Posts tagged as “O(n)”

花花酱 LeetCode 1689. Partitioning Into Minimum Number Of Deci-Binary Numbers

By zxi on December 13, 2020

A decimal number is called deci-binary if each of its digits is either 0 or 1 without any leading zeros. For example, 101 and 1100 are deci-binary, while 112 and 3001 are not.

Given a string n that represents a positive decimal integer, return the minimum number of positive deci-binary numbers needed so that they sum up to n.

Example 1:

Input: n = "32"
Output: 3
Explanation: 10 + 11 + 11 = 32

Example 2:

Input: n = "82734"
Output: 8

Example 3:

Input: n = "27346209830709182346"
Output: 9

Constraints:

1 <= n.length <= 10⁵
n consists of only digits.
n does not contain any leading zeros and represents a positive integer.

Solution: Return the max digit

Proof: For a given string, we find the maximum number m, we create m binary strings.
for each one, check each digit, if it’s greater than 0, we mark 1 at that position and decrease the digit by 1.

e.g. 21534
max is 5, we need five binary strings.
1. 11111: 21534 -> 10423
2. 10111: 10423 -> 00312
3: 00111: 00312 -> 00201
4: 00101: 00201 -> 00100
5: 00100: 00100 -> 00000

We can ignore the leading zeros.

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

C++

// Author: Huahua

class Solution {

public:

int minPartitions(string n) {

return *max_element(begin(n), end(n)) - '0';

}

};

花花酱 LeetCode 1685. Sum of Absolute Differences in a Sorted Array

By zxi on December 12, 2020

You are given an integer array nums sorted in non-decreasing order.

Build and return an integer array result with the same length as nums such that result[i] is equal to the summation of absolute differences between nums[i] and all the other elements in the array.

In other words, result[i] is equal to sum(|nums[i]-nums[j]|) where 0 <= j < nums.length and j != i (0-indexed).

Example 1:

Input: nums = [2,3,5]
Output: [4,3,5]
Explanation: Assuming the arrays are 0-indexed, then
result[0] = |2-2| + |2-3| + |2-5| = 0 + 1 + 3 = 4,
result[1] = |3-2| + |3-3| + |3-5| = 1 + 0 + 2 = 3,
result[2] = |5-2| + |5-3| + |5-5| = 3 + 2 + 0 = 5.

Example 2:

Input: nums = [1,4,6,8,10]
Output: [24,15,13,15,21]

Constraints:

2 <= nums.length <= 10⁵
1 <= nums[i] <= nums[i + 1] <= 10⁴

Solution: Prefix Sum

Let s[i] denote sum(num[i] – num[j]) 0 <= j <= i
s[i] = s[i – 1] + (num[i] – num[i – 1]) * i
Let l[i] denote sum(nums[j] – nums[i]) i <= j < n
l[i] = l[i + 1] + (nums[i + 1] – num[i]) * (n – i – 1)
ans[i] = s[i] + l[i]

e.g. 1, 3, 7, 9
s[0] = 0
s[1] = 0 + (3 – 1) * 1 = 2
s[2] = 2 + (7 – 3) * 2 = 10
s[3] = 10 + (9 – 7) * 3 = 16
l[3] = 0
l[2] = 0 + (9 – 7) * 1 = 2
l[1] = 2 + (7 – 3) * 2 = 10
l[0] = 10 + (3 – 1) * 3 = 16

ans = [16, 12, 12, 16]

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

C++

// Author: Huahua

class Solution {

public:

vector<int> getSumAbsoluteDifferences(vector<int>& nums) {

const int n = nums.size();

vector<int> ans(n);

for (int i = 1, sum = 0; i < n; ++i)

ans[i] += (sum += (nums[i] - nums[i - 1]) * i);

for (int i = n - 2, sum = 0; i >= 0; --i)

ans[i] += (sum += (nums[i + 1] - nums[i]) * (n - i - 1));

return ans;

}

};

花花酱 LeetCode 1629. Slowest Key

By zxi on October 24, 2020

A newly designed keypad was tested, where a tester pressed a sequence of n keys, one at a time.

You are given a string keysPressed of length n, where keysPressed[i] was the i^th key pressed in the testing sequence, and a sorted list releaseTimes, where releaseTimes[i] was the time the i^th key was released. Both arrays are 0-indexed. The 0^th key was pressed at the time 0, and every subsequent key was pressed at the exact time the previous key was released.

The tester wants to know the key of the keypress that had the longest duration. The i^thkeypress had a duration of releaseTimes[i] - releaseTimes[i - 1], and the 0^th keypress had a duration of releaseTimes[0].

Note that the same key could have been pressed multiple times during the test, and these multiple presses of the same key may not have had the same duration.

Return the key of the keypress that had the longest duration. If there are multiple such keypresses, return the lexicographically largest key of the keypresses.

Example 1:

Input: releaseTimes = [9,29,49,50], keysPressed = "cbcd"
Output: "c"
Explanation: The keypresses were as follows:
Keypress for 'c' had a duration of 9 (pressed at time 0 and released at time 9).
Keypress for 'b' had a duration of 29 - 9 = 20 (pressed at time 9 right after the release of the previous character and released at time 29).
Keypress for 'c' had a duration of 49 - 29 = 20 (pressed at time 29 right after the release of the previous character and released at time 49).
Keypress for 'd' had a duration of 50 - 49 = 1 (pressed at time 49 right after the release of the previous character and released at time 50).
The longest of these was the keypress for 'b' and the second keypress for 'c', both with duration 20.
'c' is lexicographically larger than 'b', so the answer is 'c'.

