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Binary Strings

Logical Breakdown

  • [ ] Subproblem Identification:
  • [ ] Optimal Substructure:
  • [ ] Constraint Handling: (e.g., Modulo \(10^9 + 7\))
  • [ ] Optimization: (Matrix Exponentiation if 'A' is huge)

Mathematical Rigor

State Definition

Let \(dp[i]\) be the state for the \(i\)-th subproblem.

Recurrence Relation

\[ dp[i] = ... \]

Visualization

graph TD A["Current State"] --> B["Previous State 1"] A --> C["Previous State 2"]

Complexity Analysis

Approach Time Complexity Space Complexity
Iterative DP \(O(N)\) \(O(1)\)

Code Reference

package com.dsa.heaps;

/ * ============================================================================================ * PROBLEM: Binary Strings (Minimum Flips of Length B) * ============================================================================================ * * --------------------- * 1. Logical Breakdown & Mathematical Formulation * --------------------- * We want to transform string A into all '1's using flips of fixed length B. * * Step 1: The Greedy Choice * - Scan from left to right. Whenever we encounter a '0', we MUST flip the window * starting at that index to turn that '0' into a '1'. * - There is no other way to fix that '0' without affecting elements to its left * (which are already '1's). * * Step 2: Efficient Range Flipping (Difference Array / Sliding Window) * - Manually flipping B elements takes O(B), leading to O(NB) total. * - To achieve O(N), we track the "current flip state" using a variable currentFlips. * - If currentFlips is odd, the actual value of A[i] is toggled. * - We use an auxiliary array isFlipped[i] to mark where a flip window ends, * allowing us to decrement currentFlips when the window slides past. * * --------------------- * 2. Complexity Analysis * --------------------- * Time Complexity: * O(N) * - We traverse the string once, performing constant time operations at each bit. * * Space Complexity: * O(N) * - To maintain the isFlipped array to track active flip windows. * * --------------------- * 3. Base Case & Failure Condition * --------------------- * - If we encounter a '0' within the last B-1 indices, it's impossible to start * a full window of length B. In this case, return -1. * ============================================================================================ / public class BinaryStrings { public int solve(String number, int A) { // TODO: Implement solution return 0; } }