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花花酱 LeetCode 1535. Find the Winner of an Array Game

Given an integer array arr of distinct integers and an integer k.

A game will be played between the first two elements of the array (i.e. arr[0] and arr[1]). In each round of the game, we compare arr[0] with arr[1], the larger integer wins and remains at position 0 and the smaller integer moves to the end of the array. The game ends when an integer wins k consecutive rounds.

Return the integer which will win the game.

It is guaranteed that there will be a winner of the game.

Example 1:

Input: arr = [2,1,3,5,4,6,7], k = 2
Output: 5
Explanation: Let's see the rounds of the game:
Round |       arr       | winner | win_count
  1   | [2,1,3,5,4,6,7] | 2      | 1
  2   | [2,3,5,4,6,7,1] | 3      | 1
  3   | [3,5,4,6,7,1,2] | 5      | 1
  4   | [5,4,6,7,1,2,3] | 5      | 2
So we can see that 4 rounds will be played and 5 is the winner because it wins 2 consecutive games.

Example 2:

Input: arr = [3,2,1], k = 10
Output: 3
Explanation: 3 will win the first 10 rounds consecutively.

Example 3:

Input: arr = [1,9,8,2,3,7,6,4,5], k = 7
Output: 9

Example 4:

Input: arr = [1,11,22,33,44,55,66,77,88,99], k = 1000000000
Output: 99

Constraints:

  • 2 <= arr.length <= 10^5
  • 1 <= arr[i] <= 10^6
  • arr contains distinct integers.
  • 1 <= k <= 10^9

Solution 1: Simulation with a List

Observation : if k >= n – 1, ans = max(arr)

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

C++

Solution 2: One pass

Since smaller numbers will be put to the end of the array, we must compare the rest of array before meeting those used numbers again. And the winner is monotonically increasing. So we can do it in one pass, just keep the largest number seen so far. If we reach to the end of the array, arr[0] will be max(arr) and it will always win no matter what k is.

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

C++

Java

Python3

花花酱 LeetCode 1534. Count Good Triplets

Given an array of integers arr, and three integers ab and c. You need to find the number of good triplets.

A triplet (arr[i], arr[j], arr[k]) is good if the following conditions are true:

  • 0 <= i < j < k < arr.length
  • |arr[i] - arr[j]| <= a
  • |arr[j] - arr[k]| <= b
  • |arr[i] - arr[k]| <= c

Where |x| denotes the absolute value of x.

Return the number of good triplets.

Example 1:

Input: arr = [3,0,1,1,9,7], a = 7, b = 2, c = 3
Output: 4
Explanation: There are 4 good triplets: [(3,0,1), (3,0,1), (3,1,1), (0,1,1)].

Example 2:

Input: arr = [1,1,2,2,3], a = 0, b = 0, c = 1
Output: 0
Explanation: No triplet satisfies all conditions.

Constraints:

  • 3 <= arr.length <= 100
  • 0 <= arr[i] <= 1000
  • 0 <= a, b, c <= 1000

Solution: Brute Force

Time complexity: O(n^3)
Space complexity: O(1)

C++

花花酱 LeetCode 1513. Number of Substrings With Only 1s

Given a binary string s (a string consisting only of ‘0’ and ‘1’s).

Return the number of substrings with all characters 1’s.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: s = "0110111"
Output: 9
Explanation: There are 9 substring in total with only 1's characters.
"1" -> 5 times.
"11" -> 3 times.
"111" -> 1 time.

Example 2:

Input: s = "101"
Output: 2
Explanation: Substring "1" is shown 2 times in s.

Example 3:

Input: s = "111111"
Output: 21
Explanation: Each substring contains only 1's characters.

Example 4:

Input: s = "000"
Output: 0

Constraints:

  • s[i] == '0' or s[i] == '1'
  • 1 <= s.length <= 10^5

Solution: DP / Prefix Sum

dp[i] := # of all 1 subarrays end with s[i].
dp[i] = dp[i-1] if s[i] == ‘1‘ else 0
ans = sum(dp)
s=1101
dp[0] = 1 // 1
dp[1] = 2 // 11, *1
dp[2] = 0 // None
dp[3] = 1 // ***1
ans = 1 + 2 + 1 = 5

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

C++

dp[i] only depends on dp[i-1], we can reduce the space complexity to O(1)

C++

Java

Python3


花花酱 LeetCode 1531. String Compression II

Run-length encoding is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "aabccc" we replace "aa" by "a2" and replace "ccc" by "c3". Thus the compressed string becomes "a2bc3".

