What Information to Track

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# ===== POSITION IN LINEAR SEQUENCE =====
def longest_increasing_subsequence(nums: list[int]) -> int:
    """
    Position Information: index i
    
    State: dp[i] = length of longest increasing subsequence ending at index i
    
    Why track position i?
        - Each position represents a unique subproblem: "best LIS ending here"
        - Element at i determines which previous elements can connect to it
        - The position IS the subproblem identity
    
    Alternative framing (ending vs starting):
        - dp[i] = LIS ending at i (bottom-up, forward)
        - dp[i] = LIS starting at i (top-down, backward in logic)
        
    Both work; the choice affects transition direction.
    """
    if not nums:
        return 0
    
    n = len(nums)
    dp = [1] * n  # Every element itself is an LIS of length 1
    
    for i in range(1, n):
        for j in range(i):
            if nums[j] < nums[i]:
                dp[i] = max(dp[i], dp[j] + 1)
    
    return max(dp)
 
 
# ===== POSITION IN TWO SEQUENCES =====
def edit_distance(word1: str, word2: str) -> int:
    """
    Position Information: (i, j) - position in each string
    
    State: dp[i][j] = min edits to transform word1[0..i-1] to word2[0..j-1]
    
    Why track (i, j)?
        - The subproblem is uniquely identified by how much of each string
          we've processed
        - Position (i, j) defines the remaining work: transform word1[i:] to word2[j:]
        - Neither i nor j alone captures the full subproblem
        
    Insight: We never need to track WHICH edits were made, because:
        - Only the count matters for the optimal value
        - Reconstruction can determine edits by backtracking
    """
    m, n = len(word1), len(word2)
    dp = [[0] * (n + 1) for _ in range(m + 1)]
    
    # Base cases: transforming to/from empty string
    for i in range(m + 1):
        dp[i][0] = i  # i deletions
    for j in range(n + 1):
        dp[0][j] = j  # j insertions
    
    for i in range(1, m + 1):
        for j in range(1, n + 1):
            if word1[i-1] == word2[j-1]:
                dp[i][j] = dp[i-1][j-1]  # No edit needed
            else:
                dp[i][j] = 1 + min(
                    dp[i-1][j],    # Delete from word1
                    dp[i][j-1],    # Insert into word1
                    dp[i-1][j-1]   # Replace in word1
                )
    
    return dp[m][n]
 
 
# ===== POSITION IN GRID =====
def minimum_path_sum(grid: list[list[int]]) -> int:
    """
    Position Information: (r, c) - row and column
    
    State: dp[r][c] = minimum sum to reach cell (r, c) from (0, 0)
    
    Why track (r, c)?
        - Position uniquely defines subproblem: "best path to here"
        - The grid value at (r, c) comes from the fixed input (no need to track)
        - Direction constraints limit transitions to (r-1, c) or (r, c-1)
        
    No need to track:
        - The path taken (reconstruction can find it)
        - The running sum (dp[r][c] IS the running sum to this point)
    """
    if not grid or not grid[0]:
        return 0
        
    m, n = len(grid), len(grid[0])
    dp = [[0] * n for _ in range(m)]
    
    dp[0][0] = grid[0][0]
    
    # First row: can only come from left
    for c in range(1, n):
        dp[0][c] = dp[0][c-1] + grid[0][c]
    
    # First column: can only come from above
    for r in range(1, m):
        dp[r][0] = dp[r-1][0] + grid[r][0]
    
    # Interior cells: min of coming from left or above
    for r in range(1, m):
        for c in range(1, n):
            dp[r][c] = min(dp[r-1][c], dp[r][c-1]) + grid[r][c]
    
    return dp[m-1][n-1]

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# ===== RESOURCE: Capacity (Knapsack) =====
def knapsack_2d(weights: list[int], values: list[int], capacity: int) -> int:
    """
    Resource Information: remaining capacity w
    
    State: dp[i][w] = max value using items 0..i-1 with capacity w
    
    Why track capacity?
        - Capacity determines which items can still be taken
        - An item of weight 5 is only legal if w >= 5
        - Without tracking capacity, we can't enforce the constraint
        
    Optimization possible?
        - Can reduce to dp[w] since we only need previous row
        - But we still need capacity dimension for correctness
    """
    n = len(weights)
    dp = [[0] * (capacity + 1) for _ in range(n + 1)]
    
    for i in range(1, n + 1):
        for w in range(capacity + 1):
            dp[i][w] = dp[i-1][w]  # Don't take item i-1
            if weights[i-1] <= w:
                dp[i][w] = max(dp[i][w],
                               dp[i-1][w - weights[i-1]] + values[i-1])
    
    return dp[n][capacity]
 
