I have some code here for finding only left first small number from an array.
public void Test() {
int[] numbers = { 121, 124, 131, 135, 219, 287, 361, 367, 382, 420 };
var onlyfirstNumbersLessThanGiven = numbers.TakeWhile(n => n < 135);
Console.WriteLine("First numbers less than ");
foreach (var n in onlyfirstNumbersLessThanGiven)
{
Console.WriteLine(n);
}
}
How to find only 131 from above array? Please help me
How about this?
Ordered array:
numbers.LastOrDefault(x => x < 135);
Unordered array:
numbers.Where(x => x < 135).Max();
Although Renan's answer is perfectly correct, why not just use First(), edited or TakeWhile() with Last()?
e.g.
int[] numbers = { 9, 34, 65, 92, 87, 435, 3, 54,
83, 23, 87, 435, 67, 12, 19 };
int first = numbers.First(number => number > 80);
int firstSmaller = numbers.TakeWhile(number => number < 80).Last(); // returns 65 since its the first one smaller than 80 in the series
Console.WriteLine(first);
And to match it with your situation:
public void Test() {
int[] numbers = { 121, 124, 131, 135, 219, 287, 361, 367, 382, 420 };
var onlyfirstNumbersLessThanGiven = numbers.TakeWhile(n => n < 135).Last();
Console.WriteLine("First numbers less than ");
Console.WriteLine(onlyfirstNumbersLessThanGiven);
}
There are restrictions: the array is assumed to be sorted, and must be in ascending order.
Thanks to Ivan Stoev for pointing out my mistake
First(n => n < 135) in your example will return 121 while OP is expecting 131
Find the number itself, since you know the number you want, you do not need any range.
numbers.Where(x => x == 131);
This example helps to clear.
public void Test()
{
int[] numbers = { 121, 324, 431, 135, 119, 287, 361, 367, 382, 420 };
var OFNLTG = numbers.Where(x => x < 121).Max();
Console.WriteLine("First Max Number:");{ Console.WriteLine(OFNLTG);
}
}
