'list index out of range' in while loop designed to return two values from list that adds to a specific sum

that is what i changed:

• array[len(array)-1] -> len(array)-1 (that's what caused your IndexError)
• indx_Dict: i changed it such that indx_Dict[sorted_index] = original_index
• sum -> sum_: sum is a built-in. it is never a good idea to use one of those as variable name! (yes, the new name could be better)

this is the final code:

def two_sum(num_array, sum_): '''1.twoSum Given an array of integers, return indices of the two numbers that add up to a specific target. ''' array = sorted(num_array) l = 0 r = len(array)-1 indx_Dict = {array.index(val): index for index, val in enumerate(num_array)} ## while (l < r) : if (array[l] + array[r]) == sum_: return [indx_Dict[l], indx_Dict[r]] elif array[l] + array[r] < sum_: l += 1 else: r -= 1 

here is a discussion about this problem: Find 2 numbers in an unsorted array equal to a given sum (which you seem to be aware of - looks like what you are trying to do). this is a python version of just that:

def two_sum(lst, total): sorted_lst = sorted(lst) n = len(lst) for i, val0 in enumerate(sorted_lst): for j in range(n-1, i, -1): val1 = sorted_lst[j] s = val0 + val1 if s < total: break if s == total: return sorted((lst.index(val0), lst.index(val1))) return None 

this version is based on looping over the indices i and j.

now here is a version that i feel is more pythonic (but maybe a little bit harder to understand; but it does the exact same as the one above). it ignores the index j completely as it is not really needed:

from itertools import islice def two_sum(lst, total): n = len(lst) sorted_lst = sorted(lst) for i, val0 in enumerate(sorted_lst): for val1 in islice(reversed(sorted_lst), n-i): s = val0 + val1 if s < total: break if s == total: return sorted((lst.index(val0), lst.index(val1))) return None 

aaaaand just for the fun of it: whenever there is a sorted list in play i feel the need to use the bisect module. (a very rudimentary benchmark showed that this may perform better for n > 10'000'000; n being the length of the list. so maybe not worth it for all practical purposes...)

def two_sum_binary(lst, total): n = len(lst) sorted_lst = sorted(lst) for i, val0 in enumerate(sorted_lst): # binary search in sorted_lst[i:] j = bisect_left(sorted_lst, total-val0, lo=i) if j >= n: continue val1 = sorted_lst[j] if val0 + val1 == total: return sorted((lst.index(val0), lst.index(val1))) else: continue return None 

for (a bit more) completeness: there is a dictionary based approach:

def two_sum_dict(lst, total): dct = {val: index for index, val in enumerate(lst)} for i, val in enumerate(lst): try: return sorted((i, dct[total-val])) except KeyError: pass return None 

i hope the code serves as its own explanation...

l and r are not your indices, but values from your array.

Say you have an array: [21,22,23,23]. l is 21, r is 23; therefore, calling array[21] is out of bounds.

Additionally, you would have a problem with your indx_Dict. You call enumerate on it, which returns [(0,21),...(3,23)]. Calling dict gives you {0:21,1:22,2:23,3:23}. There is no key equivalent to 21 or 23, which will also give you an error.

What you could try is:

def twoSum(num_array, asum): '''1.twoSum Given an array of integers, return indices of the two numbers that add up to a specific target. ''' array = sorted(num_array) l = 0 r = len(array)-1 while (l < len(array)-1) : while (r > l): if (array[l] + array[r]) == asum: return [num_array.index(array[l]),\ num_array.index(array[r])] r -= 1 r = len(array)-1 l += 1 num_array1 = [2, 7, 11, 15,1,0] target1 = 9 twoSum(num_array1, target1) 

This way, your l and r are both indices of the sorted array. It goes through every possible combination of values from the array, and returns when it either has found the sum or gone through everything. It then returns the index of the original num_array that contains the correct values.

Also, as @hiro-protagonist said, sum is a built-in function in Python already, so it should be changed to something else (asum in my example).