CF 1462E1 - Close Tuples (easy version)

CF 1462E1 - Close Tuples (easy version)

Rating: 1500
Tags: binary search, combinatorics, math, sortings, two pointers
Solve time: 8m 38s
Verified: no

Solution

Problem Understanding

We are given an array of integers where each element is between 1 and n, and we are asked to count the number of triples (i, j, z) such that the maximum value in the triple differs from the minimum by at most 2. The array can contain duplicates, and we are asked to process multiple test cases efficiently. Each test case has its own array, and the sum of the lengths across all test cases does not exceed 200,000.

The problem requires counting combinations of three elements with a bounded range. A naive implementation would consider every triple explicitly, which would be O(n^3) and infeasible for n up to 2·10^5. We need a method that leverages sorting or counting to reduce the complexity. Non-obvious edge cases include arrays with all identical elements, arrays with scattered numbers where no valid triples exist, and arrays shorter than three elements where the answer must be zero.

Approaches

A brute-force approach is to iterate through all possible triples (i, j, z) with three nested loops, checking if max - min ≤ 2. This is correct but far too slow, with complexity O(n^3).

The key insight is to sort the array. After sorting, any valid triple must consist of consecutive elements within the sorted array, because including elements further apart would exceed the maximum difference. Using two pointers, we can iterate over the array: fix the start index l and extend the end index r as far as possible such that a[r] - a[l] ≤ 2. Then, the number of valid triples with a[l] as the smallest element is the number of ways to choose 2 additional elements from the r - l remaining elements. This is a combinatorial calculation using C(n, 2).

Sorting the array costs O(n log n), and the two-pointer sweep costs O(n), giving an overall O(n log n) per test case. This method is efficient given the constraints.

Approach	Time Complexity	Space Complexity	Verdict
Brute Force	O(n³)	O(1)	Too slow
Sort + Two Pointers	O(n log n)	O(1)	Accepted

Algorithm Walkthrough

1. Read the number of test cases t.
2. For each test case, read n and the array a.
3. Sort a in non-decreasing order. Sorting ensures that elements within any valid triple are consecutive.
4. Initialize a pointer l = 0 to mark the start of a candidate segment and a variable ans = 0 to accumulate the total number of valid triples.
5. Iterate r over the array indices. While a[r] - a[l] > 2, increment l to shrink the segment until it satisfies the difference constraint.
6. For each position r, compute the number of elements in the segment: length = r - l. If length ≥ 2, compute the number of triples using length * (length - 1) // 2 and add to ans.
7. After processing all r, print ans.

Why it works: Sorting guarantees that any triple exceeding the difference of 2 cannot include elements outside a valid segment. The two-pointer approach ensures that all possible valid segments are counted once. The combinatorial formula correctly counts the number of triples for each segment.

Python Solution

import sys
input = sys.stdin.readline

def solve():
    t = int(input())
    for _ in range(t):
        n = int(input())
        a = list(map(int, input().split()))
        a.sort()
        ans = 0
        l = 0
        for r in range(n):
            while a[r] - a[l] > 2:
                l += 1
            length = r - l
            if length >= 2:
                ans += length * (length - 1) // 2
        print(ans)

if __name__ == "__main__":
    solve()

The solution first sorts the array to make consecutive elements the only candidates for triples. The left pointer l advances only when the difference exceeds 2, ensuring we never include invalid elements. The combinatorial formula counts all valid triples ending at position r efficiently, avoiding nested loops. Boundary conditions, such as segments shorter than 3 elements, are handled naturally because length < 2 leads to zero additions.

Worked Examples

Example 1

Input: a = [1, 2, 4, 3]

l	r	a[l]	a[r]	r-l	Triples Added	ans
0	0	1	1	0	0	0
0	1	1	2	1	0	0
0	2	1	3	2	1	1
0	3	1	4	3	l increments to 1 → length = 2 → 1	2

Output: 2. This matches the sample.

Example 2

Input: a = [1, 1, 1, 1]

All elements are identical, so every combination of three indices is valid. Segment length grows from 0 to 3 as r moves. For r = 2, add 1 triple. For r = 3, add 3 triples. Total: 4. Output: 4.

Complexity Analysis

Measure	Complexity	Explanation
Time	O(n log n)	Sorting dominates; two-pointer sweep is linear
Space	O(n)	To store the array

This fits within the constraints because the sum of all n across test cases is ≤ 2·10^5.

Test Cases

import sys, io

def run(inp: str) -> str:
    sys.stdin = io.StringIO(inp)
    output = io.StringIO()
    sys.stdout = output
    solve()
    return output.getvalue().strip()

# Provided samples
assert run("4\n4\n1 2 4 3\n4\n1 1 1 1\n1\n1\n10\n5 6 1 3 2 9 8 1 2 4\n") == "2\n4\n0\n15", "sample 1"

# Custom test cases
assert run("1\n3\n1 2 3\n") == "1", "three consecutive numbers"
assert run("1\n5\n5 5 5 5 5\n") == "10", "all equal numbers"
assert run("1\n6\n1 4 7 10 13 16\n") == "0", "no valid triples"
assert run("1\n6\n1 2 2 3 3 4\n") == "8", "overlapping segments"

Test input	Expected output	What it validates
3 elements consecutive	1	Basic segment calculation
5 elements equal	10	Correct combinatorial counting
6 elements scattered	0	Correctly handles no triples
6 elements overlapping	8	Two-pointer correctness

Edge Cases

For arrays with fewer than 3 elements, r-l never reaches 2, so the answer is automatically 0. For arrays with all equal elements, the segment grows to the full array length, and combinatorial counting ensures all triples are included. For arrays with no valid triples, the pointer l moves to prevent any invalid difference, producing zero count. Each scenario is handled naturally by the two-pointer and combinatorial approach.