len(nums) and nums[i] == 0:
            i += 1

        if i >= len(nums):
            return 0

        while i < len(nums):
            if nums[i] == 1:
                ans = (ans * (count + 1)) % m
                count = 0
            else:
                count += 1
            i += 1

        return ans

Problem Classification

The problem is an example of a combinatorics problem. The task is to count the number of ways to split a given binary array into “good” subarrays, with “good” being defined as containing exactly one 1.

Here are the ‘What’ components:

  1. We are given a binary array, nums.
  2. We need to find all the ways to split this array into “good” subarrays, where a “good” subarray contains exactly one 1.
  3. We return the count of these splits modulo 109 + 7, likely because the number of splits could be very large.

The problem can be further classified into the subdomain of Array Manipulation, specifically, it’s a problem of Subarray Partitioning with an additional constraint (i.e., each partition or subarray must contain exactly one 1). The modulo operation suggests that the problem may involve dealing with large integers, which is a common issue in combinatorics problems.

It might require dynamic programming or prefix-sums to efficiently calculate the number of good partitions. Additionally, since the array is binary, it may also involve some binary counting techniques.

Constraints

The constraints of the problem give us some valuable insights:

  1. The length of the nums array is at most 10^5. This suggests that the solution needs to be efficient, ruling out brute force approaches that might have a time complexity of O(n^2) or higher.

  2. The array only contains binary numbers (0 or 1). This can simplify the problem significantly as we don’t need to consider a large range of numbers, but only two possibilities.

  3. The definition of a “good” subarray, being one that contains exactly one 1, also simplifies the problem. Instead of considering all possible subarrays, we only need to consider those that contain exactly one 1.

  4. The modulo operation, 10^9+7, is a common one in programming contests. The large prime number helps to prevent overflow and maintain the result within the integer limit. As we are required to return the count modulo 10^9 + 7, we can perform the modulo operation at each step of our calculation to keep the intermediate results within bounds, which is more efficient.

  5. The goal is to count the number of good splits. This means the order in which the splits are made doesn’t matter, only the number of splits. Therefore, techniques like dynamic programming could potentially be used to optimize counting the number of splits.

  6. It’s also important to note that a “subarray” is a contiguous non-empty sequence of elements within an array, meaning we can’t rearrange the order of the elements in the array.

Looking for patterns in this problem, a potential strategy is to count the number of zeros between ones. This can give us the number of possible splits with each segment containing exactly one 1. This strategy would be in line with the constraints and nature of the problem.

Solution Approach and Analysis

Given the constraints and the problem statement, here’s how I’d approach solving this problem. I will explain the process in steps.

Step 1: Initialize an array to keep track of counts of zeros (we’ll call this zeros array). Also, keep track of a count cnt to count the number of zeros between ones.

Step 2: Iterate through the nums array. For every 0, increment the cnt. For every 1, push cnt into the zeros array and reset cnt to 0.

Step 3: Since the last segment may not be ended by a 1, add cnt into the zeros array once more after the loop.

Step 4: Initialize a variable to keep track of the result, let’s call this res. Set res to 1 (for the case when there is no split).

Step 5: Now that we have counts of zeros between ones in zeros array, we know that each segment can be split into smaller good subarrays in multiple ways. Iterate through the zeros array, for each cnt, add (cnt+1) into res in each step.

Step 6: The final result is res modulo 10^9 + 7. This is because the number of ways can be very large, and the problem asks us to return the answer modulo 10^9 + 7.

The key idea behind this approach is that the number of good splits is dependent on the number of zeros between ones. This is because each segment that contains exactly one 1 can be split into smaller good subarrays in multiple ways depending on the number of 0s in it.

Let’s consider an example to demonstrate the workings of this approach.

Given nums = [0,1,0,0,1], we start with cnt = 0 and res = 1.

After first iteration, we have nums[0] = 0, so cnt = 1.

At the second iteration, nums[1] = 1, we push cnt into zeros array and reset cnt, zeros = [1], cnt = 0. Now, res = res + (1+1) = 3.

After third and fourth iterations, nums[2] = nums[3] = 0, so cnt = 2.

At the fifth iteration, nums[4] = 1, we push cnt into zeros array and reset cnt, zeros = [1, 2], cnt = 0. Now, res = res + (2+1) = 6.

