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Posts tagged as “two pointers”

花花酱 LeetCode 76. Minimum Window Substring

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string "".

The testcases will be generated such that the answer is unique.

substring is a contiguous sequence of characters within the string.

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.

Constraints:

  • m == s.length
  • n == t.length
  • 1 <= m, n <= 105
  • s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

Solution: Hashtable + Two Pointers

Use a hashtable to store the freq of characters we need to match for t.

Use (i, j) to track a subarray that contains all the chars in t.

Time complexity: O(m + n)
Space complexity: O(m)

C++

花花酱 LeetCode 1898. Maximum Number of Removable Characters

You are given two strings s and p where p is a subsequence of s. You are also given a distinct 0-indexed integer array removable containing a subset of indices of s (s is also 0-indexed).

You want to choose an integer k (0 <= k <= removable.length) such that, after removing k characters from s using the first k indices in removablep is still a subsequence of s. More formally, you will mark the character at s[removable[i]] for each 0 <= i < k, then remove all marked characters and check if p is still a subsequence.

Return the maximum k you can choose such that p is still a subsequence of s after the removals.

subsequence of a string is a new string generated from the original string with some characters (can be none) deleted without changing the relative order of the remaining characters.

Example 1:

Input: s = "abcacb", p = "ab", removable = [3,1,0]
Output: 2
Explanation: After removing the characters at indices 3 and 1, "abcacb" becomes "accb".
"ab" is a subsequence of "accb".
If we remove the characters at indices 3, 1, and 0, "abcacb" becomes "ccb", and "ab" is no longer a subsequence.
Hence, the maximum k is 2.

Example 2:

Input: s = "abcbddddd", p = "abcd", removable = [3,2,1,4,5,6]
Output: 1
Explanation: After removing the character at index 3, "abcbddddd" becomes "abcddddd".
"abcd" is a subsequence of "abcddddd".

Example 3:

Input: s = "abcab", p = "abc", removable = [0,1,2,3,4]
Output: 0
Explanation: If you remove the first index in the array removable, "abc" is no longer a subsequence.

Constraints:

  • 1 <= p.length <= s.length <= 105
  • 0 <= removable.length < s.length
  • 0 <= removable[i] < s.length
  • p is a subsequence of s.
  • s and p both consist of lowercase English letters.
  • The elements in removable are distinct.

Solution: Binary Search + Two Pointers

If we don’t remove any thing, p is a subseq of s, as we keep removing, at some point L, p is no longer a subseq of s. e.g [0:True, 1: True, …, L – 1: True, L: False, L+1: False, …, m:False], this array is monotonic. We can use binary search to find the smallest L such that p is no long a subseq of s. Ans = L – 1.

For each guess, we can use two pointers to check whether p is subseq of removed(s) in O(n).

Time complexity: O(nlogn)
Space complexity: O(n)

C++

花花酱 LeetCode 1871. Jump Game VII

You are given a 0-indexed binary string s and two integers minJump and maxJump. In the beginning, you are standing at index 0, which is equal to '0'. You can move from index i to index j if the following conditions are fulfilled:

  • i + minJump <= j <= min(i + maxJump, s.length - 1), and
  • s[j] == '0'.

Return true if you can reach index s.length - 1 in s, or false otherwise.

Example 1:

Input: s = "011010", minJump = 2, maxJump = 3
Output: true
Explanation:
In the first step, move from index 0 to index 3. 
In the second step, move from index 3 to index 5.

Example 2:

Input: s = "01101110", minJump = 2, maxJump = 3
Output: false

Constraints:

  • 2 <= s.length <= 105
  • s[i] is either '0' or '1'.
  • s[0] == '0'
  • 1 <= minJump <= maxJump < s.length

Solution 1: TreeSet /Dequq + Binary Search

Maintain a set of reachable indices so far, for each ‘0’ index check whether it can be reached from any of the elements in the set.

Time complexity: O(nlogn)
Space complexity: O(n)

C++/set

C++/deque

Solution 2: Queue

Same idea, we can replace the deque in sol1 with a queue, and only check the smallest element in the queue.

C++/set

Time complexity: O(n)
Space complexity: O(n)

花花酱 LeetCode 1855. Maximum Distance Between a Pair of Values

You are given two non-increasing 0-indexed integer arrays nums1ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹ and nums2ā€‹ā€‹ā€‹ā€‹ā€‹ā€‹.

A pair of indices (i, j), where 0 <= i < nums1.length and 0 <= j < nums2.length, is valid if both i <= j and nums1[i] <= nums2[j]. The distance of the pair is j - iā€‹ā€‹ā€‹ā€‹.

Return the maximum distance of any valid pair (i, j). If there are no valid pairs, return 0.

An array arr is non-increasing if arr[i-1] >= arr[i] for every 1 <= i < arr.length.

Example 1:

Input: nums1 = [55,30,5,4,2], nums2 = [100,20,10,10,5]
Output: 2
Explanation: The valid pairs are (0,0), (2,2), (2,3), (2,4), (3,3), (3,4), and (4,4).
The maximum distance is 2 with pair (2,4).

Example 2:

Input: nums1 = [2,2,2], nums2 = [10,10,1]
Output: 1
Explanation: The valid pairs are (0,0), (0,1), and (1,1).
The maximum distance is 1 with pair (0,1).

Example 3:

Input: nums1 = [30,29,19,5], nums2 = [25,25,25,25,25]
Output: 2
Explanation: The valid pairs are (2,2), (2,3), (2,4), (3,3), and (3,4).
The maximum distance is 2 with pair (2,4).

Example 4:

Input: nums1 = [5,4], nums2 = [3,2]
Output: 0
Explanation: There are no valid pairs, so return 0.

Constraints:

  • 1 <= nums1.length <= 105
  • 1 <= nums2.length <= 105
  • 1 <= nums1[i], nums2[j] <= 105
  • Both nums1 and nums2 are non-increasing.

Solution: Two Pointers

For each i, find the largest j such that nums[j] >= nums[i].

Time complexity: O(n + m)
Space complexity: O(1)

C++

花花酱 LeetCode 1793. Maximum Score of a Good Subarray

You are given an array of integers nums (0-indexed) and an integer k.

The score of a subarray (i, j) is defined as min(nums[i], nums[i+1], ..., nums[j]) * (j - i + 1). A good subarray is a subarray where i <= k <= j.

Return the maximum possible score of a good subarray.

Example 1:

Input: nums = [1,4,3,7,4,5], k = 3
Output: 15
Explanation: The optimal subarray is (1, 5) with a score of min(4,3,7,4,5) * (5-1+1) = 3 * 5 = 15. 

Example 2:

Input: nums = [5,5,4,5,4,1,1,1], k = 0
Output: 20
Explanation: The optimal subarray is (0, 4) with a score of min(5,5,4,5,4) * (4-0+1) = 4 * 5 = 20.

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 2 * 104
  • 0 <= k < nums.length

Solutions: Two Pointers

maintain a window [i, j], m = min(nums[i~j]), expend to the left if nums[i – 1] >= nums[j + 1], otherwise expend to the right.

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

C++