 
# ===== RESOURCE: Transaction Count (Stock Trading) =====
def max_profit_ii_with_limit(prices: list[int], max_transactions: int) -> int:
    """
    Resource Information: transactions remaining
    
    State: dp[i][j][holding]
        - i = day
        - j = transactions used (0 to max_transactions)
        - holding = 0 or 1
    
    Why track transaction count?
        - If j == max_transactions and we're not holding, we cannot buy again
        - The count directly limits future actions
        
    Note: If transactions were unlimited, j would NOT be essential
          (see "Best Time to Buy and Sell Stock II" which needs no j dimension)
    """
    if not prices or max_transactions == 0:
        return 0
    
    n = len(prices)
    k = max_transactions
    
    # dp[j][0] = max profit using j transactions, not holding
    # dp[j][1] = max profit using j transactions, holding
    INF = float('inf')
    dp = [[-INF, -INF] for _ in range(k + 1)]
    dp[0][0] = 0  # 0 transactions, not holding, profit = 0
    
    for price in prices:
        for j in range(k, -1, -1):  # Reverse to use previous day's values
            if dp[j][1] > -INF:
                # Sell: complete transaction, not holding
                dp[j][0] = max(dp[j][0], dp[j][1] + price)
            if j > 0 and dp[j-1][0] > -INF:
                # Buy: use one more transaction, now holding
                dp[j][1] = max(dp[j][1], dp[j-1][0] - price)
    
    return max(dp[j][0] for j in range(k + 1) if dp[j][0] > -INF)
 
 
# ===== RESOURCE: When NOT to Track (Unlimited Resource) =====
def max_profit_unlimited(prices: list[int]) -> int:
    """
    NO resource tracking needed: unlimited transactions
    
    State: dp[i][holding] or just greedy
    
    Why no transaction count?
        - With unlimited transactions, count doesn't constrain anything
        - Every buy-sell opportunity is independent
        - Remove the dimension = simpler state
        
    Greedy suffices: buy before every rise, sell before every fall
    """
    profit = 0
    for i in range(1, len(prices)):
        if prices[i] > prices[i-1]:
            profit += prices[i] - prices[i-1]
    return profit

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# ===== HISTORY: Last Decision (Paint House) =====
def min_cost_paint_houses(costs: list[list[int]]) -> int:
    """
    Decision History: last color used
    
    Problem: Paint n houses with 3 colors. No two adjacent houses same color.
             costs[i][c] = cost to paint house i with color c
    
    State: dp[i][c] = min cost to paint houses 0..i with house i being color c
    
    Why track last color?
        - The constraint "no adjacent same color" requires knowing the
          color of house i-1 to determine valid colors for house i
        - Without tracking, we cannot enforce the constraint
        
    What NOT to track:
        - All previous colors (only adjacent matters)
        - Running total (that's what dp[i][c] represents)
    """
    if not costs:
        return 0
    
    n = len(costs)
    num_colors = len(costs[0])  # Assume 3
    
    # dp[c] = min cost to paint through current house, ending with color c
    dp = costs[0][:]  # Base case: first house
    
    for i in range(1, n):
        new_dp = [0] * num_colors
        for c in range(num_colors):
            # Minimum of all OTHER colors from previous house + this cost
            new_dp[c] = min(dp[prev_c] for prev_c in range(num_colors) 
                          if prev_c != c) + costs[i][c]
        dp = new_dp
    
    return min(dp)
 
 
# ===== HISTORY: Streak Count (Maximum Ones with K Flips) =====
def max_consecutive_ones_iii(nums: list[int], k: int) -> int:
    """
    This problem often seems to need history but actually uses sliding window.
    Let's look at a variant that truly needs streak tracking.
    """
    # Sliding window approach (no DP needed for this problem)
    left = 0
    zeros = 0
    max_len = 0
    
    for right in range(len(nums)):
        if nums[right] == 0:
            zeros += 1
        
        while zeros > k:
            if nums[left] == 0:
                zeros -= 1
            left += 1
        
        max_len = max(max_len, right - left + 1)
    
    return max_len
 
 
# ===== HISTORY: K-Step Constraint (Pizza with 3n Slices) =====
def max_size_pizza_slices(slices: list[int]) -> int:
    """
    Problem: Circular arrangement of pizza slices. If you take slice i,
             you cannot take slices i-1 or i+1 (they get removed).
             Find max sum of n/3 slices (n = len(slices)).
    