Notice that in this problem, we are not adding '1' after single characters.

Given a string s and an integer k. You need to delete at most k characters from s such that the run-length encoded version of s has minimum length.

Find the minimum length of the run-length encoded version of s after deleting at most k characters.

Example 1:

Input: s = "aaabcccd", k = 2
Output: 4
Explanation: Compressing s without deleting anything will give us "a3bc3d" of length 6. Deleting any of the characters 'a' or 'c' would at most decrease the length of the compressed string to 5, for instance delete 2 'a' then we will have s = "abcccd" which compressed is abc3d. Therefore, the optimal way is to delete 'b' and 'd', then the compressed version of s will be "a3c3" of length 4.

Example 2:

Input: s = "aabbaa", k = 2
Output: 2
Explanation: If we delete both 'b' characters, the resulting compressed string would be "a4" of length 2.

Example 3:

Input: s = "aaaaaaaaaaa", k = 0
Output: 3
Explanation: Since k is zero, we cannot delete anything. The compressed string is "a11" of length 3.

Constraints:

  • 1 <= s.length <= 100
  • 0 <= k <= s.length
  • s contains only lowercase English letters.

Solution 0: Brute Force DFS (TLE)

Time complexity: O(C(n,k))
Space complexity: O(k)

C++

Solution1: DP

State:
i: the start index of the substring
last: last char
len: run-length
k: # of chars that can be deleted.

base case:
1. k < 0: return inf # invalid
2. i >= s.length(): return 0 # done

Transition:
1. if s[i] == last: return carry + dp(i + 1, last, len + 1, k)

2. if s[i] != last:
return min(1 + dp(i + 1, s[i], 1, k, # start a new group with s[i]
dp(i + 1, last, len, k -1) # delete / skip s[i], keep it as is.

Time complexity: O(n^3*26)
Space complexity: O(n^3*26)

C++

State compression

dp[i][k] := min len of s[i:] encoded by deleting at most k charchters.

dp[i][k] = min(dp[i+1][k-1] # delete s[i]
encode_len(s[i~j] == s[i]) + dp(j+1, k – sum(s[i~j])) for j in range(i, n)) # keep

Time complexity: O(n^2*k)
Space complexity: O(n*k)

C++

Java

Python3

花花酱 LeetCode 1529. Bulb Switcher IV

There is a room with n bulbs, numbered from 0 to n-1, arranged in a row from left to right. Initially all the bulbs are turned off.

Your task is to obtain the configuration represented by target where target[i] is ‘1’ if the i-th bulb is turned on and is ‘0’ if it is turned off.

You have a switch to flip the state of the bulb, a flip operation is defined as follows:

  • Choose any bulb (index i) of your current configuration.
  • Flip each bulb from index i to n-1.

When any bulb is flipped it means that if it is 0 it changes to 1 and if it is 1 it changes to 0.

Return the minimum number of flips required to form target.

Example 1:

Input: target = "10111"
Output: 3
Explanation: Initial configuration "00000".
flip from the third bulb:  "00000" -> "00111"
flip from the first bulb:  "00111" -> "11000"
flip from the second bulb:  "11000" -> "10111"
We need at least 3 flip operations to form target.

Example 2:

Input: target = "101"
Output: 3
Explanation: "000" -> "111" -> "100" -> "101".

Example 3:

Input: target = "00000"
Output: 0

Example 4:

Input: target = "001011101"
Output: 5

Constraints:

  • 1 <= target.length <= 10^5
  • target[i] == '0' or target[i] == '1'

Solution: XOR

Flip from left to right, since flipping the a bulb won’t affect anything in the left.
We count how many times flipped so far, and that % 2 will be the state for all the bulb to the right.
If the current bulb’s state != target, we have to flip once.

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

C++