Algorithm to Divide a list of numbers into 2 equal sum lists

Question

There is a list of numbers.

The list is to be divided into 2 equal sized lists, with a minimal difference in sum. The sums have to be printed.

#Example:
>>>que = [2,3,10,5,8,9,7,3,5,2]
>>>make_teams(que)
27 27

Is there an error in the following code algorithm for some case?

How do I optimize and/or pythonize this?

def make_teams(que):
    que.sort()
    if len(que)%2: que.insert(0,0)
    t1,t2 = [],[]
    while que:
    val = (que.pop(), que.pop())
    if sum(t1)>sum(t2):
        t2.append(val[0])
        t1.append(val[1])
    else:
        t1.append(val[0])
        t2.append(val[1])
    print min(sum(t1),sum(t2)), max(sum(t1),sum(t2)), "\n"

Question is from http://www.codechef.com/problems/TEAMSEL/

Answer 1

Dynamic programming is the solution you're looking for.

Example with [4, 3, 10, 3, 2, 5]:

X-Axis: Reachable sum of group.        max = sum(all numbers) / 2    (rounded up)
Y-Axis: Count elements in group.       max = count numbers / 2       (rounded up)

      1  2  3  4  5  6  7  8  9 10 11 12 13 14
 1  |  |  |  | 4|  |  |  |  |  |  |  |  |  |  |       //  4
 2  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |
 3  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |

      1  2  3  4  5  6  7  8  9 10 11 12 13 14
 1  |  |  | 3| 4|  |  |  |  |  |  |  |  |  |  |       //  3
 2  |  |  |  |  |  |  | 3|  |  |  |  |  |  |  |
 3  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |

      1  2  3  4  5  6  7  8  9 10 11 12 13 14
 1  |  |  | 3| 4|  |  |  |  |  |10|  |  |  |  |       // 10
 2  |  |  |  |  |  |  | 3|  |  |  |  |  |10|10|
 3  |  |  |  |  |  |  |  |  |  |  |  |  |  |  |

      1  2  3  4  5  6  7  8  9 10 11 12 13 14
 1  |  |  | 3| 4|  |  |  |  |  |10|  |  |  |  |       //  3
 2  |  |  |  |  |  | 3| 3|  |  |  |  |  |10|10|
 3  |  |  |  |  |  |  |  |  |  | 3|  |  |  |  |

      1  2  3  4  5  6  7  8  9 10 11 12 13 14
 1  |  | 2| 3| 4|  |  |  |  |  |10|  |  |  |  |       //  2
 2  |  |  |  |  | 2| 3| 3|  |  |  |  | 2|10|10|
 3  |  |  |  |  |  |  |  | 2| 2| 3|  |  |  |  |

      1  2  3  4  5  6  7  8  9 10 11 12 13 14
 1  |  | 2| 3| 4| 5|  |  |  |  |10|  |  |  |  |       //  5
 2  |  |  |  |  | 2| 3| 3| 5| 5|  |  | 2|10|10|
 3  |  |  |  |  |  |  |  | 2| 2| 3| 5| 5|  |  |
                                       ^

12 is our lucky number! Backtracing to get the group:

12 - 5 = 7        {5}
 7 - 3 = 4        {5, 3}
 4 - 4 = 0        {5, 3, 4}

The other set can then be calculated: {4,3,10,3,2,5} - {5,3,4} = {10,3,2}

All fields with a number are possible solutions for one bag. Choose the one that is furthest in the bottom right corner.

BTW: It's called the knapsack-problem.

If all weights (w1, ..., wn and W) are nonnegative integers, the knapsack problem can be solved in pseudo-polynomial time using dynamic programming.