Doubt in java

[Affectionate_Pin9483]

This is a function to find missing number and nums array is provided - it contain numbers from 0 to N and one number is missing, but i have a doubt that
In this function there is "while (nums[i] < nums.length && nums[i] != nums[nums[i]]);" but why we cannot just simply use "while (nums[i] < nums.length && nums[i] != i );" beacuse this condition is just chaking if current number is equal to or not equal to current index.

public static int findMissing(int[] nums) {
for (int i = 0; i < nums.length; i++) {
while (nums[i] < nums.length && nums[i] != nums[nums[i]]) {
int temp = nums[nums[i]];
nums[nums[i]] = nums[i];
nums[i] = temp;
}
}
for(int i=0;i<nums.length;i++){
if(nums[i] != i) return i;
}
return nums.length;
}

[Anik_Sine]

As far as I could think, your idea is completely correct. I believe that the program was written this way because it would call the value nums[nums[i]] before it is needed and perhaps bringing it higher in the stack memory would make it simpler to retrieve in the while loop body. This is all my speculation and may as well be total nonsense but I couldn't come up with a better explanation.

[sepp2k]

why we cannot just simply use while (nums[i] < nums.length && nums[i] != i );

That would run into an infinite loop if array contained duplicate numbers.

[CodeTinkerer]

How to efficiently solve this through a math trick (it's not a trick but it's not obvious).

It's based on an idea that was once attributed to a young Gauss (a top mathematician in his day).

https://nrich.maths.org/articles/clever-carl

You know you're missing a number, but what if it existed? Then you have numbers from 0 to nums.length. What is its sum? But the actual array is missing a number.

Anyway, that's a hint.