function solution(A) {
let numbers = {};
let n = A.length;
let sumAll = 0;
let sumN = n * (n + 1) / 2; // sum(1~N):
for (let i = 0; i < n; i++) {
if (numbers[A[i]]) {
return 0; // duplicated
} else {
numbers[A[i]] = true;
}
sumAll += A[i];
}
return sumAll === sumN ? 1 : 0;
}
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