Add Kotlin implementation to find missing number in a list

[Balaviknesh]

No description provided.

[super30admin]

Strengths:

• You have chosen an efficient algorithm (binary search) for the problem, which is good for sorted arrays.
• The code is clean and well-organized with proper variable names.
• You handled the case when no number is missing by returning -1.

Areas for Improvement:

• The solution does not account for cases where the missing number is the first element (1) or the last element (n). For example, if the input is [2,3,4,5], your code should return 1, but it currently does not.
• The condition if(list[0] + list.size - 1 == list[list.size - 1]) return -1 is incorrect for detecting no missing number. This condition checks if the last element is equal to the first element plus the length minus one. However, this only works if the array starts at 1. If the array starts at a number greater than 1, this condition will fail. For example, [2,3,4,5] has length 4, and 2 + 4 - 1 = 5, which equals the last element, so your code would return -1, but actually, 1 is missing.
• The binary search logic is based on the difference between list[mid] and list[0], which is not the standard approach. The standard approach is to compare the value at mid with the expected value (which would be mid + 1 if the array started at 1). However, since the array might not start at 1, you should use the index to compute the expected value. For instance, at index i, the expected value is i + 1 (if the array should start at 1). But if the array does not start at 1, then the expected value might be different. Actually, the problem states that the array contains integers from 1 to n with one missing, so the array should ideally start at 1 and end at n. However, if the missing number is 1, the array will start at 2. Similarly, if the missing number is n, the array will end at n-1.

To fix this, you should use the fact that in a complete array without duplicates, the value at index i should be i + 1. So, you can check if the value at mid is equal to mid + 1. If it is, then the missing number is to the right; otherwise, it's to the left. This is a common approach.

Here is a corrected version of the binary search:

var low = 0
var high = list.size - 1

while (low <= high) {
    val mid = low + (high - low) / 2
    if (list[mid] == mid + 1) {
        low = mid + 1
    } else {
        high = mid - 1
    }
}
return low + 1

This will work for arrays that start at 1. However, if the array does not start at 1 (meaning the missing number is 1), then the first element will be 2. At index0, the value is 2, which is not equal to 1 (0+1). So, the binary search would correctly identify that the missing number is on the left. But note: if the array does not start at 1, then the entire array is shifted. Actually, the problem assumes the array should contain numbers from 1 to n, so if the first element is not 1, then 1 is missing. So, you can first check if the first element is 1. If not, return 1.

Similarly, if the last element is not n (which is list.size + 1), then the missing number is n. So, you should check these edge cases first.

Revised plan:

1. If the list is empty, return -1.
2. If the first element is not 1, return 1.
3. If the last element is not list.size + 1, return list.size + 1.
4. Otherwise, perform binary search to find the missing number.

Example:
For [2,3,4,5]: first element is 2 != 1, so return 1.
For [1,2,3,4]: last element is 4, but n should be 5 (since size=4, n=5), so return 5.

Then, inside the binary search, we assume that the missing number is between the first and last. The binary search checks for the point where the value is not equal to index+1.

So, the code should be:

fun searchMissingNumber(list: List<Int>): Int {
    val n = list.size + 1
    if (list.isEmpty()) return -1
    if (list[0] != 1) return 1
    if (list[list.size-1] != n) return n

    var low = 0
    var high = list.size - 1
    while (low <= high) {
        val mid = low + (high - low) / 2
        if (list[mid] == mid + 1) {
            low = mid + 1
        } else {
            high = mid - 1
        }
    }
    return low + 1
}

This should handle all cases.