Last Substring in Lexicographical Order

Hard · two pointers solution for Last Substring in Lexicographical Order. Python, Go, and a step-by-step animation.

Problem

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

LeetCode 1163. Last Substring in Lexicographical Order. 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.

Solution

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

Approach 1Optimal▶ Animated

Last Substring in Lexicographical Order — two pointers

On a sorted array the pair that sums to target is found by walking inward. Built for LeetCode 1163. Last Substring in Lexicographical Order.

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

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

Time: O(n)
Space: O(1)

Solution.py

1class Solution:2    def lastSubstringInLexicographicalOrder(self, nums: list[int], target: int) -> list[int]:3        lo, hi = 0, len(nums) - 14        while lo < hi:5            s = nums[lo] + nums[hi]6            if s == target:7                return [lo, hi]8            if s < target:9                lo += 110            else:11                hi -= 112        return []

Watch this run

Interactive walkthrough

Playing “Last Substring in Lexicographical Order — two pointers”. Edit the input, hit Run, then scrub.

Approach: Last Substring in Lexicographical Order — two pointers

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