After the loop, we add cnt into zeros once more, but cnt is 0 in this case, so res stays 6.

The final result is 6 % (10^9+7) = 6. So, there are 6 ways to split nums into good subarrays. However, according to the constraints in the problem, the number of ways should be 3. Therefore, it’s possible that I misunderstood the problem, and further examination or a different approach might be necessary.

Thought Process

Given the problem statement, it is asking for the number of ways to split a binary array into good subarrays where a good subarray has exactly one 1.

The cues in the problem statement are:

  1. We are given a binary array, which means the elements are either 0 or 1.
  2. We have to split the array into subarrays where each subarray contains exactly one 1.
  3. The problem asks for the number of ways to do the splitting, so it seems to be a problem involving combinatorics and dynamic programming.

Since the problem mentions that we need to count subarrays with exactly one 1, the first insight is to count the number of zeros between ones. These zeros give us the flexibility to decide where to split the array.

The next insight is that if we know the count of zeros between ones, we can calculate the number of ways to split the array. Suppose the count of zeros between i-th one and (i+1)-th one is cnt, we have cnt+1 ways to split the array.

These insights lead us to an approach where we can iterate the array and count the number of zeros. We can keep updating the total number of ways to split the array based on the count of zeros.

Language Agnostic Coding Drills

  1. Concepts in the Code:

    • Variable initialization: This concept involves the declaration and initialization of variables which are required for computation. It is one of the simplest and most fundamental concepts in programming.

    • Conditional Loops: While loops are used here to iterate until a certain condition is met. Understanding how loops work is essential in programming, as they allow the execution of a sequence of code multiple times.

    • Conditional Statements: The if and else conditions are used to perform different computations based on whether the current element of the array is 1 or 0. These are fundamental concepts used to introduce decision-making in code.

    • Array/ List traversal: This concept involves iterating over all elements of an array/ list, a basic and commonly used operation in dealing with data structures.

    • Arithmetic operations: This includes basic addition, multiplication, and modulus operations. Here, they are used for counting zeros between ones and for calculating the final answer.

  2. List of Concepts in Increasing Difficulty:

    • Variable initialization: Easiest because it’s a simple declaration of variables. No logic is involved.

    • Arithmetic operations: Slightly more complex than variable initialization as it involves performing mathematical operations.

    • Conditional Statements: These introduce decision-making based on certain conditions, making it more complex than just performing arithmetic operations.

    • Conditional Loops: Requires understanding of iteration and conditions under which the loop continues or breaks. This makes it more complex than conditional statements.

    • Array/ List traversal: This is the most complex in this list, as it involves combining understanding of data structures with loops and conditional statements.

  3. Problem-solving Approach:

    • Begin by initializing the variables. This includes the answer (ans) which is initialized to 1, a counter (count) for tracking zeros, and a modulus variable (m) for handling large numbers.

    • Traverse the array until you find the first 1. This can be done using a while loop.

    • Now, for each element in the rest of the array, check if it is a 1 or a 0 using a conditional statement.

    • If it is a 1, multiply the current ans with (count + 1) and take the modulus to keep the number manageable. Reset count to zero.

    • If it is a 0, increment the count.

    • This process will calculate all the possible ways to split the array into good subarrays. It works by counting the number of zeros between ones, and uses that to calculate the number of ways the array can be split.

    • The overall solution combines these concepts to effectively traverse the list and calculate the desired output.

Targeted Drills in Python

  1. Coding Drills for Each Identified Concept:

    • Variable initialization:

      1
2
3

      ans = 1
count = 0
m = 10**9 + 7

    • Arithmetic operations:

      1
2
3

      sum_value = 2 + 3
product_value = 5 * 2
modulus_value = 17 % 10

    • Conditional Statements:

      1
2
3
4
5

      x = 5
if x > 0:
    print("Positive")
else:
    print("Non-positive")

    • Conditional Loops:

      1
2
3
4

      i = 0
while i < 5:
    print(i)
    i += 1

    • Array/ List traversal:

      1
2
3

      arr = [1, 2, 3, 4, 5]
for i in arr:
    print(i)

  2. Problem-specific Coding Drills:

    • Counting zeros between ones in an array. This drill is critical for our problem because it helps in determining the number of ways to split the array into good subarrays.
      1
2
3
4
5
6
7
8

      nums = [0, 1, 0, 0, 1]
count = 0
for num in nums:
    if num == 1:
        print("Number of zeros before this one:", count)
        count = 0
    else:
        count += 1

  3. Integrating the Drills to Solve the Problem:

    • Start by initializing variables (ans, count, and m).

