Problem

Submissions

3

Avg. time to solve

15 min

Success Rate

85%

Problem Statement

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1. 'SEQUENCE' may contain more than one equilibrium indices.
2. If there are no equilibrium indices, return an empty sequence.
3. Consider the sum of elements lower than the first index and higher than the last index to be 0 (zero).
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```
The first line of input contains an integer ‘T’ denoting the number of test cases.
The first line of each test case contains an integer ‘N’ denoting the number of elements in the given array/list 'SEQUENCE'.
The second line of each test case contains ‘N’ space-separated integers denoting the elements in the 'SEQUENCE'.
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For each test case, return a sequence of equilibrium indices. If no such index exists, return an empty sequence.
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You don't need to print the output, it has already been taken care of. Just implement the given function.
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```
1 <= T <= 50
1 <= N <= 10^4
-10^9 <= SEQUENCE[i] <= 10^9
Where 'SEQUENCE[i]' denotes the elements of the array 'SEQUENCE'.
Time limit: 1 sec
```

```
2
7
-7 1 5 2 -4 3 0
5
1 2 3 4 5
```

```
3 6
(Empty Sequence)
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```
In test case 1, in the given sequence of indices, index 3 (consider 0 based indexing) is the equilibrium index, because the sum of the elements present at the indices lower than 3 i.e [-7 + 1 + 5 = 1 ] is equal to sum of all the elements present at indices higher than 3 i.e [ -4 + 3 + 0 = 1]. Similarly, for index = 6 the sum of elements at lower indices [-7+1+5+2+(-4)+3] is equal to the sum of higher indices(since it is the last element, the sum of higher indices is 0 as explained above). Hence we return the sequence [3, 6] which are the equilibrium indices.
In test case 2, for each index, we find the sum of all the values present at indices lower than the index and greater than the index. Consider index = 2, sum of elements present at indices lower than 2 is (1+ 2 = 3), and the sum of elements present at indices higher than 2 is (4 + 5 = 9), Because the sum doesn’t match, the index 2 is not at equilibrium. We can check in a similar way for all the indices of the sequence and conclude no index satisfies the equilibrium condition, therefore we return an empty sequence.
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```
2
7
-2 1 9 2 -6 3 0
4
100 -99 2 1
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```
2
2
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```
In test case 1, in the given sequence of indices, index 2 (consider 0 based indexing) is the equilibrium index, because the sum of the elements present at the indices lower than 2 i.e [-2 + 1 = -1 ] is equal to sum of all the elements present at indices higher than 2 i.e [ 2 + -6 + 3 + 0 = -1]. Hence we return the sequence [2] which is the equilibrium index.
In test case 2, for each index, we find the sum of all the values present at indices lower than the index and greater than the index. Consider index = 2, Sum of elements present at indices lower than 2 is (100 - 99 = 1), and the sum of elements present at indices higher than 2 is (1). Hence we return the sequence [2] which is the equilibrium index
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