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04. Counting Elements. PermCheck.swift
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04. Counting Elements. PermCheck.swift
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import Foundation
import Glibc
// Solution @ Sergey Leschev, Belarusian State University
// 04. Counting Elements. PermCheck.
// A non-empty array A consisting of N integers is given.
// A permutation is a sequence containing each element from 1 to N once, and only once.
// For example, array A such that:
// A[0] = 4
// A[1] = 1
// A[2] = 3
// A[3] = 2
// is a permutation, but array A such that:
// A[0] = 4
// A[1] = 1
// A[2] = 3
// is not a permutation, because value 2 is missing.
// The goal is to check whether array A is a permutation.
// Write a function:
// class Solution { public int solution(int[] A); }
// that, given an array A, returns 1 if array A is a permutation and 0 if it is not.
// For example, given array A such that:
// A[0] = 4
// A[1] = 1
// A[2] = 3
// A[3] = 2
// the function should return 1.
// Given array A such that:
// A[0] = 4
// A[1] = 1
// A[2] = 3
// the function should return 0.
// Write an efficient algorithm for the following assumptions:
// N is an integer within the range [1..100,000];
// each element of array A is an integer within the range [1..1,000,000,000].
public func solution(_ A: inout [Int]) -> Int {
var sum = 0
var set = Set<Int>()
for e in A {
if set.contains(e) { return 0 }
set.insert(e)
sum += e
}
let count = A.count
let gaussSum = (count * (count + 1)) / 2
let result = gaussSum - sum
return result == 0 ? 1 : 0
}