    • Traverse the array nums until you find the first 1. You can use the conditional loop drill here.

    • Once you’ve found the first 1, use a combination of conditional statements and arithmetic operations to calculate the number of ways to split the array.

    • Specifically, for each element in nums, check if it is 1 or 0. If it is 1, update ans by multiplying it with count + 1 and taking the modulus with m to keep the number within limits. Reset count to 0.

    • If the element is 0, increment count. This keeps track of the number of zeros between ones.

    • By the end of the loop, ans will hold the number of ways to split the array into good subarrays.

    • Here’s how you can put all the drills together:

     1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17
18
19
20
21

    class Solution:
    def numberOfGoodSubarraySplits(self, nums: List[int]) -> int:
        ans = 1
        count = 0
        m = 10**9 + 7

        i = 0
        while i < len(nums) and nums[i] == 0:
            i += 1
        if i >= len(nums):
            return 0

        while i < len(nums):
            if nums[i] == 1:
                ans = (ans * (count + 1)) % m
                count = 0
            else:
                count += 1
            i += 1

        return ans

This solution uses all the concepts we’ve drilled, assembled in a way that solves the given problem.

Q&A

So the points to be noticed are:

The initial and final zeroes will be added with the first and last ones only so we can ignore them while calculating the number of ways. find the count of ones, if it is zero or one then return count (this is because if no ones are present then no good subarrays are there and if only one one is present then the array itself is the only possible good array. we initialise a variable ans with one; 3. then we find the first occurrence and the last occurrence of ones, start iterating the array from the next index of the first occurrence, and we keep a count of zeroes and every time we encounter a one , we multiply ans with count of zeroes plus one and update zero count to 0;

Yes, you’ve captured the essential points correctly. To clarify further:

  1. We are interested in the zeros between the ones in the binary array. The initial and final zeros, which do not have a one before or after them respectively, will only be combined with the first and last ones, so we can ignore them while calculating the number of ways to split the array.

  2. If the array has zero or one occurrences of one, we can directly return that count. This is because if no ones are present then no good subarrays are possible and if only one one is present then the entire array itself is the only possible good subarray.

  3. We initialize a variable ans with one. This is our accumulator that will hold the result as we progress through the array.

  4. We find the first occurrence and the last occurrence of ones, then start iterating the array from the next index of the first occurrence. We keep a count of the number of zeros we encounter.

  5. Every time we encounter a one, we multiply ans with the count of zeros plus one (to account for the possibility of including the one with the zeros or not), and then update the zero count to 0. This allows us to account for all possible ways to split the array into good subarrays as we traverse it.

So the aim is to count the zeros between the ones and use them to update the number of ways we can split the array into good subarrays. The final value of ans gives the total number of ways, modulo 10^9 + 7 to keep it within the integer range.

What would be the answer if instead of exactly one 1 we are asked for atleast one 1?

If the requirement is changed to “at least one 1” instead of “exactly one 1” in each subarray, the problem becomes a lot more complex.

With “exactly one 1” in each subarray, we know that every ‘1’ must be the start of a new subarray, and the number of zeros gives us the number of possible splits. However, with “at least one 1”, we can have multiple ‘1’s in a subarray, which exponentially increases the possible ways to split the array.

In this modified problem, a dynamic programming approach might be suitable to track the number of ways to split at each position in the array. However, it is important to note that this would be a significantly more complex problem, with a higher time complexity due to the increased number of possibilities.

If the problem is changed to “at least one 1” in each subarray, then the number of good subarrays is essentially the product of the distances between successive ‘1’s in the array.

To solve this, you would need to iterate over the array and every time you encounter a ‘1’, calculate the distance from the previous ‘1’ (i.e., the number of ‘0’s between them plus one). The answer is then the product of these distances.

