Given a list of integers and a single sum value, return the first two values (parse from the left please) in order of appearance that add up to form the sum.
Example:
sum_pairs([11, 3, 7, 5], 10)
# ^--^ 3 + 7 = 10
== [3, 7]
sum_pairs([4, 3, 2, 3, 4], 6)
# ^-----^ 4 + 2 = 6, indices: 0, 2 *
# ^-----^ 3 + 3 = 6, indices: 1, 3
# ^-----^ 2 + 4 = 6, indices: 2, 4
# * entire pair is earlier, and therefore is the correct answer
== [4, 2]
sum_pairs([0, 0, -2, 3], 2)
# there are no pairs of values that can be added to produce 2.
== None/nil/undefined (Based on the language)
sum_pairs([10, 5, 2, 3, 7, 5], 10)
# ^-----------^ 5 + 5 = 10, indices: 1, 5
# ^--^ 3 + 7 = 10, indices: 3, 4 *
# * entire pair is earlier, and therefore is the correct answer
== [3, 7]
My code:
def sum_pairs(L,I):
past = {}
#set the first value in L as my key for dict
past[L[0]] = 1
idx = 1
while idx < len(L):
needed = I - L[idx]
if needed in past:
return [needed, L[idx]]
past[L[idx]] = 0
idx += 1
return None
This should be faster since its O(n) rather than O(n^2) algorithm of your post
def sum_pairs(arr, s):
" O(n) algorithm "
# Reverse dictionary lookup for values (only keep first entry)
# It produces a list of indices for a given value
# Example arr = (10, 5, 5) => d = {10: [0], 5: [1, 2]}
d = {}
for i, v in enumerate(arr):
if not v in d:
d[v] = [i]
elif len(d[v]) == 1:
# Only take first two values (i.e. arr = [5]*1e7
# will only produce {5: [0, 1]}
d[v].append(i)
# Find all possible pairs
# For each v in dictionary d, we can only use it in a pair if s-v is also in the dictionary
# so we need to find elements such that both v and s-v are in d (which would be array arr)
result = [[(x, y) for x in d.get(v, []) for y in d.get(s-v, []) if x < y] for v in d]
flatten_result = [item for sublist in result for item in sublist]
if flatten_result:
# Find the earliest indexes
min1 = min(flatten_result, key = lambda x: max(x))
# Return values with earliest indexes
return arr[min1[0]], arr[min1[1]], min1
else:
return None
print(sum_pairs([11, 3, 7, 5], 10)) # (3, 7, (1, 2))
print(sum_pairs([4, 3, 2, 3, 4], 6)) # (4, 2, (0, 2))
print(sum_pairs([0, 0, -2, 3], 2)) # None
print(sum_pairs([10, 5, 2, 3, 7, 5], 10)) # (3, 7, (3, 4))
print(sum_pairs([5, 9, 13, -3], 10)) # (13, -3, (2, 3))
print(sum_pairs([10, 5, 5], 10))
Performance Testing
import time
N = 10000000 # 10 million elements
# Test 1 -- repeating elements
arr = [5]*N
t_start = time.time()
ans = sum_pairs(arr, 10)
print(f'Processing time: {time.time() - t_start:.2f} seconds')
#>> Processing time: 11.81 seconds
from random import randint
arr = [randint(0, 20) for _ in range(N)]
t_start = time.time()
ans = sum_pairs(arr, 10)
print(f'Processing time: {time.time() - t_start:.2f} seconds')
#>> Processing time: 9.12 seconds
Result Less than 12 seconds for 10 million samples using on line IDE from https://repl.it/languages/python3