Two Pointers

What is it?

Keep two positions in a sequence and advance them according to what you observe, so that each element is visited a bounded number of times. The pointers may start at opposite ends and close in, or start together and move in the same direction. Either way the win is the same: a decision that would need a nested scan becomes a single pass, because every move permanently rules out a region you will never need to revisit.

The core idea

Template

The shape, not a solution. Adapt the condition and the state to the problem in front of you.

1def opposite_ends(nums, target):2    """Pointers start apart and close in. Requires sorted input."""3    left, right = 0, len(nums) - 14    while left < right:5        total = nums[left] + nums[right]6        if total == target:7            return [left, right]8        if total < target:9            # Everything left of 'left' is even smaller, so it can never10            # reach the target with any partner. Discard it.11            left += 112        else:13            right -= 114    return []15 16 17def same_direction(nums):18    """Slow marks where the next kept item goes; fast scans ahead."""19    slow = 020    for fast in range(len(nums)):21        if keep(nums[fast]):22            nums[slow] = nums[fast]23            slow += 124    return slow  # length of the compacted prefix

Common mistakes

• Using `while left <= right` when the two pointers must reference distinct elements — this lets an element pair with itself.
• Forgetting to skip duplicates after recording a match, which produces the same answer several times.
• Sorting when the problem requires original indices, then returning positions from the sorted copy.
• Moving both pointers on every iteration, which can step straight over the answer.