curriculum/challenges/english/blocks/rosetta-code-challenges/5987fd532b954e0f21b5d3f6.md
An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence $A$:
<ul style='list-style: none;'> <li><big>$A_0 = -7$</big></li> <li><big>$A_1 = 1$</big></li> <li><big>$A_2 = 5$</big></li> <li><big>$A_3 = 2$</big></li> <li><big>$A_4 = -4$</big></li> <li><big>$A_5 = 3$</big></li> <li><big>$A_6 = 0$</big></li> </ul>3 is an equilibrium index, because:
6 is also an equilibrium index, because:
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence $A$.
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
equilibrium should be a function.
assert(typeof equilibrium === 'function');
equilibrium([-7, 1, 5, 2, -4, 3, 0]) should return [3,6].
assert.deepEqual(
equilibrium(globalThis.equilibriumTests[0]),
globalThis.ans[0]
);
equilibrium([2, 4, 6]) should return [].
assert.deepEqual(
equilibrium(globalThis.equilibriumTests[1]),
globalThis.ans[1]
);
equilibrium([2, 9, 2]) should return [1].
assert.deepEqual(
equilibrium(globalThis.equilibriumTests[2]),
globalThis.ans[2]
);
equilibrium([1, -1, 1, -1, 1, -1, 1]) should return [0,1,2,3,4,5,6].
assert.deepEqual(
equilibrium(globalThis.equilibriumTests[3]),
globalThis.ans[3]
);
equilibrium([1]) should return [0].
assert.deepEqual(
equilibrium(globalThis.equilibriumTests[4]),
globalThis.ans[4]
);
equilibrium([]) should return [].
assert.deepEqual(
equilibrium(globalThis.equilibriumTests[5]),
globalThis.ans[5]
);
globalThis.equilibriumTests =
[[-7, 1, 5, 2, -4, 3, 0], // 3, 6
[2, 4, 6], // empty
[2, 9, 2], // 1
[1, -1, 1, -1, 1, -1, 1], // 0,1,2,3,4,5,6
[1], // 0
[] // empty
];
globalThis.ans = [[3, 6], [], [1], [0, 1, 2, 3, 4, 5, 6], [0], []];
function equilibrium(a) {
}
function equilibrium(a) {
let N = a.length,
i,
l = [],
r = [],
e = [];
for (l[0] = a[0], r[N - 1] = a[N - 1], i = 1; i < N; i++)
{ l[i] = l[i - 1] + a[i], r[N - i - 1] = r[N - i] + a[N - i - 1]; }
for (i = 0; i < N; i++)
{ if (l[i] === r[i]) e.push(i); }
return e;
}