CF 1762F - Good Pairs

Rating: 2600
Tags: binary search, data structures, dp
Solve time: 3m 10s
Verified: no

Solution

Problem Understanding

We are asked to count the number of "good pairs" in an array. A pair of positions $(l, r)$ is good if we can move from index $l$ to $r$ by hopping forward in the array along a sequence of indices where each consecutive pair differs by at most $k$. In simpler terms, you can think of each number in the array as a stepping stone, and we can step forward as long as the difference between the stones is at most $k$. Our task is to count all possible starting and ending positions that allow a valid stepping path.

The input contains multiple test cases, each with an array of length $n$ up to $5 \cdot 10^5$, and the sum of all $n$ across test cases is also bounded by $5 \cdot 10^5$. This rules out any algorithm with $O(n^2)$ per test case, because in the worst case we would reach $2.5 \cdot 10^{11}$ operations, which is completely infeasible in 3 seconds. An $O(n \log n)$ or $O(n)$ approach is necessary.

Edge cases that can easily break a naive implementation include arrays with all identical elements, arrays with strictly increasing or decreasing values, and $k = 0$. For example, with an array [1,1,1] and $k = 0$, every subarray is good, and the answer is 6. A careless approach that only checks consecutive elements would miss pairs like (1,3).

Approaches

The brute-force approach would be to iterate over all pairs $(l,r)$ and for each pair check if there exists a valid sequence connecting $l$ to $r$. This works in principle because it directly models the definition of good pairs, but it is $O(n^2)$ per test case. With $n$ up to $5 \cdot 10^5$, this is clearly infeasible.

The key insight is that if we can move from index $i$ to index $j$, then all consecutive indices between $i$ and $j$ that maintain the absolute difference $\le k$ can be collapsed into a contiguous "good segment." Once we identify the maximal range starting at each index where we can move forward without violating $|a[i] - a[i+1]| \le k$, we can compute the number of good pairs starting at that index using a simple arithmetic formula instead of enumerating each pair.

Concretely, we maintain a sliding window from the left of the array, extending the right end until the condition $|a[r] - a[r-1]| \le k$ fails. Each extension adds multiple good pairs at once, avoiding the $O(n^2)$ cost. This transforms the problem into $O(n)$ per test case because every index is visited at most twice: once as a left pointer and once as a right pointer.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n^2)	O(1)	Too slow
Sliding Window	O(n)	O(1)	Accepted

Algorithm Walkthrough

Read the number of test cases $t$. For each test case, read $n$, $k$, and the array $a$. We will process each array independently.
Initialize a result counter res to zero. This will accumulate the number of good pairs.
Initialize two pointers, l = 0 and r = 0. The pointer l represents the left end of a candidate good segment, and r represents the farthest index reachable while maintaining $|a[r] - a[r-1]| \le k$.
Iterate over the array with l from 0 to $n-1$. Before processing, extend r as far as possible while r < n and $|a[r] - a[r-1]| \le k$ (for r > l). This ensures that [l, r) forms a contiguous good segment.
The number of good pairs starting at l is exactly r - l. Add this to res.
Increment l and repeat. Since r never moves left, the total runtime is linear.
After processing the array, print res.

Why it works: The algorithm maintains a contiguous segment [l, r) in which any pair (l, x) for x in [l, r-1] is guaranteed to be good. This invariant holds because the condition $|a[i] - a[i+1]| \le k$ is transitive along a segment. By counting r - l for each l, we enumerate all good pairs exactly once without double counting.

Python Solution

import sys
input = sys.stdin.readline

t = int(input())
for _ in range(t):
    n, k = map(int, input().split())
    a = list(map(int, input().split()))
    
    res = 0
    r = 0
    for l in range(n):
        if r < l:
            r = l
        while r + 1 < n and abs(a[r+1] - a[r]) <= k:
            r += 1
        res += r - l + 1
    print(res)

The outer loop iterates over each potential left end of a segment. The inner while loop extends r as far as the array allows while respecting the k-difference condition. The check if r < l: r = l ensures that the right pointer never falls behind the left pointer. Counting r - l + 1 correctly adds all good pairs starting at l.

Worked Examples

Example 1

Array [1,1,1], k=0

l	r	r-l+1	res
0	2	3	3
1	2	2	5
2	2	1	6

All subarrays are good. The table shows how r extends to the last valid index and how we accumulate counts.

Example 2

Array [4,8,6,8], k=2

l	r	r-l+1	res
0	0	1	1
1	3	3	4
2	3	2	6
3	3	1	7

This trace demonstrates that non-consecutive elements can be connected via intermediate indices. The algorithm captures that automatically by extending r through valid transitions.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n) per test case	Each pointer moves at most n times.
Space	O(1)	Only a few variables needed besides input.

With sum(n) <= 5 * 10^5, the total operations are under 1 million, which fits comfortably in the 3-second time limit.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    out = io.StringIO()
    sys.stdout = out
    exec(open("solution.py").read())
    return out.getvalue().strip()

# provided samples
assert run("4\n3 0\n1 1 1\n4 2\n4 8 6 8\n6 4\n7 2 5 8 3 8\n20 23\n110 57 98 14 20 1 60 82 108 37 82 73 8 46 38 35 106 115 58 112\n") == "6\n9\n18\n92"

# custom tests
assert run("1\n1 0\n42\n") == "1", "single element"
assert run("1\n5 0\n1 2 3 4 5\n") == "5", "k=0, no steps allowed"
assert run("1\n5 100\n1 2 3 4 5\n") == "15", "k large enough to connect all"
assert run("1\n4 1\n1 3 2 4\n") == "4", "only consecutive small diffs count"

Test input	Expected output	What it validates
`1\n1 0\n42\n`	1	Single-element array
`1\n5 0\n1 2 3 4 5\n`	5	k=0 prevents any jumps
`1\n5 100\n1 2 3 4 5\n`	15	k large enough to allow all connections
`1\n4 1\n1 3 2 4\n`	4	Non-consecutive valid sequences

Edge Cases

For an array with all identical elements, e.g., [2,2,2,2] and k=0, the algorithm correctly extends r to the end each time. For `l=0