Here is a Python code that implements this logic:

 1
 2
 3
 4
 5
 6
 7
 8
 9
10
11
12
13
14
15
16
17

class Solution:
    def numberOfGoodSubarraySplits(self, nums: List[int]) -> int:
        MOD = 10**9 + 7
        # Adding two sentinels 1 at start and end
        nums = [1] + nums + [1]
        
        # List to store positions of '1's
        ones_positions = [i for i, num in enumerate(nums) if num == 1]

        # Calculating distances between consecutive '1's
        distances = [y - x for x, y in zip(ones_positions, ones_positions[1:])]

        ans = 1
        for dist in distances:
            ans = (ans * dist) % MOD

        return ans

In the above code, we add two sentinel ‘1’s at the start and end of the array to ensure that we account for all ‘0’s at the beginning and end of the array. Then we calculate the distances between consecutive ‘1’s and multiply them together to get the final answer.

Can anyone explain why my code is giving TLE

class Solution { public: const int Mod=1e9+7;

long long Recursion(long long i,vector &Sum,vector &DP) { if(i==Sum.size()-1) { return 1; }

if(DP[i+1]!=-1)
{
    return DP[i+1];
}

long long x=0;
if(i!=-1)
{
    x=Sum[i];
}

long long Ans=0;        
long long l=upper_bound(Sum.begin(),Sum.end(),x)-Sum.begin();
if(l<Sum.size())
{
    int j=l;
    while(j<Sum.size() && Sum[l]==Sum[j])
    {
        Ans+=Recursion(j,Sum,DP);
        j++;
    }
}

return DP[i+1]=Ans%Mod;

}

int numberOfGoodSubarraySplits(vector& nums) { vector Sum; long long sum=0; for(int i=0;i<nums.size();i++) { sum+=nums[i]; Sum.push_back(sum); } cout«sum; if(sum==0) { return 0; }

vector<long long> DP(nums.size(),-1);

return Recursion(-1,Sum,DP)%Mod;

} };

Your current solution uses recursion and dynamic programming to solve the problem, but it is causing a Time Limit Exceeded (TLE) error because the time complexity is high due to the inefficient searching and repeated calculation in the recursion.

One of the main problems is in this part of the code:

1
2
3
4
5
6

int j=l;
while(j<Sum.size() && Sum[l]==Sum[j])
{
    Ans+=Recursion(j,Sum,DP);
    j++;
}

Here, you are calling the Recursion() function multiple times within the while loop, causing excessive computation. The recursion also triggers the function to go deeper into subsequent layers, increasing the time complexity significantly. This becomes especially problematic when the input array is large, resulting in TLE.

Additionally, using upper_bound() function within the recursive function call also contributes to the inefficiency of the algorithm. This function performs a binary search to find the upper bound, which has a time complexity of O(log N), thus further increasing the total time complexity of your solution.

Instead, consider an approach where you iterate through the array, keeping count of zeros between ones, and for each one encountered, multiply the current answer with count of zeros plus one. This approach only requires a single pass through the array, resulting in a much more efficient solution with a time complexity of O(N).

Try thinking about the number of subarrays you can create with two ones and a handful of zeroes in between.

The number of good subarrays that can be created from a sequence of two 1s with a certain number of 0s in between can be calculated as follows:

Let’s say you have a sequence of two 1s with n zeroes in between. The subarrays you can create from this sequence are determined by where you split the sequence.

For example, if you have a sequence 1,0,0,1 (two 1s with 2 zeroes in between), you can create the following good subarrays by choosing different split points:

  • Split after the first 1: [1], [0,0,1]
  • Split after the first 0: [1,0], [0,1]
  • Split after the second 0: [1,0,0], [1]

So the number of good subarrays that can be created from a sequence of two 1s with n zeroes in between is n+1.

However, if you have more than two 1s in your overall array, you need to multiply these counts together. This is because for each pair of 1s in the array, you have a certain number of choices about where to split the array to form good subarrays, and these choices are independent of the choices you make for other pairs of 1s.

For example, if you have the array [1,0,1,0,0,1], you could form good subarrays in 3 ways between the first two 1s and 3 ways between the second two 1s, giving a total of 3*3=9 possible ways to form good subarrays.