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Posts published in “Two pointers”

花花酱 LeetCode 240. Search a 2D Matrix II

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

  • Integers in each row are sorted in ascending from left to right.
  • Integers in each column are sorted in ascending from top to bottom.

Example:

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Solution 1: Two Pointers

Start from first row + last column, if the current value is larger than target, –column; if smaller then ++row.

e.g.
1. r = 0, c = 4, v = 15, 15 > 5 => –c
2. r = 0, c = 3, v = 11, 11 > 5 => –c
3. r = 0, c = 2, v = 7, 7 > 5 => –c
4. r = 0, c = 1, v = 4, 4 < 5 => ++r
5. r = 1, c = 1, v = 5, 5 = 5, found it!

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

C++

花花酱 LeetCode 209. Minimum Size Subarray Sum

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguoussubarray of which the sum ≥ s. If there isn’t one, return 0 instead.

Example: 

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n). 

Solution 1: Two Pointers (Sliding Window)

Maintain a sliding window [l, r) such that sum(nums[l:r)) >= s, then move l to l + 1, and move r accordingly to make the window valid.

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

C++

花花酱 LeetCode 125. Valid Palindrome

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

Note: For the purpose of this problem, we define empty string as valid palindrome.

Example 1:

Input: "A man, a plan, a canal: Panama"
Output: true

Example 2:

Input: "race a car"
Output: false

Solution: Two pointers

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

C++

花花酱 LeetCode 1004. Max Consecutive Ones III

Given an array A of 0s and 1s, we may change up to K values from 0 to 1.

Return the length of the longest (contiguous) subarray that contains only 1s. 

Example 1:

Input: A = [1,1,1,0,0,0,1,1,1,1,0], K = 2
Output: 6
Explanation: 
[1,1,1,0,0,1,1,1,1,1,1]
Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.

Example 2:

Input: A = [0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,1], K = 3
Output: 10
Explanation: 
[0,0,1,1,1,1,1,1,1,1,1,1,0,0,0,1,1,1,1]
Bolded numbers were flipped from 0 to 1.  The longest subarray is underlined.

Note:

  1. 1 <= A.length <= 20000
  2. 0 <= K <= A.length
  3. A[i] is 0 or 1 

Solution : Sliding Window

Maintain a window that has at most K zeros

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

C++

花花酱 LeetCode 992. Subarrays with K Different Integers

Given an array A of positive integers, call a (contiguous, not necessarily distinct) subarray of A good if the number of different integers in that subarray is exactly K.

(For example, [1,2,3,1,2] has 3 different integers: 12, and 3.)

Return the number of good subarrays of A.

Example 1:

Input: A = [1,2,1,2,3], K = 2
Output: 7
Explanation: Subarrays formed with exactly 2 different integers: [1,2], [2,1], [1,2], [2,3], [1,2,1], [2,1,2], [1,2,1,2].

Example 2:

Input: A = [1,2,1,3,4], K = 3
Output: 3
Explanation: Subarrays formed with exactly 3 different integers: [1,2,1,3], [2,1,3], [1,3,4].

Note:

  1. 1 <= A.length <= 20000
  2. 1 <= A[i] <= A.length
  3. 1 <= K <= A.length

Solution: Two pointers + indirection

Let f(x) denote the number of subarrays with x or less distinct numbers.
ans = f(K) – f(K-1)
It takes O(n) Time and O(n) Space to compute f(x)

C++

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