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Subsequence After One Replacement

MediumTwo pointersString
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Medium · two pointers solution for Subsequence After One Replacement. Python, Go, and a step-by-step animation.

01

Problem

Read the prompt, flip through the examples, then steal the idea.

LeetCode 3983. Subsequence After One Replacement. The copyrighted prompt is on LeetCode. This page is a working two pointers solution for this problem: editable input, Python, Go, and a step-by-step animation.

Constraints

  • —Player inputs are capped so the animation stays readable.
  • —Match LeetCode's own constraints when you submit there.

Examples

Input

nums =
[2, 7, 11, 15]
target =
9

Output

[0, 1]

2 + 7 = 9. The inward walk finds the pair in one comparison.

02

Solution

Python and Go. The walkthrough below runs this same code.

Approach 1Optimal▶ Animated

Subsequence After One Replacement — two pointers

On a sorted array the pair that sums to target is found by walking inward. Built for LeetCode 3983. Subsequence After One Replacement.

If the current sum is too small, the only way to grow it is to advance lo. If it is too big, retreat hi. Each step discards an index forever, so the walk is linear.

Algorithm

  1. Set lo = 0, hi = n−1.
  2. While lo < hi, compare nums[lo] + nums[hi] to target.
  3. Equal → return the indices. Too small → lo++. Too big → hi--.
Time
O(n)
Space
O(1)

Assumes the array is sorted.

Solution.py
class Solution:    def subsequenceAfterOneReplacement(self, nums: list[int], target: int) -> list[int]:        lo, hi = 0, len(nums) - 1        while lo < hi:            s = nums[lo] + nums[hi]            if s == target:                return [lo, hi]            if s < target:                lo += 1            else:                hi -= 1        return []
Watch this run

03

Interactive walkthrough

Playing “Subsequence After One Replacement — two pointers”. Edit the input, hit Run, then scrub.

Approach: Subsequence After One Replacement — two pointers

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