Binary Search - Time Complexity

Time Complexity of Binary Search Algorithm Algorithm: Int BinarySearch(int Array[], int value, int left, int right){ Int mid=(left+right)/2 If(Array[mid]== value){ Return mid; }elif(left>right){ Return -1; }elif(Array[mid] > value){ BinarySearch(Array,left,mid-1); }else{ BinarySearch(Array, mid+1,right); } End if } Let N be the Size of the Binary tree T(N) = C + T(N/2) ------ 1 T(N/2) = C + T (N/4) ------2 Substitute 2 à 1 T(N) = 2C + T(N/4) -------3 T(N/4) = C + T(N/8) --------4 Substitute 4 à 3 T(N) = 3C + T(N/8) ------5 pattern T(N) = iC + T(N/2 i ) where I is an integer At some point , as N/2 i diminishes , we reach only 1 element, So ...