Given an array of integers, find an equilibrium index. An equilibrium index of an array is an index such that the sum of elements at lower indexes is equal to the sum of elements at higher indexes.

Edit 2017.07.01 Added three more example input & output. Fixed an off-by-one error in the Ruby implementation.

For example, consider the following array A.

       A: -7  1  5  2 -4  3  0
position:  0  1  2  3  4  5  6

Index 3 is an equilibrium index.

A[0] + A[1] + A[2] = A[4] + A[5] + A[6]

Index 6 is also an equilibrium index, because sum of zero elements is zero.

A[0] + A[1] + A[2] + A[3] + A[4] + A[5] = 0

Index 7 is not an equilibrium index, because it is not a valid index of array A.

Input. The input file consists of one or more arrays, each on a single line. Each array consists of a sequence of integers, each pair separated by a single space. The input file is terminated by an EOF character. Consider the following example.

-7  1  5  2 -4  3  0
1

1 1
1 0
1 0 -1
0 1 -1
1 -1 0

Output. The output file consists of an output line for each array in the input. Each output line consists of an equilibrium index (if any) or -1 when no equilibrium index exists. For example, the following output corresponds to the previous example input.

3
0
-1
-1
0
-1
0
2

Solution

The proposed solution is O(n).

Just consider each position of a given array, from left to right. Start with a variable l for the left sum set to zero and a variable r for the right sum set to the sum of all elements except the first. Then, as you go from left to right, add the element that becomes part of the left side to l and subtract from r the element that leaves the right side. Return the current index when l and r are equal.

Ruby implementation

#!/usr/bin/env ruby

def equilibrium_index nn
  return -1 if nn.length == 0
  return 0 if nn.length == 1
  ls = 0
  rs = nn[1..-1].reduce(0, &:+)
  return 0 if rs == 0
  nn[1..-1].each_with_index do |n, i|
    ls += nn[i]
    rs -= n
    return i + 1 if ls == rs
  end
  return -1
end

while true
  nn = readline.strip.split(' ').map(&:to_i) rescue break
  puts equilibrium_index nn
end

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