Search in Rotated Sorted Array

No matter where a rotated array is cleaved in half, at least one of the two halves remains strictly sorted. Checking whether the target falls within the boundary values of that ordered half tells you whether to search inside it or discard it completely. That single branching rule preserves the logarithmic halving property at every step.

Full solution

Direct one-pass binary search: at each mid, one of the two halves is guaranteed sorted, so check the target against that half's boundary values to decide which side to keep. This is the answer an interview expects over locating the pivot first as a separate phase.

def search(nums: list[int], target: int) -> int:
    low, high = 0, len(nums) - 1
    while low <= high:
        mid = (low + high) // 2
        if nums[mid] == target:
            return mid
        if nums[low] <= nums[mid]:
            # Left half is sorted.
            if nums[low] <= target < nums[mid]:
                high = mid - 1
            else:
                low = mid + 1
        else:
            # Right half is sorted.
            if nums[mid] < target <= nums[high]:
                low = mid + 1
            else:
                high = mid - 1
    return -1

function search(nums, target) {
  let low = 0;
  let high = nums.length - 1;
  while (low <= high) {
    const mid = Math.floor((low + high) / 2);
    if (nums[mid] === target) return mid;
    if (nums[low] <= nums[mid]) {
      // Left half is sorted.
      if (nums[low] <= target && target < nums[mid]) {
        high = mid - 1;
      } else {
        low = mid + 1;
      }
    } else {
      // Right half is sorted.
      if (nums[mid] < target && target <= nums[high]) {
        low = mid + 1;
      } else {
        high = mid - 1;
      }
    }
  }
  return -1;
}

The theory behind it

Arrays — the ground this problem stands on. All Arrays problems

What Arrays is

An array is a row of fixed boxes laid side by side in computer memory, like numbered lockers in a hallway. Because each box occupies identical space and sits directly next to its neighbors, jumping to locker zero or locker ten thousand takes the exact same tiny fraction of time. Every box holds an item of the same type, addressed by an offset number called an index.

When to reach for it

Reach for an array when items arrive in a known sequence and need immediate retrieval by position number. Problems asking for running totals, prefix accumulations, cyclic rotations, or in-place rearrangements signal array mechanics. Whenever constraints require constant-time random lookups or contiguous cache scans across fixed collections, a flat sequence is the default container.

How the pattern works

Visualize a tape with zero-indexed slots stretching from start to end. Keep track of write and read cursors when modifying contents without allocating helper buffers. For running computations, maintain an invariant such as having processed all elements left of the current index while pending elements wait to the right. When modifying entries in place, consider scanning backwards from the end so unread data is not overwritten.