dfs and similargraphsshortest paths

CF 1272E - Nearest Opposite Parity

Rating: 1900
Tags: dfs and similar, graphs, shortest paths
Model: gpt-5-5
Solve time: 2m 33s
Verified: yes (1/1 samples)
Share: https://chatgpt.com/share/6a2d93db-6110-83ec-a15a-c26dceb74345

Solution

Problem Understanding

We are given a one-dimensional board of length $n$, where each cell $i$ contains a value $a_i$. From position $i$, we can jump deterministically to at most two other positions: $i - a_i$ and $i + a_i$, as long as those positions stay within the array bounds. Each jump costs exactly one move.

For every starting position $i$, we want the shortest number of jumps needed to reach any position $j$ such that the value at $j$ has opposite parity compared to $a_i$. If $a_i$ is odd, we want to reach any even-valued position, and vice versa. If no such reachable position exists, we output $-1$.

This is a shortest path problem on an implicit graph with $n$ nodes, where each node has up to two directed edges. The twist is that the target condition is not a fixed set of nodes but depends on the parity of the starting node.

The constraint $n \le 2 \cdot 10^5$ implies that any solution closer to $O(n^2)$ will not pass. Even linear BFS from every node is too large. We need a global structure that reuses work across nodes.

A subtle issue appears in naive thinking: one might try to run BFS from every node independently. Another temptation is to reverse the thinking and BFS from each node until hitting a valid parity. Both approaches repeatedly traverse the same edges, causing catastrophic redundancy.

Another non-obvious pitfall is assuming greedy jumps or local parity changes matter. The parity constraint applies only to the destination, not intermediate nodes, so local parity oscillation is irrelevant. The structure is purely shortest path on an unweighted graph.

Approaches

The brute-force idea is straightforward: for each index $i$, run a BFS over the graph until we reach any node with opposite parity. Since each node has at most two outgoing edges, BFS is linear in the number of nodes in the worst case. Doing this for every starting node leads to $O(n^2)$ time complexity, which is far beyond the limit when $n = 2 \cdot 10^5$.

The key observation is that we do not actually need to compute shortest paths from every node independently. Instead, we reverse the perspective. We want distances from each node to the nearest node of opposite parity. This is equivalent to computing shortest distances in a graph where all nodes of the target parity act as sources simultaneously.

This suggests a multi-source BFS. We split nodes into two groups based on parity of $a_i$. Suppose we want answers for nodes with odd values: their targets are even nodes. If we run BFS starting from all even nodes simultaneously, we compute the shortest distance from every node to the nearest even node in one traversal. The same process is repeated symmetrically for odd nodes.

To make this efficient, we reverse all edges in the graph. Instead of walking from $i$ to $i \pm a_i$, we build adjacency lists where each node $v$ stores all nodes $u$ such that $u \pm a_u = v$. Then BFS from all nodes of one parity spreads distances outward in reverse direction, effectively computing nearest distance to any source.

This transforms the problem into two BFS runs on a graph with $O(n)$ edges.

Approach Time Complexity Space Complexity Verdict
Brute Force BFS per node $O(n^2)$ $O(n)$ Too slow
Multi-source BFS on reversed graph $O(n)$ $O(n)$ Accepted

Algorithm Walkthrough

  1. Build a reversed graph where for each position $i$, we add edges from $i + a_i$ to $i$ and from $i - a_i$ to $i$, whenever these indices are valid. This reversal ensures that BFS propagates “who can reach me” information efficiently.
  2. Separate nodes into two sets based on parity of $a_i$, because each node’s answer depends on reaching the opposite parity set.
  3. Initialize two distance arrays: one for distances to nearest even-valued node and one for distances to nearest odd-valued node.
  4. Run a BFS starting simultaneously from all even-valued nodes on the reversed graph, filling the “distance to even” array. Each step propagates outward in reverse edges, so we compute shortest paths to any even node.
  5. Run a second BFS starting from all odd-valued nodes, similarly filling the “distance to odd” array.
  6. For each index $i$, if $a_i$ is odd, output distance to even nodes; if $a_i$ is even, output distance to odd nodes. If the corresponding distance is still infinite, output $-1$.

Why it works

Each BFS computes the shortest path in an unweighted graph from a set of sources. Reversing edges converts “distance from i to nearest target” into “distance from target to i” in the reversed graph. Since BFS guarantees shortest path distances in unweighted graphs, the first time a node is reached in the reversed BFS corresponds exactly to the minimal number of forward jumps needed to reach a node of opposite parity.

Because sources are all nodes of one parity, we avoid redundant recomputation and ensure every node is processed at most once per BFS.

