zoek voor een reeks integers de eerste reeks waar in de som gelijk is aan de gegeven waarde en leg de moeilijkheid uit.

Question

Manu · Accepted Answer

public static int findSubArrayThatSumsTo(int[] A, int S) {
	int sum = 0;
	int j = 0;
	for (int i = 0; i  S) {
		sum-=A[j];
		j++;
	    }
	    if (sum == S) {
		System.out.format("Sum (%d) found between indices %d and %d
", S, j, i);
		break;
	    }
	}
	return j; // return start-index
    }

Manu · Answer

Complexity of above is O(n). From the look of it (given two loops), it might seem that complexity is higher. But actually every element is accessed utmost twice. So, O(2n) = O(n).

Anonymous · Answer

The answers with keeping the running sum whilst iterating over the array, and advancing the beginning of the sequence pointer if the running sum is greater than the desired sum, only work for sequences of all positive number, right? For arrays that allow negative numbers, we need something 'smarter' like e.g. calculating the array of partial sum and finding the pair with the desired difference - O(n log n).

Counter example - look for 5 in the sequence of {1, 7, -3}. The above solutions would find none, whereas the right answer is {1, 7, -3}

Nazir - For the problem Convert a string of hexadecimal characters to an array of nibles · Answer

use strict;
use warnings;

my $str = "0F3A45JA";

my @input = split (//, $str);
my $bin;
my @out;

foreach my $in (@input) {
        next, unless ($in =~ m/[0-9A-Fa-f]/);
        $bin = sprintf( "%b", hex( $in) );
        push (@out, $bin .= 0 x (4 - length($bin)));
}
foreach (@out) {
        print "$_
";
}

shelling · Answer

#!/usr/bin/env perl
use Modern::Perl;
use Data::Dumper;

my $array = [1,3,4,2,6,7,4];

sub find_seq_with_sum {
  my ($n, $list) = @_;
  my ($result, $s) = ([], 0);

  for (@$list) {
    while ($s > $n) { $s -= shift($result); }

    last if $s == $n;

    # $s < $n
    push($result, $_);
    $s += $_;
  }

  return $result;
}

say Dumper(find_seq_with_sum(13, $array));

Alina · Answer

sub s_sum {
	my ( $arr, $total ) = @_;
	for ( my $i = 0; $i [$j];
			push @find, $arr->[$j];
			return \@find if $sum == $total;
		}
	}
}

Alina · Answer

sub s_sum {
	my ( $arr, $total ) = @_;
	for ( my $i = 0; $i [$j];
			return [ ( @$arr )[$i..$j] ] if $sum == $total;
		}
	}
}

Alina · Answer

sub s_sum_1 {
	my ( $arr, $total ) = @_;
	for ( my $i = 0; $i [$j];
			return [ ( @$arr )[$i..$j] ] if $sum == $total;
		}
	}
}

Anonymous · Answer

Could you tell us what other questions they asked you?

Bafniz · Answer

Python answer:

def sumSeq(seq,target):
    #return list of couples of indices that can be used as slices of seq that sum to t
    for a in range(0,len(seq)):
        for b in range(a+1,len(seq)):
            t=sum(seq[a:b])
            if t>target:
                break
            if t==target:
                return (a,b)
    return None

,,,and I will spare you the extensive tests.

Anonymous · Answer

Integer[] input = new Integer[] {3, 3, 2, 1, 2, 45, 32, 1, 23, 4}; // same array to test
	Set uniques = new LinkedHashSet();
	for(int i=0; i

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