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Code logic is not clear

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#1

The logic in the code is not clear

int left = 0;

// the initial value for right index is the number of elements in the array

int right = nums.length;

// left + 1 >= right will finish while loop

while (left + 1 < right) {

  int mid = (right + left) / 2;

       

  if (nums[mid] == mid && left < nums[left]) {

    // mid is the index to answer

    return mid;

Please help me understand

#2

Hello,

I hope you’re doing well.

This function is designed to find a “fixed point” in an array nums, which is an index i such that nums[i] == i. Here’s a step-by-step explanation of what the code does:

  1. Initialize two variables:
    • left to 0, indicating the leftmost index.
    • right to the number of elements in the array, indicating the rightmost index.
  2. Enter a while loop with the condition left + 1 < right, which means the loop continues until there are at least two elements to search between.
  3. Inside the loop:
    • Calculate the middle index mid as (right + left) / 2.
    • Check if nums[mid] is equal to mid and left is less than nums[left]. If true, return mid because it’s the fixed point.
    • If nums[mid] < mid, it means that we need to look for the fixed point in the right half of the array, so update left = mid.
    • Otherwise, if nums[mid] > mid, it means we need to look for the fixed point in the left half of the array, so update right = mid.
  4. After the loop, if left points to a fixed point (nums[left] == left), return left.
  5. Also, if right is within the bounds of the array and it points to a fixed point (nums[right] == right), return right.
  6. If no fixed point is found, return -1 to indicate that there’s no such element in the array.

The algorithm use binary search to locate the fixed point within the array, using the fact that the array is sorted and the target value (the index) exhibits a specific relationship with the values in the array.