Asked in: UBER
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All Testcases Passed ✔
Solution
from functools import lru_cache
def countBitwisePalindromicPaths (N, M, grid):
mod = 10**9+7
@lru_cache(None)
// ... rest of solution available after purchaseExplanation
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To solve the problem of counting the number of bitwise palindromic paths from the top-left corner to the bottom-right corner of a binary grid, where movement is restricted to right or down steps, you need to carefully consider both the nature of palindromes and the constraints of the grid traversal.
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Step 1: Understand the Path and Palindrome Structure
- A path is defined by a sequence of moves starting at (0,0) and ending at (N-1, M-1).
- You can move only right or down, so every path corresponds to a unique sequence of bits from cells visited.
- The total length of any path is fixed: it will always have exactly (N + M - 1) bits because you start at (0,0) and move (N-1) steps down and (M-1) steps right in some order.
- The palindrome condition means this sequence of bits must read the same forwards and backwards.
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Step 2: Observing the Palindromic Path Property
- Since the path length is fixed, the palindrome condition requires symmetry in the bits from the start and end positions.
- Imagine the path sequence indexed from 0 to L-1 (where L = N + M - 1).
- For the path to be palindromic, the bit at index i must be equal to the bit at index L - 1 - i.
- This symmetry means the bit collected at the ith step from the start corresponds to the bit collected at the ith step from the end.
- This crucial insight allows us to think of the problem in a bidirectional manner.
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Step 3: Two-Pointer / Two-Ends Approach
- Instead of enumerating all paths from start to end (which can be exponential), think about simultaneously traversing from the start cell and from the end cell moving towards the middle.
- At each step, you keep track of pairs of positions:
- One position moving forward (starting from top-left)
- One position moving backward (starting from bottom-right)
- For the path to be palindromic, the bits at these two positions must be the same.
- The two positions are always chosen such that they correspond to symmetric indices in the path sequence.
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Step 4: Dynamic Programming State Definition
- Define a DP state that records the number of ways to reach a pair of positions (r1, c1) and (r2, c2), where (r1, c1) is a cell reachable from the start, and (r2, c2) is a cell reachable from the end.
- The positions satisfy that the number of steps from start to (r1, c1) plus the number of steps from (r2, c2) to end equals L - 1.
- This means these positions are "mirrored" in the path.
- You maintain a count of how many paths can reach these two positions with bits matching, ensuring palindrome property.
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Step 5: Transitions in the DP
- From a current pair of positions, you consider moving the start position either right or down (if possible).
- Similarly, you move the end position either left or up (if possible).
- For each possible move, if the bits at the new positions are equal, you add the count from the current state to the new state.
- This way, you propagate counts of palindromic path fragments inward from both ends.
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Step 6: Base Cases
- The initial DP state corresponds to the pair of positions both at the start and end cells: (0,0) and (N-1,M-1).
- If the bits at these cells are equal, you initialize that DP state with count 1.
- If they are not equal, the answer is immediately zero.
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Step 7: Ending Condition and Result
- The process continues until the two pointers meet or cross.
- When the pointers meet (they may be at the same cell or adjacent cells, depending on odd or even path length), the DP states represent fully palindromic paths.
- Sum all DP counts for valid meeting positions to get the total number of palindromic paths.
- Return the count modulo 10^9+7 as per the problem constraints.
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Step 8: Complexity and Feasibility
- Since N and M are up to 15, the total grid size is small enough to allow a DP over pairs of positions.
- The number of states is roughly O(N^2 * M^2), which is manageable.
- Efficient implementation and pruning (only states where bits match are considered) helps performance.
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Step 9: Summary of the Approach
- Think about the palindrome property as symmetry between start and end positions.
- Use two pointers: one advancing from the top-left, one retreating from the bottom-right.
- Use dynamic programming to count the number of ways to reach each pair of positions with matching bits.
- Accumulate counts until pointers meet.
- Return the total count modulo the large prime.
By approaching the problem as a symmetric traversal with two pointers and applying DP over pairs of positions, you transform an exponential path counting problem into a tractable and elegant dynamic programming solution that leverages the palindrome constraints naturally.
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