/**
 *
 * Ashish Patel
 * e: ashishsushilPatel@gmail.com
 * w: https://ashish.me
 *
 */

/*
 * Given a non-empty array of integers, return the third maximum number
 * in this array. If it does not exist, return the maximum number. The
 * time complexity must be in O(n).
 *
 * Example 1:
 * Input: [3, 2, 1]
 * Output: 1
 * Explanation: The third maximum is 1.
 */

function thirdMaxmiumNumber(value) {
  value = Array.from(new Set(value))
  if (value.length < 3) return Math.max(...value)
  let firstMax = -Infinity,
    secondMax = -Infinity,
    thirdMax = -Infinity
  for (let index = 0; index < value.length; index++) {
    const currentNumber = value[index]
    if (currentNumber < secondMax && currentNumber > thirdMax) {
      thirdMax = currentNumber
    }
    if (currentNumber < firstMax && currentNumber > thirdMax) {
      thirdMax = secondMax
      secondMax = currentNumber
    }
    if (currentNumber > firstMax) {
      thirdMax = secondMax
      secondMax = firstMax
      firstMax = currentNumber
    }
  }
  return thirdMax
}

test('third Maxmium Number', () => {
  expect(thirdMaxmiumNumber([5, 2, 2])).toEqual(5)
  expect(thirdMaxmiumNumber([3, 2, 1])).toEqual(1)
  expect(thirdMaxmiumNumber([1, 2])).toEqual(2)
  expect(thirdMaxmiumNumber([2, 2, 3, 1])).toEqual(1)
  expect(thirdMaxmiumNumber([1, 2, -2147483648])).toEqual(-2147483648)
})