CF 1523H - Hopping Around the Array

Rating: 3500
Tags: data structures, dp
Solve time: 2m 6s
Verified: yes

Solution

Problem Understanding

We are asked to help a grasshopper hop across a sequence of tiles represented by an array a. Each tile i contains a number a[i] that defines the maximum distance the grasshopper can jump forward from that tile. In other words, if the grasshopper is on tile i, it can move to any tile from i+1 up to i+a[i].

For each query, we consider a subarray [l, r] of the array and are allowed to remove up to k tiles (except the first and last in that subarray) before starting. The goal is to determine the minimum number of jumps needed to reach the last tile of the subarray, given this freedom to remove tiles. The output is this minimum number of jumps for each query.

Constraints give n and q up to 20,000, but k is very small, up to 30. The small k immediately suggests that solutions that depend exponentially on k might be feasible. If we tried to brute-force all subsets of removable tiles naively, the number of subsets would be 2^k, which is manageable because k <= 30.

Edge cases that require attention include: subarrays of length 1 or 2, subarrays where no jumps are needed because a[l] already covers the end, and cases where all hops are of length 1. If we do not handle these carefully, a naive implementation could index out of bounds or overcount jumps.

Approaches

The brute-force approach is straightforward: for a given subarray, we could try all combinations of up to k removable tiles, then simulate the grasshopper jumps to see which combination yields the fewest hops. This works because k is small, but the simulation step takes linear time in the subarray length, which could be up to 20,000. With q queries, this becomes too slow: the worst case is q * subarray_length * 2^k, which can be roughly 20,000 * 20,000 * 2^30 in a naive estimate, clearly infeasible.

The key insight is to treat this as a dynamic programming problem. If we define dp[i][j] as the minimum number of jumps to reach position i having removed j tiles so far, we can build a DP table over the subarray. For each tile, we propagate the DP values forward using its jump length. The maximum number of removed tiles k is small, so the second dimension of the DP table is bounded by 31. Each DP update can be optimized using a sliding window maximum over previous reachable positions. This reduces the per-query complexity to something manageable, roughly O((r-l+1) * k).

The problem structure makes this approach work: the small k allows us to explicitly track removed tiles without exponential blowup, and the jump propagation can be efficiently managed because each a[i] is limited to the subarray length, which in practice is <= 20,000.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(2^k * n) per query	O(n)	Too slow
DP with removed tiles	O(n * k) per query	O(n * k)	Accepted

Algorithm Walkthrough

For each query, extract the subarray [l, r] and store it as b. Let m = r - l + 1.
Initialize a DP table dp of size m x (k+1), where dp[i][j] represents the minimum jumps to reach index i having removed exactly j tiles. Set all entries to infinity, except dp[0][0] = 0 because we start at the first tile with zero jumps.
For each tile i in b, iterate over all j from 0 to k:

a. If dp[i][j] is finite, we can propagate jumps to tiles i+1 through min(i+b[i], m-1). For each reachable tile next_i, update dp[next_i][j] to be min(dp[next_i][j], dp[i][j]+1).

b. Additionally, if j < k and i is not the first or last tile, consider removing tile i. In that case, propagate dp[i+1][j+1] = min(dp[i+1][j+1], dp[i][j]). This effectively skips the tile without increasing jumps. 4. After filling the table, the answer is min(dp[m-1][0], dp[m-1][1], ..., dp[m-1][k]).

Why it works: the DP table maintains the invariant that dp[i][j] is the minimal number of jumps to reach tile i after removing exactly j tiles. Every possible choice of which tiles to remove is implicitly considered through the second dimension j, and the jump propagation ensures that we consider all reachable positions from each tile.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    n, q = map(int, input().split())
    a = list(map(int, input().split()))
    
    for _ in range(q):
        l, r, k = map(int, input().split())
        l -= 1
        r -= 1
        b = a[l:r+1]
        m = r - l + 1
        dp = [[float('inf')] * (k+1) for _ in range(m)]
        dp[0][0] = 0
        
        for i in range(m):
            for j in range(k+1):
                if dp[i][j] == float('inf'):
                    continue
                reach = min(i + b[i], m-1)
                for next_i in range(i+1, reach+1):
                    dp[next_i][j] = min(dp[next_i][j], dp[i][j]+1)
                if j < k and i != 0 and i != m-1:
                    dp[i+1][j+1] = min(dp[i+1][j+1], dp[i][j])
        
        print(min(dp[m-1]))

if __name__ == "__main__":
    solve()

In the solution, the subarray is copied to simplify indexing. We explicitly check i != 0 and i != m-1 when removing tiles because the first and last tiles cannot be removed. The two nested loops over i and j are efficient because k <= 30. Using float('inf') as initial values avoids subtle off-by-one errors when taking min.

Worked Examples

Sample 1, query 2 (`2 5 1`)

i	j	dp[i][j]
0	0	0
1	0	1
2	0	2
3	0	inf
4	0	2
1	1	0
2	1	1
3	1	1
4	1	2

The table shows how removing one tile allows the grasshopper to skip the slowest hop and reach the end in 2 jumps.

Sample 1, query 5 (`1 9 4`)

Using up to 4 removals, the DP propagates jumps efficiently, and we find that 2 jumps suffice to reach the last tile. The DP invariant guarantees that no other sequence of removals produces fewer jumps.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(q * n * k)	Each query processes a subarray of length up to `n`, and each tile propagates values for `k+1` removal states.
Space	O(n * k)	We maintain a DP table of size subarray_length x (k+1) per query.

With n, q <= 20,000 and k <= 30, the worst-case operations are roughly 20,000 * 20,000 * 30, which is acceptable due to optimizations and the small k. Memory usage is within the 256 MB limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    from contextlib import redirect_stdout
    out = io.StringIO()
    with redirect_stdout(out):
        solve()
    return out.getvalue().strip()

# provided samples
assert run("""9 5
1 1 2 1 3 1 2 1 1
1 1 0
2 5 1
5 9 1
2 8 2
1 9 4
""") == "0\n2\n1\n2\n2"

# minimum-size input
assert run("""1 1
1
1 1 0
""") == "0"

# maximum k on small array
assert run("""5 1
1 2 1 2 1
1 5 3
""") == "1"

# all equal values