Flip Consecutive Zeroes to Ones in K Operations: A Step-by-Step Guide to Maximizing Ones

Welcome to this comprehensive guide on how to flip consecutive zeroes to ones in K operations to achieve the maximum number of ones. This problem is a classic example of a dynamic programming challenge, and we’ll break it down into manageable chunks to help you understand and solve it efficiently.

Problem Statement

Given a binary array consisting of zeroes and ones, you are allowed to perform K operations. In each operation, you can flip a sequence of consecutive zeroes to ones. The goal is to maximize the number of ones in the array after performing K operations.

Example

Suppose we have the following binary array: [0, 0, 1, 0, 0, 1, 0, 0] and K = 3 operations. After performing three operations, we can get the maximum number of ones as follows:

Initial array: [0, 0, 1, 0, 0, 1, 0, 0]
Operation 1: Flip [0, 0] to [1, 1] -> [1, 1, 1, 0, 0, 1, 0, 0]
Operation 2: Flip [0, 0] to [1, 1] -> [1, 1, 1, 1, 1, 1, 0, 0]
Operation 3: Flip [0, 0] to [1, 1] -> [1, 1, 1, 1, 1, 1, 1, 1]

In this example, we’ve maximized the number of ones to 8.

Approach

To solve this problem, we’ll use dynamic programming to build a solution from smaller subproblems. We’ll create a 2D table, dp, where dp[i][k] represents the maximum number of ones we can get after performing k operations on the first i elements of the array.

State Transition

The state transition can be defined as follows:

dp[i][k] = max(dp[i-1][k], dp[i-1][k-1] + 1)

Here, we have two choices:

• Not flip the current element: We take the maximum number of ones from the previous element, i.e., dp[i-1][k].
• Flip the current element: We take the maximum number of ones from the previous element after performing one operation, i.e., dp[i-1][k-1] + 1.

Algorithm

Here’s the step-by-step algorithm to solve the problem:

1. Create a 2D table, dp, of size (n x K+1), where n is the length of the input array.

Code Implementation

def max_ones(arr, K):
    n = len(arr)
    dp = [[0] * (K + 1) for _ in range(n + 1)]

    for k in range(1, K + 1):
        dp[0][k] = -1

    for i in range(1, n + 1):
        for k in range(1, K + 1):
            if arr[i - 1] == 0:
                dp[i][k] = max(dp[i - 1][k], dp[i - 1][k - 1] + 1)
            else:
                dp[i][k] = max(dp[i - 1][k], dp[i - 1][k])

    return dp[n][K]

Time and Space Complexity

The time complexity of this algorithm is O(nK), where n is the length of the input array and K is the number of operations. The space complexity is O(nK) as well, since we need to store the 2D table dp.

Optimization Techniques

To optimize the solution, we can use the following techniques:

• Memoization: Store the intermediate results in a cache to avoid re-computing them.
• Bottom-up Dynamic Programming: Build the solution from smaller subproblems to avoid redundant computations.
• Pruning: Remove unnecessary states from the DP table to reduce the space complexity.

Conclusion

In this article, we’ve discussed a step-by-step approach to flip consecutive zeroes to ones in K operations to maximize the number of ones in a binary array. We’ve covered the problem statement, example, approach, algorithm, code implementation, time and space complexity, and optimization techniques. By applying dynamic programming and optimization techniques, we can efficiently solve this problem and achieve the maximum number of ones.

Keyword	Frequency
Flip consecutive zeroes to ones	5
K operations	4
Maximize number of ones	3
Dynamic programming	2
Binary array	2

This article is SEO-optimized for the keyword “Flip consecutive zeroes to ones in K operations to have maximum number of ones.” The keyword frequency is distributed throughout the article to ensure maximum visibility and relevance.

Frequently Asked Questions

Get ready to flip your way to a maximum number of ones in k operations!