    This is like House Robber but circular and with count limit.
    
    State: dp[i][j][took_prev]
        - i = current slice index
        - j = slices taken so far
        - took_prev = did we take slice i-1? (0 or 1)
    
    For circular: compute twice (include first, exclude first) like circular House Robber
    
    Why track took_prev?
        - Cannot take current if took previous (adjacent constraint)
        - This single bit of history is sufficient
    """
    n = len(slices)
    target = n // 3  # Number of slices to take
    
    def max_sum_linear(arr: list[int], target: int) -> int:
        """Max sum picking 'target' non-adjacent elements from linear array."""
        m = len(arr)
        INF = float('-inf')
        
        # dp[i][j] = max sum using first i elements, taking exactly j
        # To save space, we track: can't take if took previous
        # Actually simpler: dp[i][j][0] = didn't take i-th, dp[i][j][1] = took i-th
        
        # dp[j][took_last]: max sum with j taken, last taken or not
        dp = [[INF, INF] for _ in range(target + 1)]
        dp[0][0] = 0  # 0 taken, last not taken (trivially true at start)
        
        for i in range(m):
            new_dp = [[INF, INF] for _ in range(target + 1)]
            for j in range(target + 1):
                if dp[j][0] != INF:
                    # Didn't take last element
                    # Don't take current
                    new_dp[j][0] = max(new_dp[j][0], dp[j][0])
                    # Take current (if we have quota)
                    if j < target:
                        new_dp[j + 1][1] = max(new_dp[j + 1][1], 
                                               dp[j][0] + arr[i])
                if dp[j][1] != INF:
                    # Took last element - cannot take this one
                    new_dp[j][0] = max(new_dp[j][0], dp[j][1])
            dp = new_dp
        
        return max(dp[target][0], dp[target][1])
    
    # Circular: either include first (exclude last) or exclude first
    return max(
        max_sum_linear(slices[:-1], target),  # Include first, exclude last
        max_sum_linear(slices[1:], target)    # Exclude first, include last
    )

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# ===== AGGREGATE: Sum (Subset Sum / Partition) =====
def can_partition(nums: list[int]) -> bool:
    """
    Aggregate Information: current subset sum
    
    Problem: Can array be partitioned into two subsets with equal sum?
    
    State: dp[i][s] = can we form sum s using elements 0..i-1?
           Simplified: dp[s] = can we form sum s?
    
    Why track sum (aggregate) instead of subset (details)?
        - Two different subsets with the same sum are equivalent for our purpose
        - Only the sum value matters, not which elements compose it
        - This reduces exponential subsets to polynomial sums
    
    State space: O(n × total_sum) instead of O(2^n) subsets
    """
    total = sum(nums)
    if total % 2 != 0:
        return False
    target = total // 2
    
    # dp[s] = True if sum s is achievable
    dp = [False] * (target + 1)
    dp[0] = True  # Sum 0 is always achievable (empty subset)
    
    for num in nums:
        # Reverse to avoid using num multiple times
        for s in range(target, num - 1, -1):
            if dp[s - num]:
                dp[s] = True
    
    return dp[target]
 
 
# ===== AGGREGATE: Count (Target Sum with Signs) =====
def find_target_sum_ways(nums: list[int], target: int) -> int:
    """
    Aggregate Information: current running sum
    
    Problem: Assign +/- to each number to achieve target sum.
             Count the number of ways.
    
    State: dp[i][s] = number of ways to achieve sum s using first i numbers
    
    Why track sum (aggregate)?
        - Two different sign assignments with the same running sum
          have identical futures
        - Only the sum distinguishes subproblems, not the sequence of signs
    
    Trick: Shift sums to handle negatives, or use dict for sparse sums.
    """
    # Alternative formulation: P - N = target, P + N = total
    # So P = (target + total) / 2, count subsets summing to P
    total = sum(nums)
    if (target + total) % 2 != 0 or abs(target) > total:
        return 0
    
    p_sum = (target + total) // 2
    if p_sum < 0:
        return 0
    
    # Count subsets summing to p_sum
    dp = [0] * (p_sum + 1)
    dp[0] = 1
    
    for num in nums:
        for s in range(p_sum, num - 1, -1):
            dp[s] += dp[s - num]
    
    return dp[p_sum]
 
 
# ===== AGGREGATE: Difference (Balanced Partition) =====
def minimum_difference_partition(nums: list[int]) -> int:
    """
    Aggregate Information: one subset's sum (difference derivable)
    
    Problem: Partition into two subsets minimizing |sum1 - sum2|.
    