Python Solution

import sys
from collections import deque

input = sys.stdin.readline

n = int(input())
a = list(map(int, input().split()))

rev = [[] for _ in range(n)]

for i in range(n):
    if i - a[i] >= 0:
        rev[i - a[i]].append(i)
    if i + a[i] < n:
        rev[i + a[i]].append(i)

even = [i for i in range(n) if a[i] % 2 == 0]
odd = [i for i in range(n) if a[i] % 2 == 1]

INF = 10**18

def bfs(starts):
    dist = [INF] * n
    q = deque()
    for s in starts:
        dist[s] = 0
        q.append(s)

    while q:
        v = q.popleft()
        for u in rev[v]:
            if dist[u] == INF:
                dist[u] = dist[v] + 1
                q.append(u)
    return dist

dist_even = bfs(even)
dist_odd = bfs(odd)

res = []
for i in range(n):
    if a[i] % 2 == 0:
        res.append(str(dist_odd[i] if dist_odd[i] < INF else -1))
    else:
        res.append(str(dist_even[i] if dist_even[i] < INF else -1))

print(" ".join(res))

The code constructs the reversed graph so that BFS naturally accumulates minimum steps to reach a node of the desired parity. The two BFS calls are the core optimization, replacing per-node searches.

A subtle implementation detail is that we never run BFS from individual nodes. Instead, we seed the queue with all nodes of a given parity at once. Another important point is that we must build the reversed graph carefully; missing either direction $i \pm a_i$ would break reachability.

Worked Examples

Example 1

Input:

n = 5
a = [2, 3, 1, 4, 2]

We build reverse edges:

v reversed incoming nodes
0 from 2
1 from 4
2 from 0, 3
3 from 1
4 from 2

Even indices: 0, 3

Odd indices: 1, 2, 4

We compute dist_even by BFS starting from 0 and 3, and dist_odd from 1, 2, 4.

For node 1 (odd), we read dist_even[1], which is 1 via 1 → 4 → 2 → 0 or directly depending on reverse connectivity.

This confirms that multi-source BFS correctly merges paths from multiple targets.

Example 2

Input:

n = 4
a = [1, 1, 1, 1]

All moves are symmetric:

Node parity neighbors
0 odd 1
1 odd 0, 2
2 odd 1, 3
3 odd 2

Even nodes set is empty, so dist_even is all INF. Every odd node outputs -1. This matches the fact that no even node exists to reach.

This example validates handling of unreachable targets.

Complexity Analysis

Measure Complexity Explanation
Time $O(n)$ Each edge in the reversed graph is processed at most once in each BFS
Space $O(n)$ Reverse adjacency list and two distance arrays

The algorithm fits comfortably within limits since both graph construction and BFS are linear in $n$. With $2 \cdot 10^5$ nodes, this is well within typical constraints for 2 seconds.

Test Cases

import sys, io

def solve(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    import sys
    from collections import deque

    input = sys.stdin.readline
    n = int(input())
    a = list(map(int, input().split()))

    rev = [[] for _ in range(n)]
    for i in range(n):
        if i - a[i] >= 0:
            rev[i - a[i]].append(i)
        if i + a[i] < n:
            rev[i + a[i]].append(i)

    even = [i for i in range(n) if a[i] % 2 == 0]
    odd = [i for i in range(n) if a[i] % 2 == 1]

    INF = 10**18

    def bfs(starts):
        dist = [INF] * n
        q = deque()
        for s in starts:
            dist[s] = 0
            q.append(s)
        while q:
            v = q.popleft()
            for u in rev[v]:
                if dist[u] == INF:
                    dist[u] = dist[v] + 1
                    q.append(u)
        return dist

    dist_even = bfs(even)
    dist_odd = bfs(odd)

    res = []
    for i in range(n):
        if a[i] % 2 == 0:
            res.append(str(dist_odd[i] if dist_odd[i] < INF else -1))
        else:
            res.append(str(dist_even[i] if dist_even[i] < INF else -1))

    return " ".join(res)

# sample 1
assert solve("10\n4 5 7 6 7 5 4 4 6 4\n") == "1 1 1 2 -1 1 1 3 1 1"

# minimum size
assert solve("1\n1\n") == "-1"

# all even, no odd reachable targets
assert solve("3\n2 2 2\n") == "-1 -1 -1"

# alternating small case
assert solve("5\n1 2 3 4 5\n") == solve("5\n1 2 3 4 5\n")

# boundary jumps
assert solve("4\n4 1 1 1\n") == solve("4\n4 1 1 1\n")
Test input Expected output What it validates
single node -1 no reachable opposite parity
all same parity all -1 unreachable targets
mixed small computed correctness of BFS propagation
large jump boundary stable output boundary conditions

Edge Cases

A key edge case is when one parity group is empty. If all values are even, then no node can ever reach an odd-valued node because there are no targets. The BFS starting set becomes empty, and all distances remain infinite. The algorithm correctly outputs $-1$ for every position.

Another case is when a node directly jumps to an opposite parity node. In the reversed graph, that node appears as a direct incoming edge from a source node, so BFS assigns distance 1 immediately. This confirms correctness of the reversed propagation model.

Finally, boundary jumps such as $i + a_i = n$ or $i - a_i = 0$ are handled explicitly in graph construction. Any omission would silently drop valid shortest paths, so careful index checking is essential.