answer:To get the maximum possible sum, we should pick all the items with 1s and as many items with 0s as possible, because picking a -1 would decrease the sum. However, we can only pick at most k items, so if there are more than k items with 1s and 0s combined, we need to prioritize picking the items with 1s and then as many items with 0s as possible without exceeding k. We can calculate the maximum possible sum as follows: * If k is less than or equal to the number of items with 1s, then we can pick all of them and get a sum of numOnes. * Otherwise, we can pick all the items with 1s and (k - numOnes) items with 0s, which gives us a sum of numOnes + (k - numOnes) * 0 = numOnes. Here's the code to implement the above approach: ```python class Solution: def maxSum(self, numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int: return min(k, numOnes) ``` This solution has a time complexity of O(1), as it only performs constant operations to calculate the answer.,To solve this problem, we can use dynamic programming. Let's create an array `dp` of size `k` where `dpi` will represent the maximum sum we can achieve by choosing `i` items. We will fill the `dp` array in decreasing order of numbers on items. 1. If `i` equals to `k` or is greater than `k`, we will choose the maximum possible number which is `1` and transfer it to `dp` array. - `dpi - 1 = dpi + 1` - Then move on to the next `i` 2. If `i` is equal to `0` (smallest number), we will choose the maximum possible number available which is `-1`. - `dp0 = -1` - Try the next `i`, `i - 1` from `dp1`. 3. For any other `i`, if `dpi` is already calculated, then we will add the maximum number available to it. - `dpi = max(dpi, dpi - 1) + 1` - Try the next smaller `i`. Let's go through an example, say: numOnes = 5, numZeros = 7, numNegOnes = 2, k = 3 dp values for the example are: -1, 5, 19, 29, 39, 49 Let's fill the `dp` array: 1. First, we will first fill the smallest `i` which is `0` as it will give us the maximum value, i.e., `-1`. 2. Then, we will take the value of the maximum value which is `1` because we have `numOnes` items with the number `1` written on them. - So `dp1 = dp0 + 1` - Since `dp1` is already filled, we will compare it with `dp0` (maximum value when you have all the maximum number values in your selection). - If `dp1` is greater than or equal to `dp0`, then we will take the `dp1` value. - Note that we can assign `dp1` to `dp0`. - In our case, `dp1 = dp0 + 1 = -1 + 1 = 0`. 3. Now, we have to take the maximum value possible which is `2`. - So `dpi = max(dpi - 1, dp0 + 1)`. In this case, `dp2 = max(dp1, 0 + 1) = max(0, 1) = 1`. - Again, we will take the value of `dp2` which is `1` and assign it to `dp0`. 4. We will now check if at index `3`, if `dp3` is possible, then we will take the value of `dp3` which is `max(2, 1) + 1 = 3`. - If `dp3` is `3`, then we will assign `dp3 = 4`. 5. If we have the value of `dp3`, then we will compare it with `dp4`, and assign the maximum value to it. - Since `dp3` is `5`, then `dp4` = `5`. 6. Lastly, we will fill the maximum value of `dp3` with `k` which is 3 and end our program. Final `dp` array: -1, 5, 19, 29, 39, 49 These are the maximum possible sums of numbers written on the items.,def max_sum(numOnes: int, numZeros: int, numNegOnes: int, k: int) -> int: max_sum = 0 for i in range(k + 1): ones = min(i, numOnes) neg_ones = min(k - i, numNegOnes) max_sum = max(max_sum, ones - neg_ones) return max_sum

answer:We can solve this problem by using a sliding window approach. We initialize two pointers `i` and `j` to 0 and maintain a window from `i` to `j` (inclusive) in the array `arr`. At each step, we check if the subarray `arri..j` is equal to the subarray `targeti..j`. If it is, we increment `i` and `j` by 1. If it is not, we reverse the subarray `arri..j` and then increment `i` by 1. We continue this process until `i` is equal to `target.length`. If at the end of this process, `i` is still equal to `target.length`, it means that we were able to make `arr` equal to `target` using reversals, and we return `true`. Otherwise, we return `false`. Here is some sample code that implements this approach: <pre lang="java"> class Solution { public boolean canBeEqual(int target, int arr) { int i = 0, j = 0; while (i &lt; target.length) { if (targeti == arrj) { i++; j++; } else { // Reverse the current subarray int k = i; while (k &lt; j) { int temp = arrk; arrk = arrj; arrj = temp; k++; j--; } j++; } } return i == target.length; } } </pre> This approach has a time complexity of O(n^2) because we may need to reverse a subarray at each step, which takes O(n) time. However, this is still efficient enough to pass all the test cases. It also has a space complexity of O(1) because we do not use any additional space that scales with the input size. I hope this helps! Let me know if you have any questions.,def numberOfSubstrings(s: str) -> int: res, i, count = 0, 0, 0, 0, 0 for j, ch in enumerate(s): countord(ch) - ord('a') += 1 while count0 > 0 and count1 > 0 and count2 > 0: countord(si) - ord('a') -= 1 i += 1 res += i return res,To solve this problem, we need to find a way to transform the array `arr` into the array `target` using the allowed operation of reversing subarrays of `arr`. First, let's learn two helper functions that will be useful: 1) `isAncestor` function: This function checks if an array is the ancestor of another array. Given two arrays `t` and `a`, this function will return `true` if all prefixes of `t` are also prefixes of `a`, and `false` otherwise. For example: ``` isAncestor(1, 3, 4, 1, 9, 4, 3) -> true isAncestor(1, 3, 4, 1, 9) -> false ``` 2) `hasAncestor` function: This function returns true if `arr` has an ancestor array. This function works by looping through all possible prefixes of `arr` and checking if they are the same as any prefix of the target array `target`. Now, let's write an algorithm to solve this problem: 1) Initialize two variables `n` and `check`, where `n` represents the length of the arrays, and `check` represents whether the arrays have a common ancestor or not. 2) Use the `isAncestor` function to check if any prefix of the target array `target` is also a prefix of the array `arr`. If any prefix of `target` is a prefix of `arr`, set `check` to `true`, otherwise set it to `false`. 3) If `check` is `false` and the sum of all integers in `target` is equal to the sum of all integers in `arr`, set `check` to `true`. In other words, if the arrays have different sizes but have the same sum, `check` can also be `true`. 4) If `check` is `true`, return `true`, otherwise return `false`. Here is the completed code: ```python def reverse(arr): arr_len = len(arr) n = arr_len // 2 for i in range(1 + n): rand_idx = randint(0, n - 1) arr = arr:rand_idx + arrrand_idx:arr_len - rand_idx - 1 + arrrand_idx + 1: return arr def canConvert(arr, target): n = len(arr) check = False if arr == target: return True if n != target.length: check = isAncestor(target, arr) if not check: check = sum(arr) == sum(target) return check ```

answer:def min_operations(current, correct): current_mins = int(current:2) * 60 + int(current3:) correct_mins = int(correct:2) * 60 + int(correct3:) if current_mins > correct_mins: correct_mins += 24 * 60 diff = correct_mins - current_mins count = diff // 60 + (diff % 60) // 15 + ((diff % 60) % 15) // 5 + (((diff % 60) % 15) % 5) return count, To determine if a mountain is visible, we can check if its peak is higher than all other peaks to its right and to its left. However, we can optimize this by sorting the peaks by their x-coordinates and performing a single pass from left to right while keeping track of the highest peak to the right and to the left. First, sort the peaks by their x-coordinates. Then, initialize the highest peak to the left (hpl) and the highest peak to the right (hpr) to negative infinity. Also, initialize the number of visible mountains (result) to 0. For each peak, we do the following: * If the current peak's y-coordinate is greater than both hpl and hpr, increment the result by 1. * Update hpl to be the maximum of its current value and the y-coordinate of the current peak. * Move to the next peak. At the end, return the result. Here is some Python code to illustrate the algorithm: ``` def visibleMountains(peaks): peaks.sort() hpl, hpr, result = -float('inf'), -float('inf'), 0 for x, y in peaks: if y > hpl and y > hpr: result += 1 hpl, hpr = max(hpl, y), max(hpr, y) return result ``` This algorithm runs in O(n log n) time, where n is the number of peaks, due to the sorting step. After that, the rest of the algorithm takes O(n) time. Therefore, the overall time complexity is O(n log n). The space complexity is O(1) since we only use a constant amount of extra space.,To calculate the number of visible mountains, we can use a simple approach: 1. Create a 2D array, `maze`, with dimensions `2 * len(peaks), len(peaks)`, where each cell will represent a point on the x-axis relative to the mountain `i`. Let `max_x_axis` be `2 * len(peaks) - 1`. 2. Iterate over each mountain `i` and calculate its peak coordinates relative to `max_x_axis` and put them in the `maze`. Note that the `maze` is half-offset from the mountain due to the isosceles triangle shape. 3. Check every mountain for visibility. For every mountain `idx`, iterate over `mazemax_x_axis - xi: max_x_axis + xi, yi:`, where `peaksidx = xi, yi`, to check if its peak lies within another mountain or its sides. If it does, count it as visible. 4. The total number of visible mountains is the number of elements in the resulting array that have value 1, excluding the first element which is always 0. Here is the function in Python: ```python def numVisibleMountains(peaks): max_x_axis = 2 * len(peaks) - 1 maze = 0 * len(peaks) for _ in range(max_x_axis) for i in range(len(peaks)): xi, yi = peaksi right_start, right_end = max_x_axis - xi, max_x_axis left_start, left_end = 0, max_x_axis - xi for x in range(left_start, right_end): mazexyi = 1 for x in range(left_start, right_end): mazexyi + 1 = 1 count = 0 for i in range(1, len(peaks)): count += mazepeaksi1 + 1.count(1) > 0 return count ``` This approach takes into account the specific shape of the mountains and checks if the peak lies within another mountain or its sides to determine visibility.

answer:Here's a Python solution to the problem: ```python def is_match(num: int) -> bool: if num < 10: # Check for single digit numbers, as there is only one way they can be reversed. return num == (num // 10) * 10 + num % 10 reverse = 0 while num > 0: reverse = reverse * 10 + num % 10 num = num // 10 return reverse == num ``` This function works by simulating the act of reversing `num` and comparing it to `reverse1` and `reverse2` to determine if `reverse2` equals `num`. It starts by checking if `num` is a single digit number. For each number greater than or equal to 10, it subtracts the last (least significant) digit, multiplies it by 10, and adds it to `reverse`. Then it performs the same process for the next least significant digit and continues until all digits have been reversed. This solution has a time complexity of O(logn), where `n` is the size of the integer. It does this by first reversing `num` to `reverse` and then reversing `reverse` back to `num` which eliminates the need of a third variable. The space required is also O(1) as no large storage is needed for an integer of the given size.,def minOperations(nums): nums = sorted(set(nums)) n = len(nums) ans = n for i in range(n): j = 0 while j < n and numsj <= numsi + n - 1: j += 1 ans = min(ans, n - (j - i)) return ans, To solve this problem, we need to reverse an integer twice and check if the final result is equal to the original number. Here is a Python solution: ``` def isSameAfterReversals(num: int) -> bool: # Reverse the number once reversed1 = int(str(num)::-1) # Reverse the number again reversed2 = int(str(reversed1)::-1) # Check if reversed2 equals num return reversed2 == num ``` The time complexity of the solution is O(log(10^6)) = O(1) as the number of digits in the input number is at most 6. The space complexity is also O(1) as we are only using a constant amount of additional space. We can also optimize the solution by removing the need for an intermediate variable `reversed1` as follows: ``` def isSameAfterReversals(num: int) -> bool: return int(str(int(str(num)::-1))::-1) == num ``` This solution has the same time and space complexity as the previous one.