Posts tagged as “greedy”

花花酱 LeetCode 991. Broken Calculator

By zxi on February 10, 2019

On a broken calculator that has a number showing on its display, we can perform two operations:

Double: Multiply the number on the display by 2, or;
Decrement: Subtract 1 from the number on the display.

Initially, the calculator is displaying the number X.

Return the minimum number of operations needed to display the number Y.

Example 1:

Input: X = 2, Y = 3
Output: 2
Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.

Example 2:

Input: X = 5, Y = 8
Output: 2
Explanation: Use decrement and then double {5 -> 4 -> 8}.

Example 3:

Input: X = 3, Y = 10
Output: 3
Explanation:  Use double, decrement and double {3 -> 6 -> 5 -> 10}.

Example 4:

Input: X = 1024, Y = 1
Output: 1023
Explanation: Use decrement operations 1023 times.

Note:

1 <= X <= 10^9
1 <= Y <= 10^9

Solution: Greedy

Thinking backwards, making Y <= X by adding 1 or dividing 2.

If Y is even, (Y + 1) // 2 == Y // 2, there is no need to do the extra step
If Y is odd (Y + 1) // 2 = (Y // 2) + 1, so only do + 1 when Y is odd

Time complexity: O(log(Y-X))
Space complexity: O(1)

C++

// Author: Huahua, running time: 4 ms, 4.7 MB

class Solution {

public:

int brokenCalc(int X, int Y) {

if (X <= Y) return X - Y;

return 1 + brokenCalc(X, Y % 2 ? Y + 1 : Y / 2);

}

};

花花酱 LeetCode 984. String Without AAA or BBB

By zxi on January 27, 2019

Given two integers A and B, return any string S such that:

S has length A + B and contains exactly A 'a' letters, and exactly B 'b' letters;
The substring 'aaa' does not occur in S;
The substring 'bbb' does not occur in S.

Example 1:

Input: A = 1, B = 2
Output: "abb"
Explanation: "abb", "bab" and "bba" are all correct answers.

Example 2:

Input: A = 4, B = 1
Output: "aabaa"

Note:

0 <= A <= 100
0 <= B <= 100
It is guaranteed such an S exists for the given A and B.

C++

class Solution {

public:

string strWithout3a3b(int A, int B) {

char a = 'a';

char b = 'b';

if (B > A) {

swap(A, B);

swap(a, b);

}

string ans;

while (A || B) {

if (A > 0) { ans += a; --A; }

if (A > B) { ans += a; --A; }

if (B > 0) { ans += b; --B; }

}

return ans;

}

};

C++ / overload

class Solution {

public:

string strWithout3a3b(int A, int B, char a = 'a', char b = 'b') {

if (B > A) return strWithout3a3b(B, A, b, a);

string ans;

while (A || B) {

if (A > 0) { ans += a; --A; }

if (A > B) { ans += a; --A; }

if (B > 0) { ans += b; --B; }

}

return ans;

}

};

花花酱 LeetCode 976. Largest Perimeter Triangle

By zxi on January 14, 2019

Given an array A of positive lengths, return the largest perimeter of a triangle with non-zero area, formed from 3 of these lengths.

If it is impossible to form any triangle of non-zero area, return 0.

Example 1:

Input: [2,1,2]
Output: 5

Example 2:

Input: [1,2,1]
Output: 0

Example 3:

Input: [3,2,3,4]
Output: 10

Example 4:

Input: [3,6,2,3]
Output: 8

Note:

3 <= A.length <= 10000
1 <= A[i] <= 10^6

Solution: Greedy

Answer must be 3 consecutive numbers in the sorted array
if A[i] >= A[i+1] + A[i+2], then A[i] >= A[i+j] + A[i+k], 1 < j < k
if A[i] < A[i+1] + A[i+2], then A[i] + A[i+1] + A[i+2] is the answer

C++

class Solution {

public:

int largestPerimeter(vector<int>& A) {

sort(rbegin(A), rend(A));

for (int i = 0; i < A.size() - 2; ++i)

if (A[i] < A[i + 1] + A[i + 2])

return A[i] + A[i + 1] + A[i + 2];

return 0;

}

};

花花酱 LeetCode 936. Stamping The Sequence

By zxi on December 8, 2018

Problem

You want to form a target string of lowercase letters.

