Posts published in “Array”

花花酱 LeetCode 1051. Height Checker

By zxi on May 27, 2019

Students are asked to stand in non-decreasing order of heights for an annual photo.

Return the minimum number of students not standing in the right positions. (This is the number of students that must move in order for all students to be standing in non-decreasing order of height.)

Example 1:

Input: [1,1,4,2,1,3]
Output: 3
Explanation: 
Students with heights 4, 3 and the last 1 are not standing in the right positions.

Note:

1 <= heights.length <= 100
1 <= heights[i] <= 100

Solution: Sorting

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

C++

class Solution {

public:

int heightChecker(vector<int>& heights) {

vector<int> h(heights);

sort(begin(h), end(h));

int ans = 0;

for (int i = 0; i < h.size(); ++i)

ans += (heights[i] != h[i]);

return ans;

}

};

LeetCode 930. Binary Subarrays With Sum

By zxi on March 11, 2019

In an array A of 0s and 1s, how many non-empty subarrays have sum S?

Example 1:

Input: A = [1,0,1,0,1], S = 2
Output: 4
Explanation: 
The 4 subarrays are bolded below:
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]
[1,0,1,0,1]

Note:

A.length <= 30000
0 <= S <= A.length
A[i] is either 0 or 1.

Solution: Prefix Sum

counts[s] := # of subarrays start from 0 that have sum of s
ans += counts[s – S] if s >= S

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

C++

// Author: Huahua, 28 ms, 12.3 MB

class Solution {

public:

int numSubarraysWithSum(vector<int>& A, int S) {

const int n = A.size();

vector<int> counts(n + 1);

counts[0] = 1;

int ans = 0;

int sum = 0;

for (int a : A) {

sum += a;

if (sum >= S) ans += counts[sum - S];

++counts[sum];

}

return ans;

}

};

花花酱 1007. Minimum Domino Rotations For Equal Row

By zxi on March 9, 2019

In a row of dominoes, A[i] and B[i] represent the top and bottom halves of the i-th domino. (A domino is a tile with two numbers from 1 to 6 – one on each half of the tile.)

We may rotate the i-th domino, so that A[i] and B[i] swap values.

Return the minimum number of rotations so that all the values in A are the same, or all the values in B are the same.

If it cannot be done, return -1.

Example 1:

Input: A = [2,1,2,4,2,2], B = [5,2,6,2,3,2]
Output: 2
Explanation: 
The first figure represents the dominoes as given by A and B: before we do any rotations.
If we rotate the second and fourth dominoes, we can make every value in the top row equal to 2, as indicated by the second figure.

Example 2:

Input: A = [3,5,1,2,3], B = [3,6,3,3,4]
Output: -1
Explanation: 
In this case, it is not possible to rotate the dominoes to make one row of values equal.

Note:

1 <= A[i], B[i] <= 6
2 <= A.length == B.length <= 20000

Solution: Brute Force

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

C++

// Author: Huahua, 72 ms, 14.8 MB

class Solution {

public:

int minDominoRotations(vector<int>& A, vector<int>& B) {

int ans = INT_MAX;

for (int r = 1; r <= 6; ++r) {

bool flag = true;

int count_a = 0;

int count_b = 0;

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

if (A[i] != r && B[i] != r) {

flag = false;

break;

}

else if (A[i] == r && B[i] == r) continue;

else if (A[i] == r) {

++count_a;

} else if (B[i] == r) {

++count_b;

}

if (flag) ans = min(ans, min(count_a, count_b));

}

return ans == INT_MAX ? -1 : ans;

}

};

花花酱 LeetCode 73. Set Matrix Zeroes

By zxi on February 9, 2019

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

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

Example 2:

Input: 
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output: 
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]

Follow up:

A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m + n) space, but still not the best solution.
Could you devise a constant space solution?

Solution 1

Use two arrays to track whether the i-th row / j-th column need to be zeroed.

Time complexity: O(mn)
Space complexity: O(m+n)

C++

class Solution {

public:

void setZeroes(vector<vector<int>>& matrix) {

const int m = matrix.size();

const int n = matrix[0].size();

vector<int> rows(m);

vector<int> cols(n);

for (int i = 0; i < m; ++i)

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

rows[i] |= (matrix[i][j] == 0);

cols[j] |= (matrix[i][j] == 0);

}

for (int i = 0; i < m; ++i)

for (int j = 0; j < n; ++j)

if (rows[i] || cols[j]) matrix[i][j] = 0;

}

};

Solution 2

Use the first row / first col to indicate whether the i-th row / j-th column need be zeroed.

C++

class Solution {

public:

void setZeroes(vector<vector<int>>& matrix) {

const int m = matrix.size();

const int n = matrix[0].size();

bool col0 = false;

bool row0 = false;

for (int i = 0; i < m; ++i)

col0 |= (matrix[i][0] == 0);

for (int j = 0; j < n; ++j)

row0 |= (matrix[0][j] == 0);

for (int i = 1; i < m; ++i)

for (int j = 1; j < n; ++j)

if (matrix[i][j] == 0)

matrix[0][j] = matrix[i][0] = 0;

for (int i = 1; i < m; ++i)

for (int j = 1; j < n; ++j)

if (matrix[i][0] == 0 || matrix[0][j] == 0)

matrix[i][j] = 0;

if (row0)

for (int j = 0; j < n; ++j)

matrix[0][j] = 0;

if (col0)

for (int i = 0; i < m; ++i)

matrix[i][0] = 0;

}

};

花花酱 LeetCode 941. Valid Mountain Array

By zxi on November 18, 2018

Problem

Given an array A of integers, return true if and only if it is a valid mountain array.

Recall that A is a mountain array if and only if:

A.length >= 3
There exists some i with 0 < i < A.length - 1 such that:
- A[0] < A[1] < ... A[i-1] < A[i]
- A[i] > A[i+1] > ... > A[B.length - 1]

Example 1:

Input: [2,1]
Output: false

Example 2:

Input: [3,5,5]
Output: false

Example 3:

Input: [0,3,2,1]
Output: true

Note:

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

Solution

Use has_up and has_down to track whether we have A[i] > A[i – 1] and A[i] < A[i – 1] receptively.

return false if any of the following happened:

size(A) < 3
has_down happened before has_up
not has_down or not has_up
A[i – 1] < A[i] after has_down
A[i – 1] > A[i] before has_up

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua, 24 ms

class Solution {

public:

bool validMountainArray(vector<int>& A) {

if (A.size() < 3) return false;

bool has_up = false;

bool has_down = false;

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

if (A[i] > A[i - 1]) {

if (has_down) return false;

has_up = true;

}

else if (A[i] < A[i - 1]){

if (!has_up) return false;

has_down = true;

} else {

return false;

}

return has_up && has_down;

}

};