Posts tagged as “binary search”

花花酱 LeetCode 1208. Get Equal Substrings Within Budget

By zxi on September 29, 2019

You are given two strings s and t of the same length. You want to change s to t. Changing the i-th character of s to i-th character of t costs |s[i] - t[i]| that is, the absolute difference between the ASCII values of the characters.

You are also given an integer maxCost.

Return the maximum length of a substring of s that can be changed to be the same as the corresponding substring of twith a cost less than or equal to maxCost.

If there is no substring from s that can be changed to its corresponding substring from t, return 0.

Example 1:

Input: s = "abcd", t = "bcdf", cost = 3
Output: 3
Explanation: "abc" of s can change to "bcd". That costs 3, so the maximum length is 3.

Example 2:

Input: s = "abcd", t = "cdef", cost = 3
Output: 1
Explanation: Each charactor in s costs 2 to change to charactor in t, so the maximum length is 1.

Example 3:

Input: s = "abcd", t = "acde", cost = 0
Output: 1
Explanation: You can't make any change, so the maximum length is 1.

Constraints:

1 <= s.length, t.length <= 10^5
0 <= maxCost <= 10^6
s and t only contain lower case English letters.

Solution 1: Binary Search

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

C++

// Author: Huahua

class Solution {

public:

int equalSubstring(string s, string t, int maxCost) {

const int n = s.length();

vector<int> c(n + 1);

auto cit = begin(c) + 1;

int ans = 0;

for (int i = 0; i < n; ++i, ++cit) {

*cit = c[i] + abs(s[i] - t[i]);

int len = cit - lower_bound(begin(c), cit, *cit - maxCost);

ans = max(ans, len);

}

return ans;

}

};

Solution 2: Sliding Window

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

C++

// Author: Huahua

class Solution {

public:

int equalSubstring(string s, string t, int maxCost) {

const int n = s.length();

int ans = 0;

for (int i = 0, j = 0, cost = 0; i < n && j < n; ++i) {

while (j < n) {

int c = abs(s[j] - t[j]);

if (cost + c > maxCost) break;

cost += c;

++j;

}

ans = max(ans, j - i);

cost -= abs(s[i] - t[i]);

}

return ans;

}

};

花花酱 LeetCode 1201. Ugly Number III

By zxi on September 22, 2019

Write a program to find the n-th ugly number.

Ugly numbers are positive integers which are divisible by a or b or c.

Example 1:

Input: n = 3, a = 2, b = 3, c = 5
Output: 4
Explanation: The ugly numbers are 2, 3, 4, 5, 6, 8, 9, 10... The 3rd is 4.

Example 2:

Input: n = 4, a = 2, b = 3, c = 4
Output: 6
Explanation: The ugly numbers are 2, 3, 4, 6, 8, 9, 12... The 4th is 6.

Example 3:

Input: n = 5, a = 2, b = 11, c = 13
Output: 10
Explanation: The ugly numbers are 2, 4, 6, 8, 10, 11, 12, 13... The 5th is 10.

Example 4:

Input: n = 1000000000, a = 2, b = 217983653, c = 336916467
Output: 1999999984

Constraints:

1 <= n, a, b, c <= 10^9
1 <= a * b * c <= 10^18
It’s guaranteed that the result will be in range [1, 2 * 10^9]

Solution: Binary Search

Number of ugly numbers that are <= m are:

m / a + m / b + m / c – (m / LCM(a,b) + m / LCM(a, c) + m / LCM(b, c) + m / LCM(a, LCM(b, c))

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

C++

// Author: Huahua

class Solution {

public:

int nthUglyNumber(int n, long a, long b, long c) {

long l = 1;

long r = INT_MAX;

long ab = lcm(a, b);

long ac = lcm(a, c);

long bc = lcm(b, c);

long abc = lcm(a, bc);

while (l < r) {

long m = l + (r - l) / 2;

long k = m / a + m / b + m / c - m / ab - m / ac - m / bc + m / abc;

if (k >= n) r = m;

else l = m + 1;

}

return l;

}

};

花花酱 LeetCode 34. Find First and Last Position of Element in Sorted Array

By zxi on August 20, 2019

Given an array of integers nums sorted in ascending order, find the starting and ending position of a given target value.

Your algorithm’s runtime complexity must be in the order of O(log n).

If the target is not found in the array, return [-1, -1].

Example 1:

Input: nums = [5,7,7,8,8,10], target = 8
Output: [3,4]

Example 2:

Input: nums = [5,7,7,8,8,10], target = 6
Output: [-1,-1]

Solution: Binary Search

Basically this problem asks you to implement lower_bound and upper_bound using binary search.

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

C++

// Author: Huahua

class Solution {

public:

vector<int> searchRange(vector<int>& nums, int target) {

return {firstPos(nums, target), lastPos(nums, target)};

}

private:

int firstPos(const vector<int>& nums, int target) {

int l = 0;

int r = nums.size();

while (l < r) {

int m = l + (r - l) / 2;

if (nums[m] >= target) {

r = m;

} else {

l = m + 1;

}

if (l == nums.size() || nums[l] != target) return -1;

return l;

}

int lastPos(const vector<int>& nums, int target) {

int l = 0;

int r = nums.size();

while (l < r) {

int m = l + (r - l) / 2;

if (nums[m] > target) {

r = m;

} else {

l = m + 1;

}

// l points to the first element this is greater than target.

--l;

if (l < 0 || nums[l] != target) return -1;

return l;

}

};

花花酱 Binary Search II SP17

By zxi on April 20, 2019

Binary Search I SP5

For the given function g(x) and a range [l, r] find the smallest int number m such that g(m) is True, if not found, return r + 1.

花花酱 LeetCode 1011. Capacity To Ship Packages Within D Days

By zxi on March 16, 2019

A conveyor belt has packages that must be shipped from one port to another within D days.

The i-th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation: 
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10

Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation: 
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation: 
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Note:

1 <= D <= weights.length <= 50000
1 <= weights[i] <= 500

Solution: Binary Search

Find the smallest capacity such that can finish in D days.

Time complexity: O(n * log(sum(weights))
Space complexity: O(1)

C++

// Author: Huahua, running time: 52 ms, 11.9 MB

class Solution {

public:

int shipWithinDays(vector<int>& weights, int D) {

int l = *max_element(begin(weights), end(weights));

int r = accumulate(begin(weights), end(weights), 0) + 1;

while (l < r) {

int m = l + (r - l) / 2;

int d = 1;

int t = 0;

for (int w : weights)

if ((t += w) > m) {

t = w;

++d;

}

d > D ? l = m + 1 : r = m;

}

return l;

}

};