Posts published in “Greedy”

花花酱 LeetCode 945. Minimum Increment to Make Array Unique

By zxi on November 29, 2018

Problem

Given an array of integers A, a move consists of choosing any A[i], and incrementing it by 1.

Return the least number of moves to make every value in A unique.

Example 1:

Input: [1,2,2]
Output: 1
Explanation:  After 1 move, the array could be [1, 2, 3].

Example 2:

Input: [3,2,1,2,1,7]
Output: 6
Explanation:  After 6 moves, the array could be [3, 4, 1, 2, 5, 7].
It can be shown with 5 or less moves that it is impossible for the array to have all unique values.

Note:

0 <= A.length <= 40000
0 <= A[i] < 40000

Solution: Greedy

Sort the elements, make sure A[i] >= A[i-1] + 1, if not increase A[i] to A[i – 1] + 1

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, 56 ms

class Solution {

public:

int minIncrementForUnique(vector<int>& A) {

int ans = 0;

sort(begin(A), end(A));

for (int i = 1; i < A.size(); ++i) {

if (A[i] > A[i - 1]) continue;

ans += (A[i - 1] - A[i]) + 1;

A[i] = A[i - 1] + 1;

}

return ans;

}

};

Python3

# Author: Huahua, Running time: 196 ms

class Solution(object):

def minIncrementForUnique(self, A):

A.sort()

ans = 0

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

if A[i] > A[i - 1]: continue

ans += A[i - 1] - A[i] + 1

A[i] = A[i - 1] + 1

return ans

花花酱 LeetCode 942. DI String Match

By zxi on November 18, 2018

Problem

Given a string S that only contains “I” (increase) or “D” (decrease), let N = S.length.

Return any permutation A of [0, 1, ..., N] such that for all i = 0, ..., N-1:

If S[i] == "I", then A[i] < A[i+1]
If S[i] == "D", then A[i] > A[i+1]

Example 1:

Input: "IDID"
Output: [0,4,1,3,2]

Example 2:

Input: "III"
Output: [0,1,2,3]

Example 3:

Input: "DDI"
Output: [3,2,0,1]

Note:

1 <= S.length <= 10000
S only contains characters "I" or "D".

Solution: Greedy

“I” -> use the smallest possible number

“D” -> use the largest possible number

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua, 44 ms

class Solution {

public:

vector<int> diStringMatch(string S) {

const int n = S.length();

vector<int> ans;

int lo = 0;

int hi = n;

for (char c : S) {

if (c == 'I')

ans.push_back(lo++);

else

ans.push_back(hi--);

}

ans.push_back(lo);

return ans;

}

};

花花酱 LeetCode 55. Jump Game

By zxi on November 15, 2018

Problem

Given an array of non-negative integers, you are initially positioned at the first index of the array.

Each element in the array represents your maximum jump length at that position.

Determine if you are able to reach the last index.

Example 1:

Input: [2,3,1,1,4]
Output: true
Explanation: Jump 1 step from index 0 to 1, then 3 steps to the last index.

Example 2:

Input: [3,2,1,0,4]
Output: false
Explanation: You will always arrive at index 3 no matter what. Its maximum
             jump length is 0, which makes it impossible to reach the last index.

Solution: Greedy

If you can jump to i, then you can jump to at least i + nums[i].

Always jump as far as you can.

Keep tracking the farthest index you can jump to.

Init far = nums[0].

far = max(far, i + nums[i])

check far >= n – 1

ex 1 [2,3,1,1,4]

i nums[i] i + nums[i] far

- - - 2

0 2 2 2

1 3 4 4

2 1 3 4

3 1 4 4

4 4 8 8

ex 2 [3,2,1,0,4]

i nums[i] i + nums[i] far

- - - 3

0 3 3 3

1 2 3 3

2 1 3 3

3 0 3 3 <- can not reach index 4, break

C++

// Author: Huahua, running time 16 ms

class Solution {

public:

bool canJump(vector<int>& nums) {

const int n = nums.size();

int far = nums[0];

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

if (i > far) break;

far = max(far, i + nums[i]);

}

return far >= n-1;

}

};

花花酱 LeetCode 915. Partition Array into Disjoint Intervals

By zxi on September 30, 2018

Problem

Given an array A, partition it into two (contiguous) subarrays left and right so that:

Every element in left is less than or equal to every element in right.
left and right are non-empty.
left has the smallest possible size.

Return the length of left after such a partitioning. It is guaranteed that such a partitioning exists.

Example 1:

Input: [5,0,3,8,6]
Output: 3
Explanation: left = [5,0,3], right = [8,6]

Example 2:

Input: [1,1,1,0,6,12]
Output: 4
Explanation: left = [1,1,1,0], right = [6,12]

Note:

2 <= A.length <= 30000
0 <= A[i] <= 10^6
It is guaranteed there is at least one way to partition A as described.

Solution 1: BST

Time complexity: O(nlogn)

Space complexity: O(n)

C++

// Author: Huahua, 100 ms

class Solution {

public:

int partitionDisjoint(vector<int>& A) {

multiset<int> s(begin(A) + 1, end(A));

int left_max = A[0];

for (int i = 1; i < A.size(); ++i) {

if (*begin(s) >= left_max) return i;

s.erase(s.equal_range(A[i]).first);

left_max = max(left_max, A[i]);

}

return -1;

}

};

Solution 2: Greedy

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua, 28 ms

class Solution {

public:

int partitionDisjoint(vector<int>& A) {

int left_max = A[0];

int cur_max = A[0];

int left_len = 1;

for (int i = 1; i < A.size(); ++i) {

if (A[i] < left_max) {

left_max = cur_max;

left_len = i + 1;

} else {

cur_max = max(cur_max, A[i]);

}

return left_len;

}

};

花花酱 LeetCode 910. Smallest Range II

By zxi on September 26, 2018

Problem

Given an array A of integers, for each integer A[i] we need to choose either x = -K or x = K, and add x to A[i] (only once).

After this process, we have some array B.

Return the smallest possible difference between the maximum value of B and the minimum value of B.

Example 1:

Input: A = [1], K = 0
Output: 0
Explanation: B = [1]

Example 2:

Input: A = [0,10], K = 2
Output: 6
Explanation: B = [2,8]

Example 3:

Input: A = [1,3,6], K = 3
Output: 3
Explanation: B = [4,6,3]

Note:

1 <= A.length <= 10000
0 <= A[i] <= 10000
0 <= K <= 10000

Solution: Greedy

Sort the array and compare adjacent numbers.

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua

class Solution {

public:

int smallestRangeII(vector<int>& A, int K) {

sort(begin(A), end(A));

int ans = A.back() - A.front();

for (int i = 1; i < A.size(); ++i) {

int l = min(A.front() + K, A[i] - K);

int h = max(A.back() - K, A[i - 1] + K);

ans = min(ans, h - l);

}

return ans;

}

};

Python3

# Author: Huahua

class Solution:

def smallestRangeII(self, A, K):

A.sort()

ans = A[-1] - A[0]

for a, b in zip(A[0:-1], A[1:]):

l = min(A[0] + K, b - K)

h = max(A[-1] - K, a + K)

ans = min(ans, h - l)

return ans