Minimum Split
December 4, 2015
Today’s task is a simple exercise in list handling:
Given a list of integers, find the place where the list can be split in two that minimizes the differences in the sums of the two halves of the list. For instance, given the list [3,7,4,2,1], splitting after the second element of the list gives sums of 10 and 7 for the two halves of the list, for a difference of 3, and no other split gives a lower difference.
Your task is to write a program that splits a list as indicated above. 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 
In Python
It’s not very clear whether this means the minimum difference, or the minimum absolute value of the difference. The wording of page 2 suggests the latter, but page 1 doesn’t say so.
Haskell:
import Data.List f x = minimumBy (\(_,a) (_,b) -> compare (abs a) (abs b)) a where a = zipWith (\a b -> ((a,b), sum a - sum b)) (inits x) (tails x)Python:
def bestSplit(lst):
diff = 100000000
idx = 0
for count in range(0, len(lst)):
if diff > abs(sum(lst[:count], 0) – sum(lst[count:], count)):
diff = abs(sum(lst[:count], 0) – sum(lst[count:], count))
idx = count
return idx
fun split xs = let val sum = List.foldr (op +) 0 xs fun partial_sum [] total = [] | partial_sum (x::xs) total = let val partial = total + x in partial :: partial_sum xs partial end fun min [] i res = ( res, i ) | min (x::xs) i res = let val new_min = Int.abs(x - (sum - x)) in if new_min < res then min xs (i+1) new_min else min xs i res end in min (partial_sum xs 0) 0 (Int.abs sum) end [hunter@apollo: 12]$ poly --use minsplit.sml Poly/ML 5.2 Release > use "minsplit.sml"; val split = fn : Int.int LIST.list -> Int.int * int val it = () : unit > split [3,7,4,2,1]; val it = (3, 2) : Int.int * int > split [~1,1]; val it = (0, 0) : Int.int * int > split [1,2,3,4,5]; val it = (3, 3) : Int.int * intSolution in Python
Third attempt in posting the code :p
def list_sep(number_list): for i in xrange(1,len(number_list)): temp_diff=abs(sum(number_list[:i])-sum(number_list[i:])) if i== 1: diff=temp_diff position=i if temp_diff<diff: diff= temp_diff position=i return diff, position number_list=[3,7,4,2,1] print('Difference: %s' %list_sep(number_list)[0]) print('Split position: %s' %list_sep(number_list)[1])In Python.
from itertools import accumulate, chain def ssum(L): S = sum(L) return min((abs(S-2*a), i) for i, a in enumerate(chain([0], accumulate(L))))C++ attempt
#include “math.h”
#include
std::vector numberList;
numberList.push_back(3);
numberList.push_back(7);
numberList.push_back(4);
numberList.push_back(2);
numberList.push_back(1);
std::vector differences;
for(int split = 0; split < numberList.size(); ++split)
{
int leftSum = 0;
for(int left = 0; left < split; ++left) {
leftSum += numberList[left];
}
int rightSum = 0;
for(int right = split; right < numberList.size(); ++right) {
rightSum += numberList[right];
}
int difference = abs(rightSum – leftSum);
differences.push_back(difference);
}
int lowestDiffIndex = 0;
int lowestDiffValue = differences[0];
for(int i = 0; i < differences.size(); ++i)
{
lowestDiffIndex = (differences[i] < lowestDiffValue) ? i : lowestDiffIndex;
lowestDiffValue = (differences[i] < lowestDiffValue) ? differences[i] : lowestDiffValue;
}
// Split is at element lowestDiffIndex
A solution in Racket.