Press n or j to go to the next uncovered block, b, p or k for the previous block.
| 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 | 1x 5x 5x 5x 5x 5x 5x 12x 8x 5x 8x 3x 3x 8x 12x 6x 6x 6x 6x 12x 5x 5x 7x 7x 7x 5x 1x | // You are given an array(which will have a length of at least 3, but could be very large) containing integers.
// The array is either entirely comprised of odd integers or entirely comprised of even integers except for a single integer
// N.Write a method that takes the array as an argument and returns this "outlier" N.
//
//
// [2, 4, 0, 100, 4, 11, 2602, 36] --> 11 (the only odd number)
// [160, 3, 1719, 19, 11, 13, -21] --> 160 (the only even number)
//
//
// export function findOUtlier(integers: number[]): number {
// const isEvenNum = integers.filter(num => num % 2 == 0);
// const isOddNum = integers.filter(num => num % 2 != 0);
// return isEvenNum.length == 1 ? Number(isEvenNum) : Number(isOddNum);
// };
export function findOUtlier(integers: number[]): number {
const evenNum: number[] = [];
const oddNum: number[] = [];
const middle = Math.trunc(integers.length / 2);
let end = integers.length - 1;
let start = 0;
while (start < middle || end >= middle) {
if (start !== middle) {
if (!(integers[start] % 2)) {
evenNum.push(integers[start])
} else {
oddNum.push(integers[start])
}
}
if (!(integers[end] % 2)) {
evenNum.push(integers[end])
} else {
oddNum.push(integers[end])
}
if (evenNum.length === 1 && oddNum.length > 1 || oddNum.length === 1 && evenNum.length > 1) {
break;
}
start++;
end--;
}
return evenNum.length === 1 ? evenNum[0] : oddNum[0];
};
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