Hello @Ramasubrananiam! You can use the binary search algorithm in such situations. When you say the data is stored in a sorted list, the best approach wrt time complexity would be binary search. You can use the following code:

def binarySearch (arr, l, r, x):
# Check base case
if r >= l:
mid = l + (r - l)/2
# If element is present at the middle itself
if arr[mid] == x:
return mid
# If element is smaller than mid, then it can only be present in left subarray
elif arr[mid] > x:
return binarySearch(arr, l, mid-1, x)
# Else the element can only be present in right subarray
else:
return binarySearch(arr, mid+1, r, x)
else:
# Element is not present in the array
return -1
# Test array
arr = [ 2, 3, 4, 10, 40 ]
x = 10
# Function call
result = binarySearch(arr, 0, len(arr)-1, x)
if result != -1:
print "Element is present at index %d" % result
else:
print "Element is not present in array"