At the beginning, your sequence is target.length '?' marks. You also have a stamp of lowercase letters.

On each turn, you may place the stamp over the sequence, and replace every letter in the sequence with the corresponding letter from the stamp. You can make up to 10 * target.length turns.

For example, if the initial sequence is “?????”, and your stamp is "abc", then you may make “abc??”, “?abc?”, “??abc” in the first turn. (Note that the stamp must be fully contained in the boundaries of the sequence in order to stamp.)

If the sequence is possible to stamp, then return an array of the index of the left-most letter being stamped at each turn. If the sequence is not possible to stamp, return an empty array.

For example, if the sequence is “ababc”, and the stamp is "abc", then we could return the answer [0, 2], corresponding to the moves “?????” -> “abc??” -> “ababc”.

Also, if the sequence is possible to stamp, it is guaranteed it is possible to stamp within 10 * target.length moves. Any answers specifying more than this number of moves will not be accepted.

Example 1:

Input: stamp = "abc", target = "ababc"
Output: [0,2]
([1,0,2] would also be accepted as an answer, as well as some other answers.)

Example 2:

Input: stamp = "abca", target = "aabcaca"
Output: [3,0,1]

Note:

1 <= stamp.length <= target.length <= 1000
stamp and target only contain lowercase letters.

Solution: Greedy + Reverse Simulation

Reverse the stamping process. Each time find a full or partial match. Replace the matched char to ‘?’.

Don’t forget the reverse the answer as well.

T = “ababc”, S = “abc”

T = “ab???”, index = 2

T = “?????”, index = 0

ans = [0, 2]

Time complexity: O((T – S)*S)

Space complexity: O(T)

C++

// Author: Huahua, running time: 12 ms

class Solution {

public:

vector<int> movesToStamp(string stamp, string target) {

vector<int> ans;

vector<int> seen(target.length());

int total = 0;

while (total < target.length()) {

bool found = false;

for (int i = 0; i <= target.length() - stamp.length(); ++i) {

if (seen[i]) continue;

int l = unStamp(stamp, target, i);

if (l == 0) continue;

seen[i] = 1;

total += l;

ans.push_back(i);

found = true;

}

if (!found) return {};

}

reverse(begin(ans), end(ans));

return ans;

}

private:

int unStamp(const string& stamp, string& target, int s) {

int l = stamp.size();

for (int i = 0; i < stamp.length(); ++i) {

if (target[s + i] == '?')

--l;

else if (target[s + i] != stamp[i])

return 0;

}

if (l != 0)

std::fill(begin(target) + s, begin(target) + s + stamp.length(), '?');

return l;

}

};

花花酱 LeetCode 948. Bag of Tokens

By zxi on November 29, 2018

Problem

You have an initial power P, an initial score of 0 points, and a bag of tokens.

Each token can be used at most once, has a value token[i], and has potentially two ways to use it.

If we have at least token[i] power, we may play the token face up, losing token[i] power, and gaining 1 point.
If we have at least 1 point, we may play the token face down, gaining token[i]power, and losing 1 point.

Return the largest number of points we can have after playing any number of tokens.

Example 1:

Input: tokens = [100], P = 50
Output: 0

Example 2:

Input: tokens = [100,200], P = 150
Output: 1

Example 3:

Input: tokens = [100,200,300,400], P = 200
Output: 2

Note:

tokens.length <= 1000
0 <= tokens[i] < 10000
0 <= P < 10000

Solution: Greedy + Two Pointers

Sort the tokens, gain points from the low end gain power from the high end.

Time complexity: O(nlogn)

Space complexity: O(1)

C++

// Author: Huahua, Running time: 8 ms

class Solution {

public:

int bagOfTokensScore(vector<int>& tokens, int P) {

sort(begin(tokens), end(tokens));

int points = 0;

int ans = 0;

int i = 0;

int j = tokens.size() - 1;

while (i <= j)

if (P >= tokens[i]) {

P -= tokens[i++];

ans = max(ans, ++points);

} else if (points > 0) {

P += tokens[j--];

--points;

} else {

break;

}

return ans;

}

};