Nearest Pair
March 13, 2018
Today’s exercise is an interview question:
You are given a list of integers and must find the nearest pair that sum to a given target. For instance, given the list (1 5 3 6 4 2), if the target is 7, there are three pairs with the required sum, (1 6), (5 2) and (3 4), but pair (1 6) has two intervening numbers, pair (5 2) has three intervening numbers, and pair (3 4) has only one intervening number, so (3 4) is the nearest pair.
Your task is to write a program to find the nearest pair. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Programming Praxis 
var a = [1, 5, 3, 6, 4, 2];
var p = [];
var q = [];
for(i=0;i<a.length;i++){
for(j=i+1;j<a.length;j++){
if(a[i] + a[j] === 7){
var k = j – i;
p.push([k]);
q.push([k,[a[i],a[j]]]);
}
This being an interview question, you’d definitely want to ask if you can assume the numbers in the list are unique.
is second iteration necessary ” for(s=0;s<item.length;s++){
A Haskell version.
import Data.Function (on) import Data.List (minimumBy, tails) -- A position and value. data PV a b = PV { pos :: a, val :: b } -- The nearest pairs of values from xs that sum to s, or nothing if no values -- have that sum. nearestPair :: Real a => a -> [a] -> Maybe (a, a) nearestPair s xs = let ps = filter (sumTo s) $ pairs $ mkPVs xs in case ps of [] -> Nothing _ -> Just $ both val $ minimumBy (compare `on` dist) ps -- The equivalent list of PVs starting at position 0. mkPVs :: [a] -> [PV Int a] mkPVs = zipWith PV [0..] -- All pairs of elements from xs, where the first element appears strictly -- earlier in the argument list than the second. pairs :: [a] -> [(a, a)] pairs xs = [(x1, x2) | (x1:xs') <- tails xs, x2 <- xs'] -- True iff the values sum to s. sumTo :: (Num b, Eq b) => b -> (PV a b, PV a b) -> Bool sumTo s (x, y) = val x + val y == s -- The distance, by position, between two PVs. dist :: Num a => (PV a b, PV a b) -> a dist (x, y) = abs (pos x - pos y) -- The result of applying the function to both elements of the tuple. both :: (a -> b) -> (a, a) -> (b, b) both f (x, y) = (f x, f y) main :: IO () main = do print $ nearestPair 7 [1, 5, 3, 6, 4, 2] print $ nearestPair 2 [1, 5, 3, 6, 4, 2]it can be done in O(N) using hash tables
Here’s a solution in Python.
from collections import defaultdict # Returns the indices of the nearest pair. # (returns indices to distinguish duplicated values) def find_nearest_pair(l, target): index_lookup = defaultdict(list) for idx, x in enumerate(l): index_lookup[x].append(idx) nearest_pair = None nearest_distance = None for idx1, x in enumerate(l): y = target - x if y not in index_lookup: continue for idx2 in index_lookup[y]: if idx1 >= idx2: continue distance = idx2 - idx1 if (nearest_pair is None) or (distance < nearest_distance): nearest_pair = (idx1, idx2) nearest_distance = distance return nearest_pair l = [1, 5, 3, 6, 4, 2] print 'list: {}'.format(l) print for target in range(13): print 'target: {}'.format(target) nearest_pair = find_nearest_pair(l, target) if not nearest_pair: continue print " indices: {}".format(nearest_pair) print " values: {}".format(tuple(l[idx] for idx in nearest_pair))Output:
Line 13 can be removed from my prior solution, as the defaultdict will return an empty list.
O(n) solution:
def nearest_pair(iterable, target): best_pair = None best_span = None seen = {} for index, right_no in enumerate(iterable): left_no = target - right_no if left_no in seen: span = abs(index - seen[left_no]) if best_span is None or span < best_span: best_span = span best_pair = (left_no, right_no) seen[right_no] = index return best_pair