Swift

The problem Given an array nums containing n distinct numbers in the range [0, n], return the only number in the range that is missing from the array. Examples Input: nums = [3,0,1] Output: 2 Explanation: n = 3 since there are 3 numbers, so all numbers are in the range [0,3]. 2 is the missing number in the range since it does not appear in nums. Input: nums = [0,1] Output: 2 Explanation: n = 2 since there are 2 numbers, so all numbers are in the range [0,2]. 2 is the missing number in the range since it does not appear in nums. Input: nums = [9,6,4,2,3,5,7,0,1] Output: 8 Explanation: n = 9 since there are 9 numbers, so all numbers are in the range [0,9]. 8 is the missing number in the range since it does not appear in nums. Constraints n == nums.length 1 <= n <= 10^4 0 <= nums[i] <= n All the numbers of nums are unique. Follow up: Could you implement a solution using only O(1) extra space complexity and O(n) runtime complexity? ...

The problem Reverse bits of a given 32-bit unsigned integer. Note: Note that in some languages, such as Java, there is no unsigned integer type. In this case, both input and output will be given as a signed integer type. They should not affect your implementation, as the integer’s internal binary representation is the same, whether it is signed or unsigned. In Java, the compiler represents the signed integers using 2’s complement notation. Therefore, in Example 2 above, the input represents the signed integer -3 and the output represents the signed integer -1073741825. Examples Input: n = 00000010100101000001111010011100 Output: 964176192 (00111001011110000010100101000000) Explanation: The input binary string 00000010100101000001111010011100 represents the unsigned integer 43261596, so return 964176192, whose binary representation is 00111001011110000010100101000000. Input: n = 11111111111111111111111111111101 Output: 3221225471 (10111111111111111111111111111111) Explanation: The input binary string 11111111111111111111111111111101 represents the unsigned integer 4294967293, so return 3221225471, whose binary representation is 10111111111111111111111111111111. Constraints The input must be a binary string of length 32 Follow up: If this function is called many times, how would you optimize it? ...

The problem Given an integer n, return an array ans of length n + 1 such that for each i (0 <= i <= n), ans[i] is the number of 1s in the binary representation of i. Examples Input: n = 2 Output: [0,1,1] Explanation: 0 --> 0 1 --> 1 2 --> 10 Input: n = 5 Output: [0,1,1,2,1,2] Explanation: 0 --> 0 1 --> 1 2 --> 10 3 --> 11 4 --> 100 5 --> 101 Constraints 0 <= n <= 10^5 Follow up: ...

The problem Given a positive integer n, write a function that returns the number of set bits in its binary representation (also known as the Hamming weight). A set bit refers to a bit in the binary representation of a number that has a value of 1. Examples Input: n = 11 Output: 3 Explanation: The input binary string 1011 has a total of three set bits. Input: n = 128 Output: 1 Explanation: The input binary string 10000000 has a total of one set bit. Input: n = 2147483645 Output: 30 Explanation: The input binary string 1111111111111111111111111111101 has a total of thirty set bits. Constraints 1 <= n <= 2^31 - 1 Follow up: If this function is called many times, how would you optimize it? ...

The problem Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0s. You must do it in place. Examples Input: matrix = [[1,1,1],[1,0,1],[1,1,1]] Output: [[1,0,1],[0,0,0],[1,0,1]] Input: matrix = [[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]] Constraints m == matrix.length n == matrix[0].length 1 <= m, n <= 200 -2³¹ <= matrix[i][j] <= 2³¹ - 1 Follow up: A straightforward solution using O(m*n) 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? Brute Force Solution func setZeroes(_ matrix: inout [[Int]]) { let rows = matrix.count let cols = matrix[0].count var mark: [[Int]] = [] for r in 0 ..< rows { var row: [Int] = [] for c in 0 ..< cols { row.append(matrix[r][c]) } mark.append(row) } for r in 0 ..< rows { for c in 0 ..< cols { if matrix[r][c] == 0 { for col in 0 ..< cols { mark[r][col] = 0 } for row in 0 ..< rows { mark[row][c] = 0 } } } } for r in 0 ..< rows { for c in 0 ..< cols { matrix[r][c] = mark[r][c] } } } Explanation ...