To solve this problem efficiently, we can utilize the properties of the XOR bitwise operation. XOR has a useful property where:

This means that when you XOR all elements of an array together, elements that appear an even number of times will cancel out to zero, leaving only the element that appears an odd number of times.

First, start with a variable initialized to zero. This will be used to XOR all elements of the array.

Next, iterate through each element of the array and XOR it with the variable initialized in the previous step.

After iterating through the array, the variable will contain the value of the unpaired element because all paired elements will cancel each other out due to their even occurrences.

Let’s see this approach implemented in JavaScript:

```
function solution(A) {
let result = 0;
for (let i = 0; i < A.length; i++) {
result ^= A[i];
}
return result;
}
```

### Explanation of the Code:

`result`

is initialized to 0. This will be our accumulator for the XOR operation.

We loop through each element of the array `A`

.

We XOR (`^=`

) each element with `result`

. This effectively accumulates all XOR operations across the array.

**Return the result**: After iterating through the array,`result`

will contain the value of the unpaired element because all paired elements cancel out.

### Efficiency:

This algorithm operates in ( O(N) ) time complexity, where ( N ) is the number of elements in the array. This is optimal for the problem constraints (with ( N ) up to 1,000,000), ensuring that the solution is both time-efficient and space-efficient.

### Edge Cases:

The algorithm handles arrays of minimum size (e.g., single element) and ensures correct results for all valid inputs within the specified constraints.

This method leverages XOR’s properties to solve the problem in a straightforward and efficient manner.