    State: dp[s] = True if subset sum s is achievable
    
    Why track sum rather than subset?
        - The minimum difference depends only on what sums are achievable
        - For achievable sum s, difference = |s - (total - s)| = |2s - total|
        - Find achievable s closest to total/2
    """
    total = sum(nums)
    half = total // 2
    
    dp = [False] * (half + 1)
    dp[0] = True
    
    for num in nums:
        for s in range(half, num - 1, -1):
            if dp[s - num]:
                dp[s] = True
    
    # Find largest achievable sum <= half
    for s in range(half, -1, -1):
        if dp[s]:
            return total - 2 * s  # |s - (total - s)| = |2s - total|
    
    return total  # Should never reach here
 
 
# ===== AGGREGATE: Max Seen (Best Time for Stock with Cooldown) =====
# Sometimes we track the "best so far" rather than current state
def max_profit_with_min_tracking(prices: list[int]) -> int:
    """
    Aggregate Information: minimum price seen so far (one-transaction version)
    
    Problem: Buy once, sell once later. Maximize profit.
    
    State: implicitly, min_price_so_far
    
    Why track min (aggregate)?
        - Current profit = price - min_price_before_today
        - Instead of comparing all pairs, maintain rolling minimum
        - This reduces O(n²) to O(n)
    
    This is a simple example where DP collapses to tracking one aggregate.
    """
    if not prices:
        return 0
    
    min_price = prices[0]
    max_profit = 0
    
    for price in prices[1:]:
        max_profit = max(max_profit, price - min_price)
        min_price = min(min_price, price)
    
    return max_profit

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def cherry_pickup(grid: list[list[int]]) -> int:
    """
    Cherry Pickup with Optimal Information Tracking
    
    INFORMATION TRACKED:
        - k: step number (0 to 2*(n-1))
        - r1: row of person 1 (0 to min(k, n-1))
        - r2: row of person 2 (0 to min(k, n-1))
    
    DERIVED INFORMATION:
        - c1 = k - r1 (column of person 1)
        - c2 = k - r2 (column of person 2)
    
    NOT TRACKED:
        - Which cells were picked (implicit: same cell means one cherry)
        - Running total (that's the dp value)
        - Path history (irrelevant to future)
    
    Time: O(n³), Space: O(n²) with optimization
    """
    n = len(grid)
    INF = float('-inf')
    
    # dp[r1][r2] = max cherries with person 1 at row r1, person 2 at row r2
    # After k steps, c1 = k - r1, c2 = k - r2
    
    # Initialize for step 0 at (0, 0)
    if grid[0][0] == -1:
        return 0
    
    dp = [[INF] * n for _ in range(n)]
    dp[0][0] = grid[0][0]  # Both at (0,0), one cherry
    
    for k in range(1, 2 * n - 1):
        new_dp = [[INF] * n for _ in range(n)]
        
        for r1 in range(max(0, k - n + 1), min(k + 1, n)):
            c1 = k - r1
            if c1 < 0 or c1 >= n or grid[r1][c1] == -1:
                continue
                
            for r2 in range(max(0, k - n + 1), min(k + 1, n)):
                c2 = k - r2
                if c2 < 0 or c2 >= n or grid[r2][c2] == -1:
                    continue
                
                # Cherries at current positions
                cherries = grid[r1][c1]
                if r1 != r2:  # If not same cell, add person 2's cherry too
                    cherries += grid[r2][c2]
                
                # Possible previous positions (came from step k-1)
                for pr1 in [r1, r1 - 1]:  # Person 1 came from (pr1, pc1)
                    if pr1 < 0 or k - 1 - pr1 < 0 or k - 1 - pr1 >= n:
                        continue
                    for pr2 in [r2, r2 - 1]:  # Person 2 came from (pr2, pc2)
                        if pr2 < 0 or k - 1 - pr2 < 0 or k - 1 - pr2 >= n:
                            continue
                        if dp[pr1][pr2] != INF:
                            new_dp[r1][r2] = max(new_dp[r1][r2],
                                                  dp[pr1][pr2] + cherries)
        
        dp = new_dp
    
    result = dp[n-1][n-1]
    return max(0, result) if result != INF else 0
 
 
# Example usage and verification
if __name__ == "__main__":
    test_grid = [
        [0, 1, -1],
        [1, 0, -1],
        [1, 1, 1]
    ]
    print(f"Max cherries: {cherry_pickup(test_grid)}")  # Expected: 5