SANTHOSH0605
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Time Complexity : O(n)
- SANTHOSH0605 February 22, 2017Algorithm:
Take arr as input array
Initialise 2 pointers
value_under_test to the index position "0"
Ptr1 to the index position "value_under_test + 1"
while (Ptr1 < max_array_size)
{
If (arr[Ptr1] > arr[value_under_test])
{
Keep advancing Ptr1 to next position
}
Else
{
advance value_under_test to position "Ptr1 + 1"
advance Ptr1 to position "value_under_test + 1"
}
}
//Once this while loop is completed, we know for sure that all values lying after position "value_under_test" is greater
//now all we have to do is to ensure that anything lying behind position "value_under_test" is smaller
loop from "value_under_test - 1" to 0
{
if any item is greater than value in position "value_under_test"
{
Inform user that no such number exist in this array
}
}
//if this loop runs fully, it means that the value at position "value_under_test" is the number that we are looking for!!!
So, return arr[value_under_test]