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Posts published in “Math”

花花酱 LeetCode 1611. Minimum One Bit Operations to Make Integers Zero

Given an integer n, you must transform it into 0 using the following operations any number of times:

  • Change the rightmost (0th) bit in the binary representation of n.
  • Change the ith bit in the binary representation of n if the (i-1)th bit is set to 1 and the (i-2)th through 0th bits are set to 0.

Return the minimum number of operations to transform n into 0.

Example 1:

Input: n = 0
Output: 0

Example 2:

Input: n = 3
Output: 2
Explanation: The binary representation of 3 is "11".
"11" -> "01" with the 2nd operation since the 0th bit is 1.
"01" -> "00" with the 1st operation.

Example 3:

Input: n = 6
Output: 4
Explanation: The binary representation of 6 is "110".
"110" -> "010" with the 2nd operation since the 1st bit is 1 and 0th through 0th bits are 0.
"010" -> "011" with the 1st operation.
"011" -> "001" with the 2nd operation since the 0th bit is 1.
"001" -> "000" with the 1st operation.

Example 4:

Input: n = 9
Output: 14

Example 5:

Input: n = 333
Output: 393

Constraints:

  • 0 <= n <= 109

Solution 1: Graycode

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

Ans is the order of n in graycode.

C++

花花酱 LeetCode 1588. Sum of All Odd Length Subarrays

Given an array of positive integers arr, calculate the sum of all possible odd-length subarrays.

A subarray is a contiguous subsequence of the array.

Return the sum of all odd-length subarrays of arr.

Example 1:

Input: arr = [1,4,2,5,3]
Output: 58
Explanation: The odd-length subarrays of arr and their sums are:
[1] = 1
[4] = 4
[2] = 2
[5] = 5
[3] = 3
[1,4,2] = 7
[4,2,5] = 11
[2,5,3] = 10
[1,4,2,5,3] = 15
If we add all these together we get 1 + 4 + 2 + 5 + 3 + 7 + 11 + 10 + 15 = 58

Example 2:

Input: arr = [1,2]
Output: 3
Explanation: There are only 2 subarrays of odd length, [1] and [2]. Their sum is 3.

Example 3:

Input: arr = [10,11,12]
Output: 66

Constraints:

  • 1 <= arr.length <= 100
  • 1 <= arr[i] <= 1000

Solution 0: Brute Force

Enumerate all odd length subarrys: O(n^2), each take O(n) to compute the sum.

Total time complexity: O(n^3)
Space complexity: O(1)

Solution 1: Running Prefix Sum

Reduce the time complexity to O(n^2)

C++

Solution 2: Math

Count how many times arr[i] can be in of an odd length subarray
we chose the start, which can be 0, 1, 2, … i, i + 1 choices
we chose the end, which can be i, i + 1, … n – 1, n – i choices
Among those 1/2 are odd length.
So there will be upper((i + 1) * (n – i) / 2) odd length subarrays contain arr[i]

ans = sum(((i + 1) * (n – i) + 1) / 2 * arr[i] for in range(n))

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

C++

花花酱 LeetCode 1573. Number of Ways to Split a String

Given a binary string s (a string consisting only of ‘0’s and ‘1’s), we can split s into 3 non-empty strings s1, s2, s3 (s1+ s2+ s3 = s).

Return the number of ways s can be split such that the number of characters ‘1’ is the same in s1, s2, and s3.

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

Example 1:

Input: s = "10101"
Output: 4
Explanation: There are four ways to split s in 3 parts where each part contain the same number of letters '1'.
"1|010|1"
"1|01|01"
"10|10|1"
"10|1|01"

Example 2:

Input: s = "1001"
Output: 0

Example 3:

Input: s = "0000"
Output: 3
Explanation: There are three ways to split s in 3 parts.
"0|0|00"
"0|00|0"
"00|0|0"

Example 4:

Input: s = "100100010100110"
Output: 12

Constraints:

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

Solution: Counting

Count how many ones in the binary string as T, if not a factor of 3, then there is no answer.

Count how many positions that have prefix sum of T/3 as l, and how many positions that have prefix sum of T/3*2 as r.

Ans = l * r

But we need to special handle the all zero cases, which equals to C(n-2, 2) = (n – 1) * (n – 2) / 2

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

C++

Python3

One pass: Space complexity: O(n)

Python3

花花酱 LeetCode 1551. Minimum Operations to Make Array Equal

You have an array arr of length n where arr[i] = (2 * i) + 1 for all valid values of i (i.e. 0 <= i < n).

In one operation, you can select two indices x and y where 0 <= x, y < n and subtract 1 from arr[x] and add 1 to arr[y] (i.e. perform arr[x] -=1 and arr[y] += 1). The goal is to make all the elements of the array equal. It is guaranteed that all the elements of the array can be made equal using some operations.

Given an integer n, the length of the array. Return the minimum number of operations needed to make all the elements of arr equal.

Example 1:

Input: n = 3
Output: 2
Explanation: arr = [1, 3, 5]
First operation choose x = 2 and y = 0, this leads arr to be [2, 3, 4]
In the second operation choose x = 2 and y = 0 again, thus arr = [3, 3, 3].

Example 2:

Input: n = 6
Output: 9

Constraints:

  • 1 <= n <= 10^4

Solution: Math

1: Find the mean (final value) of the array, assuming x, easy to show x == n
2: Compute the sum of an arithmetic progression of (x – 1) + (x – 3) + … for n // 2 pairs

e.g. n = 6
arr = [1, 3, 5, 7, 9, 11]
x = (1 + 2 * n – 1) / 2 = 6 = n
steps = (6 – 1) + (6 – 3) + (6 – 5) = (n // 2) * (n – (1 + n – 1) / 2) = (n // 2) * (n – n // 2) = 3 * 3 = 9

e.g. n = 5
arr = [1,3,5,7,9]
x = (1 + 2 * n – 1) / 2 = 5 = n
steps = (5 – 1) + (5 – 3)= (n//2) * (n – n // 2) = (n // 2) * ((n + 1) // 2) = 2 * 3 = 6

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

C++

花花酱 LeetCode 1523. Count Odd Numbers in an Interval Range

Given two non-negative integers low and high. Return the count of odd numbers between low and high (inclusive).

Example 1:

Input: low = 3, high = 7
Output: 3
Explanation: The odd numbers between 3 and 7 are [3,5,7].

Example 2:

Input: low = 8, high = 10
Output: 1
Explanation: The odd numbers between 8 and 10 are [9].

Constraints:

  • 0 <= low <= high <= 10^9

Solution: Math

The count of odd numbers between [1, low – 1] is low / 2
e.g. low = 6, we have [1,3,5] in range [1, 5] and count is 6/2 = 3.
The count of odd numbers between [1, high] is (high + 1) / 2
e.g. high = 7, we have [1,3,5,7] in range [1, 7] and count is (7+1) / 2 = 4

Then the count of odd numbers in range [low, high] = count(1, high) – count(1, low-1)
e.g. in range [6, 7] we only have [7], count: 4 – 3 = 1

ans = (high + 1) / 2 – low / 2

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

C++