Example 2:

Input: releaseTimes = [12,23,36,46,62], keysPressed = "spuda"
Output: "a"
Explanation: The keypresses were as follows:
Keypress for 's' had a duration of 12.
Keypress for 'p' had a duration of 23 - 12 = 11.
Keypress for 'u' had a duration of 36 - 23 = 13.
Keypress for 'd' had a duration of 46 - 36 = 10.
Keypress for 'a' had a duration of 62 - 46 = 16.
The longest of these was the keypress for 'a' with duration 16.

Constraints:

releaseTimes.length == n
keysPressed.length == n
2 <= n <= 1000
0 <= releaseTimes[i] <= 10⁹
releaseTimes[i] < releaseTimes[i+1]
keysPressed contains only lowercase English letters.

Solution: Straightforward

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

C++

class Solution {
public:
  char slowestKey(vector& releaseTimes, string keysPressed) {
    int l = releaseTimes[0];
    char ans = keysPressed[0];
    
    for (int i = 1; i < releaseTimes.size(); ++i) {
      int t = releaseTimes[i] - releaseTimes[i - 1];
      if (t > l) { 
        ans = keysPressed[i]; 
        l = t;
      } else if (t == l) {
        ans = max(ans, keysPressed[i]);      
      }
    }
    return ans;
  }
};

花花酱 LeetCode 1413. Minimum Value to Get Positive Step by Step Sum

By zxi on April 18, 2020

Given an array of integers nums, you start with an initial positive value startValue.

In each iteration, you calculate the step by step sum of startValue plus elements in nums (from left to right).

Return the minimum positive value of startValue such that the step by step sum is never less than 1.

Example 1:

Input: nums = [-3,2,-3,4,2]
Output: 5
Explanation: If you choose startValue = 4, in the third iteration your step by step sum is less than 1.
                step by step sum
                startValue = 4 | startValue = 5 | nums
                  (4 -3 ) = 1  | (5 -3 ) = 2    |  -3
                  (1 +2 ) = 3  | (2 +2 ) = 4    |   2
                  (3 -3 ) = 0  | (4 -3 ) = 1    |  -3
                  (0 +4 ) = 4  | (1 +4 ) = 5    |   4
                  (4 +2 ) = 6  | (5 +2 ) = 7    |   2

Example 2:

Input: nums = [1,2]
Output: 1
Explanation: Minimum start value should be positive.

Example 3:

Input: nums = [1,-2,-3]
Output: 5

Constraints:

1 <= nums.length <= 100
-100 <= nums[i] <= 100

Solution: Prefix sum

Find the minimum prefix sum, ans = – min(prefix_sum, 0) + 1

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

C++

// Author: Huahua

class Solution {

public:

int minStartValue(vector<int>& nums) {

int min_sum = 0;

int sum = 0;

for (int num : nums) {

sum += num;

min_sum = min(min_sum, sum);

}

return -min_sum + 1;

}

};

花花酱 LeetCode 1404. Number of Steps to Reduce a Number in Binary Representation to One

By zxi on April 5, 2020

Given a number s in their binary representation. Return the number of steps to reduce it to 1 under the following rules:

If the current number is even, you have to divide it by 2.
If the current number is odd, you have to add 1 to it.

It’s guaranteed that you can always reach to one for all testcases.

Example 1:

Input: s = "1101"
Output: 6
Explanation: "1101" corressponds to number 13 in their decimal representation.
Step 1) 13 is odd, add 1 and obtain 14. 
Step 2) 14 is even, divide by 2 and obtain 7.
Step 3) 7 is odd, add 1 and obtain 8.
Step 4) 8 is even, divide by 2 and obtain 4.  
Step 5) 4 is even, divide by 2 and obtain 2. 
Step 6) 2 is even, divide by 2 and obtain 1.

Example 2:

Input: s = "10"
Output: 1
Explanation: "10" corressponds to number 2 in their decimal representation.
Step 1) 2 is even, divide by 2 and obtain 1.

Example 3:

Input: s = "1"
Output: 0

Constraints:

1 <= s.length <= 500
s consists of characters ‘0’ or ‘1’
s[0] == '1'

Solution: Simulation

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

C++

// Author: Huahua

class Solution {

public:

int numSteps(string s) {

int ans = 0;

int carry = 0;

// The highest bit must be 1,

// process to the 2nd highest bit

for (int i = s.length() - 1; i > 0; --i) {

if (s[i] - '0' + carry == 1) {

ans += 2; // odd: +1, even: / 2

carry = 1; // always has a carry

} else {

ans += 1; // even: / 2

// carray remains e.g. 1 + 1 = 10, or 0 + 0 = 00

}

// If there is a carry, then it's even, one more step.

return ans + carry;

}

};

Python3

# Author: Huahua

class Solution:

def numSteps(self, s: str) -> int:

ans, carry = 0, 0

for i in range(1, len(s)):

if ord(s[-i]) - ord('0') + carry == 1:

ans += 2

carry = 1

else:

ans += 1

return ans + carry