79 lines
2.1 KiB
Python
79 lines
2.1 KiB
Python
"""
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A non-empty array A consisting of N integers is given. Array A represents
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numbers on a tape.
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Any integer P, such that 0 < P < N, splits this tape into two non-empty parts:
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A[0], A[1], ..., A[P − 1] and A[P], A[P + 1], ..., A[N − 1].
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The difference between the two parts is the value of:
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|(A[0] + A[1] + ... + A[P − 1]) − (A[P] + A[P + 1] + ... + A[N − 1])|
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In other words, it is the absolute difference between the sum of the first part
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and the sum of the second part.
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For example, consider array A such that:
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A[0] = 3
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A[1] = 1
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A[2] = 2
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A[3] = 4
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A[4] = 3
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We can split this tape in four places:
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P = 1, difference = |3 − 10| = 7
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P = 2, difference = |4 − 9| = 5
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P = 3, difference = |6 − 7| = 1
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P = 4, difference = |10 − 3| = 7
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Write a function:
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class Solution { public int solution(int[] A); }
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that, given a non-empty array A of N integers, returns the minimal difference
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that can be achieved.
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For example, given:
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A[0] = 3
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A[1] = 1
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A[2] = 2
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A[3] = 4
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A[4] = 3
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the function should return 1, as explained above.
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Write an efficient algorithm for the following assumptions:
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N is an integer within the range [2..100,000];
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each element of array A is an integer within the range [−1,000..1,000].
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"""
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def solution(A):
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size = len(A) - 1
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if size == 1:
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return abs(A[0] - A[1])
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l = [A[0]]
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r = [A[-1]]
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for i in range(size):
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l.append(l[i] + A[i+1])
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r.append(r[i] + A[size-i-1])
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res = abs(l[0] - r[size-1])
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for i in range(1, size):
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res = min(res, abs(l[i] - r[size-i-1]))
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return res
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res = solution([1, 1, 3])
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assert res == 1, "got {}".format(res)
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res = solution([3, 1, 2, 4, 3])
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assert res == 1, "got {}".format(res)
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res = solution([-2, -3, -4, -1])
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assert res == 0, "got {}".format(res)
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res = solution([1, 8, -10, 8, 7, 9, 11])
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assert res == 6, "got {}".format(res)
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res = solution([-1, 1, 0, 1, -1, 0, -1, 1, -1, 0, 1, 0, 1, -1, 0])
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assert res == 0, "got {}".format(res)
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res = solution([14, 9])
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assert res == 5, "got {}".format(res)
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